Mathematics — Class 10, Chapter 3: Mensuration

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Asked 2 times
2075 Chaitra · Set CP · 2 marks
2075 Chaitra · Set EP · 2 marks

Find the area of $\triangle MNO$ in the given figure.
Figure: Triangle $MNO$ with $MN = 24$ cm, $MO = 7$ cm and $NO = 25$ cm.

Similar questions: Area of triangle with sides 4, 3 and 5 (2075 Chaitra · Set AP) · Area of triangle with sides 12, 14 and 20 (2075 Supplementary) · Area of triangle with sides 5, 12 and 13 (2075 Chaitra · Set BP)

Answer

Check: $MN^2+MO^2=24^2+7^2=576+49=625=25^2=NO^2$ , so the triangle is right-angled at $M$ (Pythagoras' converse).

$\text{Area}=\frac12\times MN\times MO=\frac12\times24\times7=\mathbf{84\ cm^2}.$

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2075 Chaitra · Set AP · 2 marks

Find the area of $\triangle XYZ$ in the given figure.
Figure: Triangle $XYZ$ with $XY = 4$ cm, $XZ = 3$ cm and $YZ = 5$ cm.

Similar questions: Area of triangle with sides 12, 14 and 20 (2075 Supplementary) · Area of triangle with sides 5, 12 and 13 (2075 Chaitra · Set BP) · Area of triangle with sides 24, 7 and 25 (2075 Chaitra · Set CP)

Answer

Here $4^2+3^2=16+9=25=5^2$ , so $\triangle XYZ$ is right-angled (at $X$ , between the sides 4 cm and 3 cm).

$\text{Area}=\frac{1}{2}\times 4\times 3=\textbf{6 cm}^2.$

(Check with Heron: $s=6$ , $\sqrt{6\times2\times3\times1}=\sqrt{36}=6$ .)

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2075 Supplementary · 2 marks

Find the area of $\triangle XYZ$ in the given figure.
Figure: Triangle $XYZ$ with $XY = 12$ cm, $XZ = 14$ cm and $YZ = 20$ cm.

Similar questions: Area of triangle with sides 4, 3 and 5 (2075 Chaitra · Set AP) · Area of triangle with sides 5, 12 and 13 (2075 Chaitra · Set BP) · Area of triangle with sides 24, 7 and 25 (2075 Chaitra · Set CP)

Answer

Sides: $a=12$ , $b=14$ , $c=20$ cm.

$s=\frac{12+14+20}{2}=23\ \text{cm}$

$\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{23\times11\times9\times3}=\sqrt{6831}=3\sqrt{759}\approx\textbf{82.65 cm}^2.$

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2075 Chaitra · Set AP · 2 marks

If the volume of a hemisphere is $9216\pi$ cubic cm, then find its radius.

Similar questions: Radius of hemisphere from volume 19404 (2075 Chaitra · Set BP) · Radius of hemisphere from volume 486 pi (2075 Chaitra · Set CP) · Radius of hemisphere from volume 3888 pi (2075 Chaitra · Set EP)

Answer

Volume of a hemisphere $=\dfrac{2}{3}\pi r^3$ .

$\frac{2}{3}\pi r^3=9216\pi\Rightarrow r^3=9216\times\frac{3}{2}=13824=24^3.$

So the radius is $r=$ 24 cm.

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2075 Supplementary · 2 marks

If the volume of a hemisphere is $144\pi$ cubic cm, then find its diameter.

Similar questions: Diameter of hemisphere from volume 3888 pi (2075 Chaitra · Set FP) · Radius of hemisphere from volume 9216 pi (2075 Chaitra · Set AP) · Radius of hemisphere from volume 19404 (2075 Chaitra · Set BP)

Answer

$\frac{2}{3}\pi r^3=144\pi\Rightarrow r^3=144\times\frac{3}{2}=216=6^3\Rightarrow r=6\ \text{cm}.$

Diameter $=2r=2\times6=$ 12 cm.

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2075 Chaitra · Set AP · 4 marks

The given figure is a combined solid made up of a cylinder and a cone. The diameter of the base of the object is $16$ cm, the length of the cylindrical part is $75$ cm and the total length of the solid object is $90$ cm. Find the total surface area of the combined solid.

Similar questions: Total surface area of cylinder and cone solid, diameter 10 cm (2075 Chaitra · Set BP) · Total surface area of cylinder and cone solid, diameter 6 cm (2075 Chaitra · Set CP) · Total surface area of cylinder 150 cm and cone, total length 180 cm (2075 Chaitra · Set DP)

Answer

Radius $r = \dfrac{16}{2} = 8$ cm. Cylinder height $h = 75$ cm.

Cone height $= 90 - 75 = 15$ cm, so its slant height is

$l = \sqrt{8^2 + 15^2} = \sqrt{289} = 17 \text{ cm}.$

The surface of the combined solid is the curved surface of the cone, the curved surface of the cylinder and the base circle of the cylinder:

$TSA = \pi r l + 2\pi r h + \pi r^2 = \pi r(l + 2h + r)$

$= \frac{22}{7} \times 8 \times (17 + 150 + 8) = \frac{22}{7} \times 8 \times 175 = 4400 \text{ cm}^2$

The total surface area is $4400$ cm $^2$ .

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2075 Supplementary · 4 marks

The given figure is of a combined solid made up of a cylinder and a cone. The radius of the base of the object is $3$ cm, the length of the cylindrical part is $20$ cm and the total length of the solid object is $24$ cm. Find the curved surface area of the solid object.

Similar questions: Total surface area of cylinder and cone solid, radius 7 cm and length 59 cm (2075 Chaitra · Set EP) · Total surface area of cylinder and cone solid, radius 6 cm and length 28 cm (2075 Chaitra · Set FP) · Total surface area of cylinder and cone solid (2075 Chaitra · Set AP)

Answer

Radius $r = 3$ cm. Cylinder height $h = 20$ cm.

Cone height $= 24 - 20 = 4$ cm, so its slant height is

$l = \sqrt{3^2 + 4^2} = 5 \text{ cm}.$

Curved surface area $=$ cylinder curved area $+$ cone curved area:

$= 2\pi r h + \pi r l = \pi r (2h + l) = \frac{22}{7} \times 3 \times (40 + 5) = \frac{22}{7} \times 135 = \frac{2970}{7} \approx 424.29 \text{ cm}^2$

The curved surface area is about $424.29$ cm $^2$ .

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2075 Chaitra · Set BP · 2 marks

In the given figure, find the area of $\triangle ABC$ .
Figure: Triangle $ABC$ with $AB = 5$ cm, $AC = 12$ cm and $BC = 13$ cm.

Similar questions: Area of triangle with sides 4, 3 and 5 (2075 Chaitra · Set AP) · Area of triangle with sides 12, 14 and 20 (2075 Supplementary) · Area of triangle with sides 24, 7 and 25 (2075 Chaitra · Set CP)

Answer

Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ , $\triangle ABC$ is right-angled at $A$ (the sides $5$ and $12$ are perpendicular).

$\text{Area} = \tfrac{1}{2}\times 5\times 12 = 30 \text{ cm}^2$

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2075 Chaitra · Set BP · 2 marks

If the volume of a hemisphere is 19404 cubic cm, then find its radius.

Similar questions: Radius of hemisphere from volume 9216 pi (2075 Chaitra · Set AP) · Radius of hemisphere from volume 486 pi (2075 Chaitra · Set CP) · Radius of hemisphere from volume 3888 pi (2075 Chaitra · Set EP)

Answer

Volume of a hemisphere $= \dfrac{2}{3}\pi r^3$ .

$\frac{2}{3}\times\frac{22}{7}\times r^3 = 19404 \Rightarrow r^3 = \frac{19404\times 21}{44} = 441\times 21 = 9261$

$r = \sqrt[3]{9261} = 21 \text{ cm}$

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2075 Chaitra · Set BP · 4 marks

The given figure is a combined solid made up of a cylinder and a cone. The diameter of the base of the object is 10 cm, the length of the cylindrical part is 128 cm and the total length of the solid object is 140 cm. Find the total surface area of the solid object.

Similar questions: Total surface area of cylinder and cone solid (2075 Chaitra · Set AP) · Total surface area of cylinder and cone solid, diameter 6 cm (2075 Chaitra · Set CP) · Total surface area of cylinder 150 cm and cone, total length 180 cm (2075 Chaitra · Set DP)

Answer

Radius $r = 5$ cm. Cylinder height $h = 128$ cm. Cone height $= 140 - 128 = 12$ cm.

Slant height of the cone: $l = \sqrt{12^2 + 5^2} = \sqrt{169} = 13$ cm.

The total surface area consists of the curved surface of the cylinder, the circular base of the cylinder and the curved surface of the cone:

$\text{TSA} = 2\pi r h + \pi r^2 + \pi r l = \pi r(2h + r + l)$

$= \frac{22}{7}\times 5\times(256 + 5 + 13) = \frac{22}{7}\times 5\times 274 = \frac{30140}{7} \approx 4305.7 \text{ cm}^2$

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2075 Chaitra · Set CP · 2 marks

If the volume of a hemisphere is $486\pi$ cubic cm, then find its radius.

Similar questions: Radius of hemisphere from volume 9216 pi (2075 Chaitra · Set AP) · Radius of hemisphere from volume 19404 (2075 Chaitra · Set BP) · Radius of hemisphere from volume 3888 pi (2075 Chaitra · Set EP)

Answer

Volume of a hemisphere $=\dfrac23\pi r^3$ .

$\frac23\pi r^3=486\pi \Rightarrow r^3=486\times\frac32=729 \Rightarrow r=\sqrt[3]{729}=9.$

The radius is 9 cm.

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2075 Chaitra · Set CP · 4 marks

The given figure is a combined solid made up of a cylinder and a cone. The diameter of the base of the solid object is 6 cm, the length of the cylindrical part is 116 cm and the total length of the solid object is 120 cm. Find the total surface area of the solid object.

Similar questions: Total surface area of cylinder and cone solid (2075 Chaitra · Set AP) · Total surface area of cylinder and cone solid, diameter 10 cm (2075 Chaitra · Set BP) · Total surface area of cylinder 150 cm and cone, total length 180 cm (2075 Chaitra · Set DP)

Answer

Diameter $=6$ cm, so radius $r=3$ cm.

Height of the cylinder $h=116$ cm. Height of the cone $=120-116=4$ cm.

Slant height of the cone:

$l=\sqrt{r^2+H^2}=\sqrt{3^2+4^2}=5\text{ cm}.$

The total surface area consists of the curved surface of the cylinder, the circular base of the cylinder and the curved surface of the cone:

$\text{TSA}=2\pi rh+\pi r^2+\pi rl=\pi r(2h+r+l)$

$=\pi\times3\times(232+3+5)=720\pi.$

Using $\pi=\dfrac{22}{7}$ :

$\text{TSA}=720\times\frac{22}{7}\approx\mathbf{2262.86\ cm^2}\ (=720\pi\ \text{cm}^2).$

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2075 Chaitra · Set EP · 2 marks

If the volume of a hemisphere is $3888\pi$ cubic cm, then find its radius.

Similar questions: Radius of hemisphere from volume 9216 pi (2075 Chaitra · Set AP) · Radius of hemisphere from volume 19404 (2075 Chaitra · Set BP) · Radius of hemisphere from volume 486 pi (2075 Chaitra · Set CP)

Answer

Volume of a hemisphere $=\dfrac23\pi r^3$ .

$\frac23\pi r^3=3888\pi \Rightarrow r^3=3888\times\frac32=5832 \Rightarrow r=\sqrt[3]{5832}=18.$

The radius is 18 cm.

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2075 Chaitra · Set EP · 4 marks

The given figure is a combined solid made up of a cylinder and a cone. The radius of the base of the solid object is 7 cm, the length of the cylindrical part is 35 cm and the total length of the solid object is 59 cm. Find the total surface area of the solid object.

Similar questions: Total surface area of cylinder and cone solid, radius 6 cm and length 28 cm (2075 Chaitra · Set FP) · Curved surface area of cylinder and cone solid (2075 Supplementary) · Total surface area of cylinder and cone solid (2075 Chaitra · Set AP)

Answer

Radius $r = 7$ cm; cylinder height $h = 35$ cm; cone height $= 59 - 35 = 24$ cm.

Slant height of the cone: $l = \sqrt{7^2 + 24^2} = \sqrt{625} = 25$ cm.

Total surface area $=$ curved surface of cone $+$ curved surface of cylinder $+$ base of cylinder

$= \pi r l + 2\pi r h + \pi r^2 = \pi r(l + 2h + r) = \frac{22}{7}\times 7\times(25+70+7)$

$= 22 \times 102 = \textbf{2244 sq. cm}$

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2075 Chaitra · Set FP · 2 marks

Find the area of $\triangle ABC$ in the given figure.
Figure: Triangle $ABC$ with $AB = 12$ cm, $AC = 8$ cm and $BC = 13$ cm.

Similar questions: Area of triangle with sides 4, 3 and 5 (2075 Chaitra · Set AP) · Area of triangle with sides 12, 14 and 20 (2075 Supplementary) · Area of triangle with sides 5, 12 and 13 (2075 Chaitra · Set BP)

Answer

Sides: $a = 13$ , $b = 8$ , $c = 12$ cm.

$s = \dfrac{13+8+12}{2} = 16.5$ cm

$s-a = 3.5$ , $s-b = 8.5$ , $s-c = 4.5$

$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{16.5\times 3.5\times 8.5\times 4.5} = \sqrt{2208.94} \approx \textbf{47 sq. cm}$

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2075 Chaitra · Set FP · 2 marks

If the volume of a hemisphere is $3888\pi$ cubic cm, then find its diameter.

Similar questions: Diameter of hemisphere from volume 144 pi (2075 Supplementary) · Radius of hemisphere from volume 9216 pi (2075 Chaitra · Set AP) · Radius of hemisphere from volume 19404 (2075 Chaitra · Set BP)

Answer

Volume of a hemisphere $= \dfrac23\pi r^3$ .

$\frac23\pi r^3 = 3888\pi \Rightarrow r^3 = 3888\times\frac32 = 5832 \Rightarrow r = 18 \text{ cm}$

Diameter $= 2r =$ 36 cm.

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2075 Chaitra · Set FP · 4 marks

The given figure is a combined solid made up of a cylinder and a cone. The radius of the base of the solid object is 6 cm, the length of the cylindrical part is 20 cm and the total length of the solid object is 28 cm. Find the total surface area of the solid object.

Similar questions: Total surface area of cylinder and cone solid, radius 7 cm and length 59 cm (2075 Chaitra · Set EP) · Curved surface area of cylinder and cone solid (2075 Supplementary) · Total surface area of cylinder and cone solid (2075 Chaitra · Set AP)

Answer

Radius $r = 6$ cm, height of cylinder $H = 20$ cm.

Height of cone $h = 28 - 20 = 8$ cm.

Slant height of cone: $l = \sqrt{h^2 + r^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10$ cm.

The surface of the solid consists of the curved surface of the cone, the curved surface of the cylinder and the circular base of the cylinder:

\text{TSA} = \pi r l + 2\pi r H + \pi r^2 = \pi r (l + 2H + r)

= \frac{22}{7} \times 6 \times (10 + 40 + 6) = \frac{22}{7} \times 6 \times 56 = 1056 \text{ cm}^2

Answer: total surface area $= 1056$ cm $^2$

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2075 Chaitra · Set GP · 2 marks

Find the area of $\triangle EFG$ in the given figure.
Figure: Triangle $EFG$ with $EF = 15$ cm, $EG = 8$ cm and $FG = 17$ cm.

Similar questions: Area of triangle with sides 4, 3 and 5 (2075 Chaitra · Set AP) · Area of triangle with sides 12, 14 and 20 (2075 Supplementary) · Area of triangle with sides 5, 12 and 13 (2075 Chaitra · Set BP)

Answer

Check: $15^2 + 8^2 = 225 + 64 = 289 = 17^2$ , so the triangle is right-angled at $E$ (by the converse of Pythagoras' theorem), with $EF$ and $EG$ perpendicular.

\text{Area} = \frac12 \times EF \times EG = \frac12 \times 15 \times 8 = 60 \text{ cm}^2

Answer: $60$ cm $^2$

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2075 Chaitra · Set GP · 2 marks

If the volume of a hemisphere is $144\pi$ cubic cm, then find its radius.

Similar questions: Radius of hemisphere from volume 9216 pi (2075 Chaitra · Set AP) · Radius of hemisphere from volume 19404 (2075 Chaitra · Set BP) · Radius of hemisphere from volume 486 pi (2075 Chaitra · Set CP)

Answer

Volume of a hemisphere $= \dfrac23 \pi r^3$ .

\frac23 \pi r^3 = 144\pi \Rightarrow r^3 = 216 \Rightarrow r = 6

Answer: radius $= 6$ cm

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2075 Chaitra · Set GP · 4 marks

The given figure is a combined solid made up of a cylinder and a cone. The diameter of the base of the object is 14 cm, the length of the cylindrical part is 45 cm and the length of the conical part is 24 cm. Find the total surface area of that solid object.

Similar questions: Total surface area of cylinder and cone solid (2075 Chaitra · Set AP) · Total surface area of cylinder and cone solid, diameter 10 cm (2075 Chaitra · Set BP) · Total surface area of cylinder and cone solid, diameter 6 cm (2075 Chaitra · Set CP)

Answer

Radius $r = \dfrac{14}{2} = 7$ cm, cylinder height $h = 45$ cm, cone height $H = 24$ cm.

Slant height of the cone: $l = \sqrt{r^2 + H^2} = \sqrt{7^2 + 24^2} = \sqrt{625} = 25 \text{ cm}$

The surfaces on the outside are the curved surface of the cylinder, the flat circular base of the cylinder, and the curved surface of the cone (the joined circle is hidden).

$\text{TSA} = 2\pi r h + \pi r^2 + \pi r l = \pi r(2h + r + l)$ $= \frac{22}{7} \times 7 \times (90 + 7 + 25) = 22 \times 122 = 2684 \text{ cm}^2$

The total surface area of the solid is $2684$ cm $^2$ .

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2075 Chaitra · Set DP · 2 marks

Find the area of $\triangle MNO$ in the given figure.
Figure: Triangle $MNO$ with $MN = 6$ cm, $MO = 7$ cm and $NO = 5$ cm.

Similar questions: Area of triangle with sides 4, 3 and 5 (2075 Chaitra · Set AP) · Area of triangle with sides 12, 14 and 20 (2075 Supplementary) · Area of triangle with sides 5, 12 and 13 (2075 Chaitra · Set BP)

Answer

Sides: $a = 6$ cm, $b = 7$ cm, $c = 5$ cm.

$s = \frac{6 + 7 + 5}{2} = 9 \text{ cm}$ $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{9 \times 3 \times 2 \times 4} = \sqrt{216} = 6\sqrt{6} \approx 14.7 \text{ cm}^2$

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2075 Chaitra · Set DP · 2 marks

If the volume of a hemisphere is $18\pi$ cubic cm, then find its radius.

Similar questions: Radius of hemisphere from volume 9216 pi (2075 Chaitra · Set AP) · Radius of hemisphere from volume 19404 (2075 Chaitra · Set BP) · Radius of hemisphere from volume 486 pi (2075 Chaitra · Set CP)

Answer

Volume of a hemisphere $= \dfrac{2}{3}\pi r^3 = 18\pi$

$r^3 = 18 \times \frac{3}{2} = 27 \Rightarrow r = 3 \text{ cm}$

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2075 Chaitra · Set DP · 4 marks

The given figure is a combined solid object made up of a cylinder and a cone. The diameter of the base of the solid object is 32 cm, the length of the cylindrical part is 150 cm and the total length of the solid object is 180 cm. Find the total surface area of the solid object.

Similar questions: Total surface area of cylinder and cone solid (2075 Chaitra · Set AP) · Total surface area of cylinder and cone solid, diameter 10 cm (2075 Chaitra · Set BP) · Total surface area of cylinder and cone solid, diameter 6 cm (2075 Chaitra · Set CP)

Answer

Radius $r = \dfrac{32}{2} = 16$ cm, cylinder height $h = 150$ cm.

Cone height $H = 180 - 150 = 30$ cm.

Slant height of the cone: $l = \sqrt{r^2 + H^2} = \sqrt{16^2 + 30^2} = \sqrt{1156} = 34 \text{ cm}$

The outer surface consists of the curved surface of the cylinder, its flat base, and the curved surface of the cone:

$\text{TSA} = \pi r(2h + r + l) = \frac{22}{7} \times 16 \times (300 + 16 + 34)$ $= \frac{22}{7} \times 16 \times 350 = 22 \times 16 \times 50 = 17600 \text{ cm}^2$

The total surface area of the solid is $17600$ cm $^2$ .

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2082 Pre-board · PABSON Set B · 1+1+2 marks

The area of the triangular surfaces of a square based pyramid is 320 cm $^2$ and the length of the side of the base is 16 cm.
(a) Write the formula to find the total surface area of a square based pyramid. (1)
(b) Find the slant height of the pyramid. (1)
(c) Find the volume of the pyramid. (2)

Similar questions: Square pyramid with base area 64 sq cm and height 3 cm (2080 Model · Set 3) · Pyramid volume and surface area (2081 Chaitra · Lumbini) · Pyramid surface area and volume (2082 Chaitra · Lumbini)

Answer

(a) Total surface area $=a^2+2al$ , where $a$ is the side of the base and $l$ is the slant height.

(b) Area of the triangular surfaces $=2al=320$ :

2\times16\times l=320\Rightarrow l=\mathbf{10\text{ cm}}

(c) Vertical height: $h=\sqrt{l^2-\left(\frac a2\right)^2}=\sqrt{100-64}=6$ cm.

V=\frac13a^2h=\frac13\times16^2\times6=\mathbf{512\text{ cm}^3}

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Most repeated · 7 of 45 exams
Asked 7 times
2075 Chaitra · Set AP · 1 mark
2075 Chaitra · Set BP · 1 mark
2075 Chaitra · Set CP · 1 mark
2075 Chaitra · Set EP · 1 mark
2075 Chaitra · Set FP · 1 mark
2075 Chaitra · Set GP · 1 mark
2075 Chaitra · Set DP · 1 mark

Write the area of an isosceles triangle having equal sides $x$ cm and third side $y$ cm.

Answer

Take $a=b=x$ and $c=y$ . The height on the base $y$ is $\sqrt{x^2-\frac{y^2}{4}}$ , so

$\text{Area}=\frac{1}{2}\times y\times\sqrt{x^2-\frac{y^2}{4}}=\frac{y}{4}\sqrt{4x^2-y^2}\ \text{cm}^2.$

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2075 Supplementary · 2 marks

In the given figure, $WXYZ$ is a parallelogram. If $\angle WMZ = 90^\circ$ , $MZ = 4$ cm and $WM = 6$ cm, find the area of the parallelogram $WXYZ$ .
Figure: Parallelogram $WXYZ$ with $WZ \parallel XY$ ; the point $M$ lies on the side $XY$ and $WM = 6$ cm, $MZ = 4$ cm.

Similar questions: Parallelogram with a right angle at A on the top side (2075 Chaitra · Set AP) · Parallelogram of area 64 with a right angle at A (2075 Chaitra · Set CP) · Parallelogram of area 60 with a right angle at M (2075 Chaitra · Set EP)

Answer

In $\triangle WMZ$ , $\angle WMZ=90^\circ$ , so its area is

$\frac12\times WM\times MZ=\frac12\times6\times4=12\ \text{cm}^2.$

$\triangle WMZ$ and the parallelogram $WXYZ$ stand on the same base $WZ$ and lie between the same parallels $WZ$ and $XY$ (as $M$ lies on $XY$ ). Therefore

$\text{Area of }WXYZ=2\times\text{Area of }\triangle WMZ=2\times12=\textbf{24 cm}^2.$

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2075 Chaitra · Set AP · 2 marks

In the given figure, $WXYZ$ is a parallelogram. If $\angle XAY = 90^\circ$ , $AY = 12$ cm and the area of the parallelogram $WXYZ$ is $120$ sq. cm, find the length of $AX$ .
Figure: Parallelogram $WXYZ$ with $WZ \parallel XY$ ; the point $A$ lies on the side $WZ$ and $AX$ , $AY$ are joined.

Similar questions: Parallelogram of area 64 with a right angle at A (2075 Chaitra · Set CP) · Parallelogram of area 60 with a right angle at M (2075 Chaitra · Set EP) · Parallelogram of area 48 with a right angle at E (2075 Chaitra · Set FP)

Answer

$\triangle AXY$ and the parallelogram $WXYZ$ stand on the same base $XY$ and lie between the same parallels $WZ$ and $XY$ (as $A$ lies on $WZ$ ). Therefore

$\text{Area of }\triangle AXY=\frac12\times120=60\ \text{cm}^2.$

Since $\angle XAY=90^\circ$ :

$\frac12\times AX\times AY=60\Rightarrow\frac12\times AX\times12=60\Rightarrow AX=\textbf{10 cm}.$

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2075 Chaitra · Set AP · 5 marks

Two pillars of height 8 feet each, whose four faces are shown in the gate of a stadium, have one pyramid of height 4 feet each on their tops, with the same base. The base of each pillar is 6 ft $\times$ 6 ft. If the pillars with the pyramids are painted at the rate of Rs. 60 per square feet, what will be the total cost? Find it.

Similar questions: Painting two pillars with pyramids at Rs 100 per sq ft (2075 Chaitra · Set BP) · Painting two pillars with pyramids at Rs 75 per sq ft (2075 Chaitra · Set CP) · Painting two pillars with pyramids at Rs 90 per sq ft (2075 Chaitra · Set EP)

Answer

Each pillar is a square prism ( $6$ ft $\times$ $6$ ft base, height $8$ ft) with a pyramid of height $4$ ft on top. The base on the ground and the top of the prism (covered by the pyramid) are not painted, so the painted surface is the four side faces of the prism and the four slant faces of the pyramid.

Prism: lateral surface area $= 4 \times 6 \times 8 = 192$ sq. ft.

Pyramid: half of the base side is $3$ ft, so slant height

$l = \sqrt{4^2 + 3^2} = 5 \text{ ft}.$

Lateral surface area $= 4 \times \dfrac{1}{2} \times 6 \times 5 = 60$ sq. ft.

Area to paint on one pillar $= 192 + 60 = 252$ sq. ft.

Area on two pillars $= 2 \times 252 = 504$ sq. ft.

$\text{Cost} = 504 \times 60 = \text{Rs. } 30240$

The total cost of painting is Rs. 30,240.

Answer checked by Claude Sonnet · Oct 21, 2026

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2075 Chaitra · Set BP · 5 marks

Two pillars of height 8 feet each with four faces shown, of the gate of a stadium, have one pyramid of height 4 feet each having the same base on their tops. The base of each pillar is 6 ft $\times$ 6 ft. If the pillars with the pyramids are painted at the rate of Rs. 100 per square feet, what will be the total cost? Find it.

Similar questions: Painting two pillars with pyramids at Rs 75 per sq ft (2075 Chaitra · Set CP) · Painting two pillars with pyramids at Rs 90 per sq ft (2075 Chaitra · Set EP) · Painting two pillars with pyramids at Rs 110 per sq ft (2075 Chaitra · Set FP)

Answer

One pillar (four lateral faces; the top is covered by the pyramid):

$4\times(6\times 8) = 192 \text{ sq. ft}$

One pyramid (base $6\times 6$ , height 4 ft). Half the base side is $3$ ft, so the slant height is

$l = \sqrt{4^2 + 3^2} = 5 \text{ ft}$

$\text{Lateral area} = 4\times\tfrac{1}{2}\times 6\times 5 = 60 \text{ sq. ft}$

Area to be painted for one pillar with its pyramid $= 192 + 60 = 252$ sq. ft.

For two pillars: $2\times 252 = 504$ sq. ft.

$\text{Total cost} = 504\times 100 = \text{Rs } 50{,}400$

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2075 Chaitra · Set CP · 2 marks

In the given figure, $WXYZ$ is a parallelogram. If $\angle XAY = 90^\circ$ , $AY = 8$ cm and the area of the parallelogram $WXYZ$ is 64 sq. cm, find the length of $AX$ .
Figure: Parallelogram $WXYZ$ with $WZ \parallel XY$ ; the point $A$ lies on the side $WZ$ and $AX$ , $AY$ are joined.

Similar questions: Parallelogram with a right angle at A on the top side (2075 Chaitra · Set AP) · Parallelogram of area 60 with a right angle at M (2075 Chaitra · Set EP) · Parallelogram of area 48 with a right angle at E (2075 Chaitra · Set FP)

Answer

$\triangle XAY$ and the parallelogram $WXYZ$ stand on the same base $XY$ and lie between the same parallels $WZ$ and $XY$ (as $A$ lies on $WZ$ ). Therefore

$\text{Area}(\triangle XAY)=\frac12\times\text{Area}(WXYZ)=\frac12\times64=32\text{ cm}^2.$

Since $\angle XAY=90^\circ$ :

$\frac12\times AX\times AY=32 \Rightarrow \frac12\times AX\times8=32 \Rightarrow AX=8.$

So $AX=$ 8 cm.

Answer checked by Claude Sonnet · Oct 21, 2026

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2075 Chaitra · Set CP · 5 marks

Two pillars of height 8 feet each with four faces shown, of the gate of a stadium, have one pyramid of height 4 feet each having the same base on their tops. The base of each pillar is 6 ft $\times$ 6 ft. If the pillars with the pyramids are painted at the rate of Rs. 75 per square feet, what will be the total cost? Find it.

Similar questions: Painting two pillars with pyramids at Rs 100 per sq ft (2075 Chaitra · Set BP) · Painting two pillars with pyramids at Rs 90 per sq ft (2075 Chaitra · Set EP) · Painting two pillars with pyramids at Rs 110 per sq ft (2075 Chaitra · Set FP)

Answer

Only the four side faces of each pillar are painted (the base rests on the ground and the top is covered by the pyramid), plus the four slant faces of each pyramid.

Pillar (base $6\times6$ ft, height $8$ ft):

$\text{Lateral surface area}=4\times6\times8=192\text{ sq. ft}.$

Pyramid (base $6\times6$ ft, height $4$ ft):

Slant height of a face (from the apex to the middle of a base edge):

$l=\sqrt{4^2+\left(\frac62\right)^2}=\sqrt{16+9}=5\text{ ft}.$

$\text{Slant surface area}=4\times\frac12\times6\times5=60\text{ sq. ft}.$

Area to be painted for one pillar with its pyramid $=192+60=252$ sq. ft.

For two pillars $=2\times252=504$ sq. ft.

$\text{Total cost}=504\times75=\textbf{Rs 37,800}.$

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2075 Chaitra · Set EP · 2 marks

In the given figure, $ABCD$ is a parallelogram. If $\angle AMD = 90^\circ$ , $DM = 10$ cm and the area of $\square ABCD$ is 60 sq. cm, find the length of $AM$ .
Figure: Parallelogram $ABCD$ with $DC \parallel AB$ ; the point $M$ lies on the side $DC$ and $AM$ is joined.

Similar questions: Parallelogram with a right angle at A on the top side (2075 Chaitra · Set AP) · Parallelogram of area 64 with a right angle at A (2075 Chaitra · Set CP) · Parallelogram of area 48 with a right angle at E (2075 Chaitra · Set FP)

Answer

Since $\angle AMD = 90^\circ$ , $AM$ is perpendicular to the base $DC$ , so $AM$ is the height of the parallelogram on the base marked $10$ cm.

Area of parallelogram $=$ base $\times$ height

$60 = 10 \times AM \Rightarrow AM = \frac{60}{10} = \textbf{6 cm}$

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2075 Chaitra · Set EP · 5 marks

Two pillars of height 8 feet each with four faces shown, of the gate of a stadium, have one pyramid of height 4 feet each having the same base on their tops. The base of each pillar is 6 ft $\times$ 6 ft. If the pillars with the pyramids are painted at the rate of Rs. 90 per square feet, what will be the total cost? Find it.

Similar questions: Painting two pillars with pyramids at Rs 100 per sq ft (2075 Chaitra · Set BP) · Painting two pillars with pyramids at Rs 75 per sq ft (2075 Chaitra · Set CP) · Painting two pillars with pyramids at Rs 110 per sq ft (2075 Chaitra · Set FP)

Answer

Pillar (4 side faces only, the top is covered by the pyramid):

Area of four faces $= 4 \times (6 \times 8) = 192$ sq. ft.

Pyramid: base $6\times 6$ ft, height $4$ ft.

Slant height $l = \sqrt{4^2 + \left(\frac{6}{2}\right)^2} = \sqrt{16+9} = 5$ ft.

Lateral area of the pyramid $= 4\times \dfrac12 \times 6 \times 5 = 60$ sq. ft.

One pillar with its pyramid: $192 + 60 = 252$ sq. ft.

Two pillars: $2 \times 252 = 504$ sq. ft.

$\text{Total cost} = 504 \times 90 = \textbf{Rs 45,360}$

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2075 Chaitra · Set FP · 2 marks

In the given figure, $ABCD$ is a parallelogram. If $\angle AEB = 90^\circ$ , $BE = 6$ cm and the area of $\square ABCD$ is 48 sq. cm, find the length of $AE$ .
Figure: Parallelogram $ABCD$ with $AB \parallel DC$ ; the point $E$ lies on the side $DC$ and $AE$ , $BE$ are joined.

Similar questions: Parallelogram with a right angle at A on the top side (2075 Chaitra · Set AP) · Parallelogram of area 64 with a right angle at A (2075 Chaitra · Set CP) · Parallelogram of area 60 with a right angle at M (2075 Chaitra · Set EP)

Answer

Triangle $ABE$ and parallelogram $ABCD$ stand on the same base $AB$ and lie between the same parallels $AB$ and $DC$ ( $E$ lies on $DC$ ). So

\text{Area of } \triangle ABE = \frac12 \times \text{Area of } \square ABCD = \frac12 \times 48 = 24 \text{ cm}^2

Since $\angle AEB = 90^\circ$ , $AE$ and $BE$ are the perpendicular sides of the triangle:

\frac12 \times AE \times BE = 24 \Rightarrow \frac12 \times AE \times 6 = 24 \Rightarrow AE = 8 \text{ cm}

Answer: $AE = 8$ cm

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2075 Chaitra · Set FP · 5 marks

Two pillars of height 8 ft each with four faces shown, of the gate of a stadium, have one pyramid of height 4 ft each having the same base on their tops. The base of each pillar is 6 ft $\times$ 6 ft. If the pillars with the pyramids are painted at the rate of Rs. 110 per square feet, what will be the total cost? Find it.

Similar questions: Painting two pillars with pyramids at Rs 100 per sq ft (2075 Chaitra · Set BP) · Painting two pillars with pyramids at Rs 75 per sq ft (2075 Chaitra · Set CP) · Painting two pillars with pyramids at Rs 90 per sq ft (2075 Chaitra · Set EP)

Answer

One pillar: a square-based cuboid of base $6 \times 6$ ft and height 8 ft. Only the four side faces are painted (the top is covered by the pyramid).

\text{Four faces} = 4 \times 6 \times 8 = 192 \text{ sq. ft}

One pyramid: base side $a = 6$ ft, height $h = 4$ ft.

\text{Slant height } l = \sqrt{h^2 + \left(\frac a2\right)^2} = \sqrt{4^2 + 3^2} = 5 \text{ ft}

\text{Four triangular faces} = 2al = 2 \times 6 \times 5 = 60 \text{ sq. ft}

Area to paint on one pillar with its pyramid $= 192 + 60 = 252$ sq. ft.

Area for the two pillars $= 2 \times 252 = 504$ sq. ft.

\text{Total cost} = 504 \times 110 = \text{Rs } 55{,}440

Answer: Rs 55,440

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2075 Chaitra · Set GP · 2 marks

In the given figure, $PQRS$ is a parallelogram. If $\angle PTQ = 90^\circ$ , $PT = 8$ cm and the area of $\square PQRS$ is 80 sq. cm, find the length of $QT$ .
Figure: Parallelogram $PQRS$ with $PS \parallel QR$ ; the point $T$ lies on the side $PS$ and $PT$ , $QT$ are joined.

Similar questions: Parallelogram with a right angle at A on the top side (2075 Chaitra · Set AP) · Parallelogram of area 64 with a right angle at A (2075 Chaitra · Set CP) · Parallelogram of area 60 with a right angle at M (2075 Chaitra · Set EP)

Answer

Here $T$ lies on the side opposite to $PQ$ (the side parallel to $PQ$ ), so $\triangle PQT$ and parallelogram $PQRS$ stand on the same base $PQ$ and lie between the same parallels. Hence

\text{Area of } \triangle PQT = \frac12 \times \text{Area of } \square PQRS = \frac12 \times 80 = 40 \text{ cm}^2

Since $\angle PTQ = 90^\circ$ , $PT$ and $QT$ are the perpendicular sides of the triangle:

\frac12 \times PT \times QT = 40 \Rightarrow \frac12 \times 8 \times QT = 40 \Rightarrow QT = 10 \text{ cm}

Answer: $QT = 10$ cm

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Most repeated · 7 of 45 exams
2075 Chaitra · Set GP · 5 marks

Two pillars of height 8 feet each with four faces shown, of the gate of a stadium, have one pyramid of height 4 feet each having the same base on their tops. The base of each pillar is 6 ft $\times$ 6 ft. If the pillars with the pyramids are painted at the rate of Rs. 80 per square feet, what will be the total cost? Find it.

Similar questions: Painting two pillars with pyramids at Rs 100 per sq ft (2075 Chaitra · Set BP) · Painting two pillars with pyramids at Rs 75 per sq ft (2075 Chaitra · Set CP) · Painting two pillars with pyramids at Rs 90 per sq ft (2075 Chaitra · Set EP)

Answer

Each pillar is a cuboid with a square base of side $6$ ft and height $8$ ft, with a square pyramid (base $6 \times 6$ , height $4$ ft) on top. Only the four side faces of each pillar and the four slant faces of each pyramid are painted.

Pillar (four faces): $4 \times 6 \times 8 = 192 \text{ sq. ft}$

Pyramid: slant height $l = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 \text{ ft}$ $\text{Slant surface} = 4 \times \tfrac{1}{2} \times 6 \times 5 = 60 \text{ sq. ft}$

Area for one pillar with its pyramid $= 192 + 60 = 252$ sq. ft.

Area for two $= 2 \times 252 = 504$ sq. ft.

$\text{Cost} = 504 \times 80 = \text{Rs } 40{,}320$

The total cost of painting is Rs 40,320.

Answer checked by Claude Sonnet · Oct 21, 2026

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2075 Chaitra · Set DP · 2 marks

In the given figure, $GOAT$ is a parallelogram. If $\angle AMT = 90^\circ$ , $MT = 16$ cm and the area of $\square GOAT$ is 160 sq. cm, find the length of $MA$ .
Figure: Parallelogram $GOAT$ with $GT \parallel OA$ ; the point $M$ lies on the side $GO$ and $MA$ , $MT$ are joined.

Similar questions: Parallelogram with a right angle at A on the top side (2075 Chaitra · Set AP) · Parallelogram of area 64 with a right angle at A (2075 Chaitra · Set CP) · Parallelogram of area 60 with a right angle at M (2075 Chaitra · Set EP)

Answer

$\triangle MAT$ and parallelogram $GOAT$ have the same base $AT$ and lie between the same parallels $GO$ and $AT$ (as $M$ lies on $GO$ ). So

$\text{Area}(\triangle MAT) = \tfrac{1}{2} \times 160 = 80 \text{ cm}^2.$

Since $\angle AMT = 90^\circ$ , $MA$ and $MT$ are the perpendicular sides of this triangle:

$\tfrac{1}{2} \times MA \times MT = 80 \Rightarrow \tfrac{1}{2} \times MA \times 16 = 80 \Rightarrow MA = 10 \text{ cm}$

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2075 Chaitra · Set DP · 5 marks

Two pillars of height 8 feet each with four faces shown, of the gate of a stadium, have one pyramid of height 4 feet each having the same base on their tops. The base of each pillar is 6 ft $\times$ 6 ft. If the pillars with the pyramids are painted at the rate of Rs. 85 per square feet, what will be the total cost? Find it.

Similar questions: Painting two pillars with pyramids at Rs 100 per sq ft (2075 Chaitra · Set BP) · Painting two pillars with pyramids at Rs 75 per sq ft (2075 Chaitra · Set CP) · Painting two pillars with pyramids at Rs 90 per sq ft (2075 Chaitra · Set EP)

Answer

Painted surface of one pillar-with-pyramid = 4 side faces of the pillar + 4 triangular faces of the pyramid (the pillar's top is covered by the pyramid).

Pillar: $4 \times (6 \times 8) = 192$ sq. ft

Pyramid: slant height $l=\sqrt{h^2+\left(\frac{a}{2}\right)^2}=\sqrt{4^2+3^2}=5$ ft

$\text{Lateral area}=4\times\frac12\times6\times5=60 \text{ sq. ft}$

Area for one pillar $=192+60=252$ sq. ft

Area for two pillars $=2\times252=504$ sq. ft

$\text{Total cost}=504\times85=\textbf{Rs 42,840}$

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Most repeated · 5 of 45 exams
2082 Chaitra · Lumbini · 1+1+2 marks

A square based pyramid has slant height 10 cm and the length of a side of base is 12 cm.
(a) How many triangular surfaces are there in a square based pyramid? Write it. (1)
(b) What is the total surface area of the pyramid? Find it. (1)
(c) What is the volume of the pyramid? Find it. (2)

Similar questions: Pyramid volume from slant height (2082 Chaitra · Madhesh) · Square pyramid with triangular surfaces 320 sq cm and base side 16 cm (2082 Pre-board · PABSON Set B) · Pyramid from triangular surface area (2081 Shrawan · Grade increment)

Answer

(a) There are 4 triangular surfaces.

(b) $a=12$ cm, $l=10$ cm.

Total surface area $=a^2+2al=144+2\times12\times10=144+240=384$ cm $^2$ .

(c) $h=\sqrt{l^2-(a/2)^2}=\sqrt{100-36}=8$ cm.

$V=\frac13a^2h=\frac13\times144\times8=384$ cm $^3$ .

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2080 Model · Set 3 · 1+1+2 marks

The base area of a square based pyramid is 64 sq. cm and its height is 3 cm.
(a) Write the formula to find the area of the triangular surfaces of a square based pyramid. (1)
(b) Find the length of the side of the base of the pyramid. (1)
(c) Find the total surface area of the pyramid. (2)

Similar questions: Square pyramid with triangular surfaces 320 sq cm and base side 16 cm (2082 Pre-board · PABSON Set B) · Pyramid volume and surface area (2081 Chaitra · Lumbini) · Pyramid volume and triangular area (2082 Chaitra · Gandaki)

Answer

(a) Area of the triangular surfaces $=4\times\dfrac12\times b\times l=2bl$ , where $b$ is the side of the base and $l$ the slant height.

(b) Base area $=b^2=64\Rightarrow b=8$ cm.

(c) Slant height: $l=\sqrt{h^2+\left(\frac b2\right)^2}=\sqrt{3^2+4^2}=5$ cm.

Area of triangular surfaces $=2\times8\times5=80$ cm $^2$ .

$\text{Total surface area}=64+80=144\text{ cm}^2.$

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2082 Chaitra · Madhesh · 1+3+1 marks

The length of the base side and slant height of a square based pyramid are 48 cm and 25 cm respectively.
(a) How many surfaces are there in square based pyramid? Write it. (1)
(b) Find the volume of the pyramid. (3)
(c) Find the area of a triangular surface of the pyramid. (1)

Similar questions: Pyramid surface area and volume (2082 Chaitra · Lumbini) · Square pyramid with triangular surfaces 320 sq cm and base side 16 cm (2082 Pre-board · PABSON Set B) · Pyramid from triangular surface area (2081 Shrawan · Grade increment)

Answer

(a) A square based pyramid has 5 surfaces (1 base and 4 triangular faces).

(b) $a=48$ cm, $l=25$ cm. Half base $=24$ cm.

Vertical height $h=\sqrt{l^2-24^2}=\sqrt{625-576}=\sqrt{49}=7$ cm.

$V=\frac13a^2h=\frac13\times48^2\times7=\frac13\times2304\times7=5376\text{ cm}^3$

(c) Area of one triangular surface $=\frac12\times a\times l=\frac12\times48\times25=600$ cm $^2$ .

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2081 Chaitra · Lumbini · 1+2+2 marks

The vertical height of a square based pyramid is 24 cm and length of the base side is 14 cm.
(a) Write the formula to find the area of a triangular surface of the pyramid. (1)
(b) Find the volume of the pyramid. (2)
(c) Find the total surface area of the pyramid. (2)

Similar questions: Pyramid volume and surface area (2081 Chaitra · Sudurpashchim) · Square pyramid with triangular surfaces 320 sq cm and base side 16 cm (2082 Pre-board · PABSON Set B) · Square pyramid with base area 64 sq cm and height 3 cm (2080 Model · Set 3)

Answer

(a) Area of a triangular surface $=\dfrac12al$ .

(b) $V=\frac13a^2h=\frac13\times14^2\times24=\frac13\times196\times24=1568$ cm $^3$ .

(c) $l=\sqrt{24^2+7^2}=\sqrt{625}=25$ cm.

Total surface area $=a^2+2al=196+2\times14\times25=196+700=896$ cm $^2$ .

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2081 Chaitra · Sudurpashchim · 1+2 marks

The vertical height of the square based pyramid is 24 cm and the length of one side of base is 20 cm.
(a) Write the formula to find the volume of the pyramid. (1)
(b) Find the total surface area of the pyramid. (2)

Similar questions: Pyramid volume and surface area (2081 Chaitra · Lumbini) · Square pyramid with triangular surfaces 320 sq cm and base side 16 cm (2082 Pre-board · PABSON Set B) · Square pyramid with base area 64 sq cm and height 3 cm (2080 Model · Set 3)

Answer

(a) $V=\dfrac13a^2h$ .

(b) $a=20$ cm, $h=24$ cm. $l=\sqrt{24^2+10^2}=\sqrt{676}=26$ cm.

Total surface area $=a^2+2al=400+2\times20\times26=400+1040=1440$ cm $^2$ .

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2082 Chaitra · Gandaki · 4 marks

The length of side of the base of a square based pyramid is 24 cm and height is 9 cm.
(a) Write the formula for finding the volume of a square based pyramid. (1)
(b) Find the volume of the given pyramid. (1)
(c) Find the area of triangular surfaces of the pyramid. (2)

Similar questions: Square pyramid with triangular surfaces 320 sq cm and base side 16 cm (2082 Pre-board · PABSON Set B) · Square pyramid with base area 64 sq cm and height 3 cm (2080 Model · Set 3)

Answer

(a) $V=\dfrac13a^2h$ .

(b) $a=24$ cm, $h=9$ cm: $V=\frac13\times24^2\times9=\frac13\times576\times9=1728$ cm $^3$ .

(c) Slant height $l=\sqrt{h^2+(a/2)^2}=\sqrt{81+144}=\sqrt{225}=15$ cm.

Area of the 4 triangular surfaces $=2al=2\times24\times15=720$ cm $^2$ .

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Most repeated · 3 of 45 exams
2082 Chaitra · Karnali · 3+2 marks

The volume and height of a square based room are 243 cubic meter and 3 meter respectively. The area occupied by a door and two windows in the room is 12 sq. meter.
(a) What is the total cost of plastering the four walls at the rate of Rs.250 per square meter? Find it. (3)
(b) If the rate of plastering per square meter is increased by one-fourth, then what will be the increment in the total cost of plastering the walls? Find it. (2)

Similar questions: Plastering a square based room of volume 87.5 cubic m (2080 Model · Set 3) · Square room plastering and carpet (2081 Chaitra · Sudurpashchim)

Answer

(a) Base area $=\dfrac{V}{h}=\dfrac{243}{3}=81$ m $^2$ , so side $=9$ m.

Area of four walls $=4\times9\times3=108$ m $^2$ .

Area to plaster $=108-12=96$ m $^2$ .

Cost $=96\times250=$ Rs 24,000.

(b) Increase in rate $=\frac14\times250=$ Rs 62.50 per m $^2$ (new rate Rs 312.50).

Increase in total cost $=96\times62.5=$ Rs 6,000.

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Most repeated · 3 of 45 exams
2081 Shrawan · Grade increment · 1+3+2 marks

The area of triangular surfaces of a square based pyramid is 80 square cm and slant height is 5 cm.
(a) How many surfaces are there in a square based pyramid? Write it. (1)
(b) Find the total surface area of the pyramid. (3)
(c) Find the volume of the pyramid. (2)

Similar questions: Pyramid surface area and volume (2082 Chaitra · Lumbini) · Pyramid volume from slant height (2082 Chaitra · Madhesh)

Answer

(a) 5 surfaces (4 triangular + 1 square base).

(b) $2al=80$ , $l=5\Rightarrow a=8$ cm.

Base area $=64$ cm $^2$ .

Total surface area $=64+80=144$ cm $^2$ .

(c) $h=\sqrt{l^2-(a/2)^2}=\sqrt{25-16}=3$ cm.

$V=\frac13a^2h=\frac13\times64\times3=64$ cm $^3$ .

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Most repeated · 3 of 45 exams
2081 Shrawan · Grade increment · 1+2+1 marks

The length of a wall is 10 m, width is 0.5 m and height is 2 m. Bricks of size 25 cm × 12 cm × 8 cm are used to build the wall.
(a) Find the volume of the wall. (1)
(b) How many bricks are required to construct the wall? Find it. (2)
(c) Estimate the cost of bricks used in the wall at the rate of Rs 15 per brick. (1)

Similar questions: Bricks needed for a wall of 15 m by 0.4 m by 4 m (2082 Pre-qualifying · PABSON Kathmandu) · Wall bricks with clay joints (2081 Chaitra · Koshi)

Answer

(a) Volume of wall $=10\times0.5\times2=$ 10 m $^3$ .

(b) $10$ m $^3=10{,}000{,}000$ cm $^3$ . Volume of a brick $=25\times12\times8=2400$ cm $^3$ .

Bricks $=\dfrac{10000000}{2400}\approx4166.67\approx$ 4,167 bricks.

(c) Cost $=4167\times15\approx$ Rs 62,505.

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2075 Supplementary · 1 mark
2078 Baisakh · Set A · 1 mark

What is the area of an equilateral triangle having each side $x$ cm? Write it.

Answer

Area $=\dfrac{\sqrt{3}}{4}x^2$ cm $^2$ .

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2079 Chaitra · Bagmati · 1 mark
2074 Model question · 1 mark

The lengths of the sides of a triangle are $x$ cm, $y$ cm and $z$ cm. What is the semi-perimeter of the triangle?

Answer

Semi-perimeter $s = \dfrac{x + y + z}{2}$ cm.

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2082 Chaitra · Koshi · 1+3+1 marks

The length of a side of the base of a square based pyramid is 20 cm, and the length of slant edge is 26 cm.
(a) How many total surfaces are there in a square base pyramid? Write it. (1)
(b) Find the total surface area of the pyramid. (3)
(c) Compare between the vertical height and slant height of the pyramid. (1)

Similar questions: Pyramid surface area and volume (2082 Chaitra · Lumbini)

Answer

(a) A square based pyramid has 5 surfaces (1 square base + 4 triangular faces).

(b) Base side $a=20$ cm, slant edge $=26$ cm.

Half of base side $=10$ cm. Slant height $l=\sqrt{26^2-10^2}=\sqrt{676-100}=\sqrt{576}=24$ cm.

Total surface area $=a^2+2al=20^2+2\times20\times24=400+960=1360$ cm $^2$ .

(c) Vertical height $h=\sqrt{l^2-10^2}=\sqrt{576-100}=\sqrt{476}\approx21.8$ cm.

Slant height $=24$ cm, vertical height $\approx21.8$ cm, so the slant height is greater than the vertical height (by about 2.2 cm), because it is the hypotenuse of the right triangle formed with the vertical height.

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2082 Chaitra · Sudurpashchim · 1+3 marks

A rectangular room has length 16 ft., breadth 14 ft. and height 10 ft. There are two square windows of dimension 3.5 ft. and one door of dimension 3 ft. $\times$ 6 ft.
(a) How much does it cost for carpeting the room at the rate of Rs. 250 per sq. ft? Find it. (1)
(b) If the cost of plastering four walls and ceiling excluding doors and windows of the room is Rs.1,18,200, find the rate of plastering per square feet. (3)

Similar questions: Room carpeting and colouring rate (2081 Chaitra · Madhesh)

Answer

(a) Floor area $=16\times14=224$ sq. ft.

Cost of carpeting $=224\times250=$ Rs 56,000.

(b) Area of four walls $=2(l+b)h=2(16+14)\times10=600$ sq. ft.

Area of ceiling $=224$ sq. ft. Total $=824$ sq. ft.

Windows: $2\times3.5\times3.5=24.5$ sq. ft; door: $3\times6=18$ sq. ft; total $=42.5$ sq. ft.

Area to plaster $=824-42.5=781.5$ sq. ft.

Rate $=\dfrac{118200}{781.5}\approx$ Rs 151.25 per sq. ft.

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2081 Chaitra · Koshi · 3+1 marks

The length of a wall is 10 m, width is 0.5 m and height is 2 m. Bricks of size 25 cm $\times$ 12 cm $\times$ 8 cm are used to build the wall. Also, $\frac{1}{10}$ part of the wall is occupied by the clay joints.
(a) How many bricks are required to construct the wall? Find it. (3)
(b) Estimate the cost of bricks used in the wall at the rate of Rs. 14500 per 1000 bricks. (1)

Similar questions: Wall bricks and cost (2081 Shrawan · Grade increment)

Answer

(a) Volume of wall $=10\times0.5\times2=10$ m $^3$ .

Joints occupy $\frac1{10}$ , so brick volume $=\frac9{10}\times10=9$ m $^3=9{,}000{,}000$ cm $^3$ .

Volume of one brick $=25\times12\times8=2400$ cm $^3$ .

Number of bricks $=\dfrac{9000000}{2400}=$ 3,750.

(b) Cost $=\dfrac{3750}{1000}\times14500=$ Rs 54,375.

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2081 Chaitra · Madhesh · 2+2 marks

The length, breadth and height of a rectangular room are 16 ft, 12 ft and 9 ft respectively. There are two square windows of dimension 4 ft and one door of dimension 6 ft × 2 ft.
(a) How much does it cost for carpeting the room at the rate of Rs. 300 per sq. ft.? Find it. (2)
(b) If the cost of coloring four walls and ceiling excluding doors and windows of the room is Rs.19,560, find the rate of coloring per square feet. (2)

Similar questions: Room carpeting and plastering rate (2082 Chaitra · Sudurpashchim)

Answer

(a) Floor area $=16\times12=192$ sq. ft.

Cost of carpeting $=192\times300=$ Rs 57,600.

(b) Four walls $=2(16+12)\times9=504$ sq. ft; ceiling $=192$ sq. ft; total $=696$ sq. ft.

Windows $=2\times(4\times4)=32$ sq. ft; door $=6\times2=12$ sq. ft; total $=44$ sq. ft.

Area to colour $=696-44=652$ sq. ft.

Rate $=\dfrac{19560}{652}=$ Rs 30 per sq. ft.

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2081 Chaitra · Sudurpashchim · 3+1+1 marks

The volume and height of a square based room are 75 cubic meter and 3 meter respectively. The area occupied by a door and two windows in the room is 6 square meter.
(a) What is the total cost of plastering the four walls without door and windows at the rate of Rs.200 per square meter? Find it. (3)
(b) If the rate of plastering per square meter is increased by one-fourth, then what will be the increment in the total cost of plastering the walls? Find it. (1)
(c) What length of carpet having its width 2 meters should be purchased from the market for carpeting the room? Find it. (1)

Similar questions: Square room plastering cost increase (2082 Chaitra · Karnali)

Answer

(a) Base area $=\dfrac{75}{3}=25$ m $^2$ , so side $=5$ m.

Four walls $=4\times5\times3=60$ m $^2$ . Excluding door and windows: $60-6=54$ m $^2$ .

Cost $=54\times200=$ Rs 10,800.

(b) Increase in rate $=\frac14\times200=$ Rs 50 per m $^2$ .

Increase in total cost $=54\times50=$ Rs 2,700.

(c) Floor area $=25$ m $^2$ . Length of carpet $=\dfrac{25}{2}=$ 12.5 m.

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2075 Chaitra · Set AP · 2 marks

In the given figure, $AB = 10$ cm, $AA' = 20$ cm, $AC = 8$ cm and the area of the rectangular surfaces of the prism is $480$ square cm. Find the length of $BC$ .
Figure: A triangular prism $ABC\text{-}A'B'C'$ whose triangular base is $ABC$ and whose length is $AA'$ .

Similar questions: Triangular prism with rectangular surfaces 480 (2075 Chaitra · Set DP)

Answer

The area of the rectangular surfaces of a prism $=$ (perimeter of the base) $\times$ (length).

Let $BC=x$ cm. The perimeter of $\triangle ABC$ is $10+8+x=18+x$ , and the length is $AA'=20$ cm.

$(18+x)\times20=480\Rightarrow 18+x=24\Rightarrow x=6$

Hence $BC=$ 6 cm.

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2080 Model · Set 2 · 1+2+2 marks

The ratio of the slant height and a side of the base of a square based pyramid is $5:6$ and its total surface area is $1536$ sq. cm.
(a) Write the relation among the base area $(A)$ , height $(h)$ and volume $(V)$ of the pyramid. (1)
(b) Compare the base area and the area of the triangular surfaces. (2)
(c) Find the volume of the pyramid. (2)

Similar questions: Square pyramid with triangular surfaces 320 sq cm and base side 16 cm (2082 Pre-board · PABSON Set B)

Answer

(a) $V=\dfrac13\times A\times h$ (one third of the base area times the height).

(b) Let the slant height be $l=5k$ and the base side be $b=6k$ .

Base area $=b^2=36k^2$ . Area of the four triangular faces $=4\times\frac12\times b\times l=2bl=2\times6k\times5k=60k^2$ .

Total surface area: $36k^2+60k^2=96k^2=1536\Rightarrow k^2=16\Rightarrow k=4$ .

So $b=24$ cm, $l=20$ cm, base area $=576$ cm $^2$ and triangular surfaces $=60\times16=960$ cm $^2$ .

$\text{Base area}:\text{triangular surfaces}=576:960=3:5.$

The triangular surfaces have the greater area.

(c) Half the side is $12$ cm, so the height is

$h=\sqrt{l^2-12^2}=\sqrt{400-144}=\sqrt{256}=16\text{ cm}.$

$V=\frac13\times576\times16=3072\text{ cm}^3.$

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2080 Model · Set 3 · 3+2 marks

The volume and height of a square based room are 87.5 cu. m and 3.5 m respectively. The area occupied by a door and two windows in the room is 6 sq. m.
(a) What is the total cost of plastering the four walls at the rate of Rs. 150 per square meter? Find it. (3)
(b) If the rate of plastering per square meter is increased by one-third, then what will be the increment in the total cost of plastering the walls? Find it. (2)

Similar questions: Square room plastering cost increase (2082 Chaitra · Karnali)

Answer

(a) Base area $=\dfrac{\text{volume}}{\text{height}}=\dfrac{87.5}{3.5}=25$ m $^2$ , so the side of the square base is $5$ m.

Area of the four walls $=4\times5\times3.5=70$ m $^2$ .

Area to be plastered $=70-6=64$ m $^2$ .

Cost $=64\times150=$ Rs 9,600.

(b) Increase in rate $=\dfrac13\times150=$ Rs 50, so the new rate is Rs 200 per m $^2$ .

New cost $=64\times200=$ Rs 12,800.

Increment in the cost $=12{,}800-9{,}600=$ Rs 3,200.

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2075 Chaitra · Set BP · 2 marks

In the figure, the area of the rectangular surfaces of the given prism is 1188 sq. cm, $\angle ABC = 90^\circ$ , $AC = 15$ cm and $B'C' = 9$ cm. Find the length of $AA'$ .
Figure: A right triangular prism $ABC\text{-}A'B'C'$ with the right angle at $B$ and length $AA'$ .

Similar questions: Triangular prism with rectangular surfaces 180 (2075 Chaitra · Set CP)

Answer

In right-angled $\triangle ABC$ (right angle at $B$ ), $AC = 15$ cm is the hypotenuse and $BC = B'C' = 9$ cm, so

$AB = \sqrt{15^2 - 9^2} = \sqrt{144} = 12 \text{ cm}$

Area of rectangular surfaces = perimeter of base $\times$ length:

$(12 + 9 + 15)\times AA' = 1188 \Rightarrow 36\times AA' = 1188 \Rightarrow AA' = 33 \text{ cm}$

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2075 Chaitra · Set CP · 2 marks

In the given figure, the area of the rectangular surfaces of the prism is 180 sq. cm, $\angle PQR = 90^\circ$ , $PR = 5$ cm and $P'Q' = 3$ cm. Find the length of $PP'$ .
Figure: A right triangular prism $PQR\text{-}P'Q'R'$ with the right angle at $Q$ and length $PP'$ .

Similar questions: Triangular prism with rectangular surfaces 1188 (2075 Chaitra · Set BP)

Answer

In the right-angled $\triangle PQR$ (right angle at $Q$ ), $PQ=P'Q'=3$ cm and $PR=5$ cm, so

$QR=\sqrt{PR^2-PQ^2}=\sqrt{25-9}=4\text{ cm}.$

Perimeter of the base $=3+4+5=12$ cm.

Area of the rectangular surfaces $=$ perimeter of base $\times$ length of prism:

$12\times PP'=180 \Rightarrow PP'=\frac{180}{12}=15.$

So $PP'=$ 15 cm.

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2075 Chaitra · Set FP · 2 marks

In the adjoining figure, the area of the rectangular surfaces of the prism is 720 sq. cm, $PP' = 20$ cm and $PQ:PR:RQ = 5:3:4$ . Find the length of $PR$ .
Figure: A triangular prism $PQR\text{-}P'Q'R'$ with length $PP'$ .

Similar questions: Triangular prism with rectangular surfaces 720, find XY (2075 Chaitra · Set GP)

Answer

Let $PQ = 5k$ , $PR = 3k$ , $RQ = 4k$ .

Perimeter of the base $= 12k$ .

Area of rectangular surfaces $=$ perimeter $\times$ length:

$12k \times 20 = 720 \Rightarrow k = 3$

$PR = 3k = 3\times 3 = \textbf{9 cm}$

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2075 Chaitra · Set GP · 2 marks

In the adjoining figure, the area of the rectangular surfaces of the prism is 720 sq. cm, $XX' = 20$ cm and $XY:XZ:YZ = 5:3:4$ . Find the length of $XY$ .
Figure: A triangular prism $XYZ\text{-}X'Y'Z'$ with length $XX'$ .

Similar questions: Triangular prism with rectangular surfaces 720 and side ratio 5:3:4 (2075 Chaitra · Set FP)

Answer

Let $XY = 5k$ , $XZ = 3k$ , $YZ = 4k$ (cm).

The three rectangular faces have a combined area equal to the perimeter of the triangular base times the length of the prism:

(5k + 3k + 4k) \times 20 = 720 \Rightarrow 240k = 720 \Rightarrow k = 3

Therefore $XY = 5k = 5 \times 3 = 15$ cm.

(Check: sides 15, 9, 12 give $15 + 9 + 12 = 36$ and $36 \times 20 = 720$ .)

Answer: $XY = 15$ cm

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2075 Chaitra · Set DP · 2 marks

In the given figure, $AW = 20$ cm, $WW' = 10$ cm, $WX = 16$ cm and the area of the rectangular surfaces of the prism is 480 square cm. Find the length of $AX$ .
Figure: A triangular prism whose base is the triangle $AWX$ and whose length is $WW'$ .

Similar questions: Triangular prism with rectangular surfaces 480 (2075 Chaitra · Set AP)

Answer

The three rectangular faces together have area (perimeter of the base triangle) $\times$ (length of prism):

$(AW + WX + AX) \times WW' = 480$ $(20 + 16 + AX) \times 10 = 480 \Rightarrow 36 + AX = 48$ $AX = 12 \text{ cm}$

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2082 Pre-qualifying · PABSON Kathmandu · 1+2+1 marks

The length of a wall is 15 m, the width is 40 cm and the height is 4 m. Bricks of size 25 cm $\times$ 12.5 cm $\times$ 7.5 cm are used to construct the wall.
(a) Find the volume of the wall. (1)
(b) How many bricks are used to construct the wall? (2)
(c) Estimate the cost of bricks used in the wall at the rate of Rs. 14,000 per 1000 bricks. (1)

Similar questions: Wall bricks and cost (2081 Shrawan · Grade increment)

Answer

(a) Convert to cm: length $=1500$ cm, width $=40$ cm, height $=400$ cm. $V=1500\times40\times400=24{,}000{,}000\text{ cm}^3=24\text{ m}^3$

(b) Volume of one brick $=25\times12.5\times7.5=2343.75\text{ cm}^3$ . $\text{Number of bricks}=\frac{24000000}{2343.75}=10240$

(c) Cost $=\dfrac{10240}{1000}\times14000=\text{Rs }1{,}43{,}360$ .

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2082 Chaitra · Koshi · 1+1+2 marks

In the figure, a solid object is formed with the combination of a cone and a hemi-sphere having the same base. Where, the total height of the solid object is 20 cm. The diameter of the hemi sphere is 14 cm.
(a) Write the formula to find the volume of the cone. (1)
(b) Find the height of the conical part in the given solid object. (1)
(c) Find the volume of the given solid object. (2)
Figure: A solid made of a cone with a hemisphere on top sharing the same circular base (the hemisphere's flat face is the cone's base); the diameter of the common base is marked 14 cm and the total height of the whole solid, from the cone's apex to the top of the hemisphere, is marked 20 cm.

Answer

(a) Volume of cone $=\frac{1}{3}\pi r^2h$ .

(b) Diameter $=14$ cm, so $r=7$ cm. Height of hemisphere $=r=7$ cm.

Height of conical part $=20-7=13$ cm.

(c) Volume of solid = volume of cone + volume of hemisphere

$V=\frac13\pi r^2h+\frac23\pi r^3=\frac13\times\frac{22}{7}\times49\times13+\frac23\times\frac{22}{7}\times343$

$=\frac{2002}{3}+\frac{2156}{3}=\frac{4158}{3}=1386\text{ cm}^3$

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2082 Chaitra · Koshi · 2+2 marks

The length, breadth and height of a wall are 20 m, 30 cm and 5 m respectively. The bricks of size 25 cm × 12 cm × 5 cm are used to construct the wall. Also 10% part of the wall is occupied by the joints of cement and sand.
(a) How many bricks are used to construct the wall? Find it. (2)
(b) 2,000 bricks are loaded in a trip and the rent of it is Rs.600. Estimate the cost of bricks used to construct the wall at the rate of Rs.13,500 per 1,000 bricks. (2)

Answer

(a) Convert all dimensions to cm: $l=20$ m $=2000$ cm, $b=30$ cm, $h=5$ m $=500$ cm.

Volume of wall $=2000\times30\times500=30{,}000{,}000$ cm $^3$ .

Joints of cement and sand occupy 10%, so the bricks occupy $90\%$ of the wall:

$\text{Volume of bricks}=\frac{90}{100}\times30{,}000{,}000=27{,}000{,}000\text{ cm}^3$

Volume of one brick $=25\times12\times5=1500$ cm $^3$ .

$\text{Number of bricks}=\frac{27{,}000{,}000}{1500}=\mathbf{18{,}000}$

(b) Cost of bricks $=\dfrac{18{,}000}{1000}\times13{,}500=18\times13{,}500=\text{Rs }2{,}43{,}000$

Number of trips needed $=\dfrac{18{,}000}{2{,}000}=9$

Rent of trips $=9\times600=\text{Rs }5{,}400$

Total cost of bricks (including transport) $=2{,}43{,}000+5{,}400=\mathbf{Rs\ 2{,}48{,}400}$

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2082 Chaitra · Madhesh · 1+2+2 marks

A water tank is made of cylinder and cone having same radii, in which the base area of cylinder is 1.54 square meter. The heights of cone and cylinder are 2.4 meter and 3.2 meter respectively.
(a) Write the relation among $l$ , $h$ and $r$ of a cone. (1)
(b) How many maximum liters of water can be stored in the tank? Find it. (2)
(c) Compare between the curved surface area of cone and base area. (2)

Answer

(a) For a cone, $l^2=h^2+r^2$ .

(b) Base area $\pi r^2=1.54$ m $^2$ , cylinder height $=3.2$ m, cone height $=2.4$ m.

Volume of tank $=\pi r^2H+\frac13\pi r^2h=1.54\times3.2+\frac13\times1.54\times2.4$

$=4.928+1.232=6.16$ m $^3$ $=6.16\times1000=$ 6160 litres.

(c) $\pi r^2=1.54\Rightarrow r^2=\dfrac{1.54\times7}{22}=0.49\Rightarrow r=0.7$ m.

$l=\sqrt{2.4^2+0.7^2}=\sqrt{5.76+0.49}=\sqrt{6.25}=2.5$ m.

Curved surface area of cone $=\pi rl=\frac{22}{7}\times0.7\times2.5=5.5$ m $^2$ .

Base area $=1.54$ m $^2$ . So the curved surface area (5.5 m $^2$ ) is greater than the base area by $3.96$ m $^2$ ; the ratio is $5.5:1.54=25:7$ .

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2082 Chaitra · Madhesh · 2+1 marks

In the adjoining figure ABCD is a land in trapezium form in which BC = 70 feet, AD = 40 feet and AF = 20 feet.
(a) How many turfs of 4 square feet are required to lay down in the land? Find it. (2)
(b) Is Rs.40,000 sufficient to lay the turf in the land at the rate of Rs.150 per turf? Calculate it. (1)
Figure: Trapezium ABCD with AD parallel to BC (marked by arrows). AD is the top side, 40 ft, with A at the left and D at the right. BC is the bottom base, 70 ft, with B at the bottom left and C at the bottom right. AF is drawn from A perpendicular to BC, meeting it at F, with a right-angle mark at F; AF = 20 ft.

Answer

(a) $AD=40$ ft, $BC=70$ ft, height $AF=20$ ft.

Area of trapezium $=\frac12(AD+BC)\times AF=\frac12(40+70)\times20=1100$ sq. ft.

Number of turfs $=\dfrac{1100}{4}=$ 275.

(b) Cost $=275\times150=$ Rs 41,250.

Since Rs 41,250 > Rs 40,000, the amount Rs 40,000 is not sufficient (short by Rs 1,250).

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2082 Chaitra · Bagmati · 1+3+1 marks

The volume of a square based pyramid is 960 cubic cm. and its vertical height is 5 cm.
(a) What does $a$ denote in the formula $V = \frac{1}{3}a^2h$ , to find the volume of the square based pyramid? Write it. (1)
(b) What is the total surface area of the pyramid? Find it. (3)
(c) Compare the area of base and the area of a triangular surface of the pyramid. (1)

Answer

(a) In $V=\frac13a^2h$ , $a$ denotes the length of a side of the square base.

(b) $960=\frac13a^2\times5\Rightarrow a^2=\dfrac{960\times3}{5}=576\Rightarrow a=24$ cm.

Slant height $l=\sqrt{h^2+(a/2)^2}=\sqrt{5^2+12^2}=13$ cm.

Total surface area $=a^2+2al=576+2\times24\times13=576+624=1200$ cm $^2$ .

(c) Area of base $=576$ cm $^2$ . Area of one triangular surface $=\frac12al=\frac12\times24\times13=156$ cm $^2$ .

The base area is greater than the area of a triangular surface by $576-156=420$ cm $^2$ (about 3.7 times).

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2082 Chaitra · Bagmati · 1+1+2 marks

In the figure, a solid object is formed with the combination of a cylinder and a hemi-sphere having the same base. The diameter of the base of the solid object is 14 cm and its total height is 27 cm.
(a) How many total external surfaces are there in the given solid object? Write it. (1)
(b) Find the height of the cylindrical part of the solid object. (1)
(c) Find the volume of the solid object. (2)
Figure: A solid made of a cylinder with a hemisphere on top (same base). The base diameter is marked 14 cm and the total height (vertical arrow at the left) is marked 27 cm.

Answer

(a) There are 3 external surfaces: the circular base, the curved surface of the cylinder and the curved surface of the hemisphere.

(b) $r=\frac{14}{2}=7$ cm. Height of hemisphere $=7$ cm.

Height of cylindrical part $=27-7=20$ cm.

(c) $V=\pi r^2h+\frac23\pi r^3=\frac{22}{7}\times49\times20+\frac23\times\frac{22}{7}\times343$

=3080+\frac{2156}{3}=3080+718.67=3798.67\text{ cm}^3

(that is $\frac{11396}{3}$ cm $^3$ ).

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2082 Chaitra · Bagmati · 2+2 marks

The dimension of motor parking place of a house owner is 16 m long and 10 m broad. Square shaped stones having 25 cm long are laid on the parking. The price of each stone is Rs.20.
(a) How many total stones are laid on the parking? Find it. (2)
(b) To lay down stones on the parking, 3 workers worked for 2 days. Every worker received Rs.1,800 as daily wages. Estimate the total cost of laying stones on the parking. (2)

Answer

(a) Area of parking $=16\times10=160$ m $^2=160\times10000=1{,}600{,}000$ cm $^2$ .

Area of one stone $=25\times25=625$ cm $^2$ .

Number of stones $=\dfrac{1600000}{625}=$ 2,560.

(b) Cost of stones $=2560\times20=$ Rs 51,200.

Wages $=3\times2\times1800=$ Rs 10,800.

Total cost $=51200+10800=$ Rs 62,000.

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2082 Chaitra · Gandaki · 6 marks

The given figure is of a toy made with the combination of a cone and hemi-sphere. The common radius of the toy is 7 cm and total height is 31 cm.
(a) Write the relation among the radius of the base of a cone ( $r$ ), vertical height ( $h$ ) and slant height ( $l$ ). (1)
(b) Find the curved surface area of the conical part of the toy. (3)
(c) By how much the volume of conical part is less or more than the volume of hemi-spherical part? Compare with calculation. (2)
Figure: A solid toy made of a cone sitting on top of a hemisphere (cone apex up, hemisphere below, sharing the same circular base). Centre O of the common base is marked with a right-angle mark, radius 7 cm is marked on the base, and the total height from the bottom of the hemisphere to the apex of the cone is marked 31 cm.

Answer

(a) $l^2=h^2+r^2$ .

(b) $r=7$ cm, total height $=31$ cm. Hemisphere height $=7$ cm, so cone height $h=31-7=24$ cm.

$l=\sqrt{24^2+7^2}=\sqrt{625}=25$ cm.

Curved surface area of cone $=\pi rl=\frac{22}{7}\times7\times25=550$ cm $^2$ .

(c) Volume of cone $=\frac13\pi r^2h=\frac13\times\frac{22}{7}\times49\times24=1232$ cm $^3$ .

Volume of hemisphere $=\frac23\pi r^3=\frac23\times\frac{22}{7}\times343=\frac{2156}{3}\approx718.67$ cm $^3$ .

Difference $=1232-718.67=513.33$ cm $^3$ .

So the volume of the conical part is more than that of the hemispherical part by about $513.33$ cm $^3$ .

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2082 Chaitra · Gandaki · 3 marks

The parking area outside of a building is in geometric shape as shown in the given figure.
(a) Find the area of the parking area. (2)
(b) Find the total cost of leveling the parking area at the rate of Rs.2500 per 100 sq. ft. (1)
Figure: Right trapezium ABCD: AD parallel to BC (arrows marked), angle at B is a right angle. AD = 20 ft (top), AB = 24 ft (left side, perpendicular height), BC = 38 ft (bottom). A is top-left, D top-right, B bottom-left, C bottom-right.

Answer

(a) The parking area is a trapezium with parallel sides $AD=20$ ft and $BC=38$ ft; the perpendicular height is $AB=24$ ft.

Area $=\frac12(20+38)\times24=\frac12\times58\times24=696$ sq. ft.

(b) Rate $=$ Rs 2500 per 100 sq. ft $=$ Rs 25 per sq. ft.

Total cost $=696\times25=$ Rs 17,400.

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2082 Chaitra · Lumbini · 1+2+1 marks

The total height of the given combined solid is 54 cm and height of the cylindrical part is 30 cm. The slant height of the conical part is 25 cm.
(a) Write the formula to find the radius of base of cone when slant height ( $l$ ) and vertical height ( $h$ ) of cone are given. (1)
(b) Find the volume of that solid object. (2)
(c) If it is melted to form a cylindrical wire of radius 3.5 mm, what is the length of the wire? Find it. (1)
Figure: A combined solid: a cone placed on top of a cylinder of the same base radius. Total height is marked 54 cm, the height of the cylindrical part is marked 30 cm, and the slant height of the conical part is marked 25 cm.

Answer

(a) $r=\sqrt{l^2-h^2}$ .

(b) Height of cone $=54-30=24$ cm, $l=25$ cm.

$r=\sqrt{25^2-24^2}=\sqrt{49}=7$ cm.

$V=\pi r^2H+\frac13\pi r^2h=\frac{22}{7}\times49\times30+\frac13\times\frac{22}{7}\times49\times24$

$=4620+1232=5852$ cm $^3$ .

(c) Wire radius $=3.5$ mm $=0.35$ cm. Let length be $L$ .

$\pi(0.35)^2L=5852\Rightarrow\frac{22}{7}\times0.1225\times L=5852\Rightarrow0.385L=5852$

$L=\dfrac{5852}{0.385}=15200$ cm $=$ 152 m.

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Chaitra · Lumbini · 3+2 marks

A rectangular water tank of length 3 meter, breath 2 meter and height 1.5 meter is built.
(a) What is the cost of plastering the floor and 4 walls of the tank at the rate of Rs.1,500 per square meter? Find it. (3)
(b) What will be the cost to fill water in the tank at the rate of 25 paisa per litre? Find it. (2)

Answer

(a) $l=3$ m, $b=2$ m, $h=1.5$ m.

Area to plaster = floor + 4 walls $=lb+2(l+b)h=3\times2+2(3+2)\times1.5=6+15=21$ m $^2$ .

Cost $=21\times1500=$ Rs 31,500.

(b) Volume $=3\times2\times1.5=9$ m $^3=9000$ litres.

Rate $=25$ paisa $=$ Rs 0.25 per litre.

Cost $=9000\times0.25=$ Rs 2,250.

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Chaitra · Karnali · 1+1+2 marks

The perimeter of a square base pyramid is 24 meter and its height is 4 meter.
(a) Write the formula to find the volume of square based pyramid. (1)
(b) What is the volume of pyramid? Find it. (1)
(c) Find the total surface area of the pyramid. (2)

Answer

(a) $V=\dfrac13a^2h$ .

(b) Perimeter $=4a=24\Rightarrow a=6$ m, $h=4$ m.

$V=\frac13\times6^2\times4=48$ m $^3$ .

(c) $l=\sqrt{h^2+(a/2)^2}=\sqrt{16+9}=5$ m.

Total surface area $=a^2+2al=36+2\times6\times5=36+60=96$ m $^2$ .

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Chaitra · Karnali · 1+2+1 marks

A water tank with circular base built to distribute drinking water for 40 household in a rural municipality is given in the figure. Which is made with the combination of cylinder and hemi-sphere. The height of the cylinder is 4.2 meter and diameter of the base is 7 meter.
(a) Write the formula to find the base area of that tank. (1)
(b) How much liter of water is required to fill the whole tank? Find it. (2)
(c) If equal amount of water is being distributed to all the household everyday then one fourth water is consumed of full tank. How much amount should a family pay for consuming water in a month at the rate of 4 paisa per liter? Find it. (1)
Figure: A water tank made of a vertical cylinder topped by a hemisphere (dome) of the same diameter. The cylinder's height is marked 4.2 m and the base diameter is marked 7 m.

Answer

(a) Base area $=\pi r^2$ (a circle of radius $r$ ).

(b) $r=\frac72=3.5$ m, cylinder height $h=4.2$ m.

$V=\pi r^2h+\frac23\pi r^3=\frac{22}{7}\times3.5^2\times4.2+\frac23\times\frac{22}{7}\times3.5^3$

$=161.7+89.83=251.53$ m $^3$ $\approx251{,}533$ litres.

(c) Water used per day by all $=\frac14\times251533.3\approx62{,}883.3$ L.

Per household per day $=\dfrac{62883.3}{40}\approx1572.08$ L.

For a month (30 days) $=1572.08\times30\approx47{,}162.5$ L.

Payment $=47162.5\times0.04\approx$ Rs 1,886.50.

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Chaitra · Sudurpashchim · 1+2+1 marks

The diagram along side is a square based pyramid. Its slant height is 20 cm. The area of the triangular surfaces of the pyramid is 1280 square cm.
(a) Write the relation among triangular surfaces of square based pyramid. (1)
(b) Find the length of base side of the pyramid. (2)
(c) Find the ratio of total surface area and area of triangular surfaces of that pyramid. (1)
Figure: Square based pyramid with the slant height marked 20 cm (an arrow from the apex to the midpoint of a base edge).

Answer

(a) All four triangular surfaces of a square based pyramid are congruent isosceles triangles, so they have equal area; each is $\frac12al$ and the total is $4\times\frac12al=2al$ .

(b) Area of triangular surfaces $=2al=1280$ , $l=20$ :

$2\times a\times20=1280\Rightarrow a=32$ cm.

(c) Area of base $=32^2=1024$ cm $^2$ .

Total surface area $=1024+1280=2304$ cm $^2$ .

Ratio $=2304:1280=9:5$ .

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2082 Chaitra · Sudurpashchim · 1+2+2 marks

The given figure is a combined solid made up of a cylinder and cone. Diameter of base is 14 cm, height of cone is 24 cm and total height of solid object is 54 cm.
(a) Write down the formula to find the volume of the cone. (1)
(b) Find the volume of that solid object. (2)
(c) Find the total surface area of the solid object. (2)
Figure: A solid made of a cone joined to a cylinder on its circular base; the cone's height is marked 24 cm, the total length of the solid is 54 cm, and the cylinder's base diameter is 14 cm.

Answer

(a) Volume of cone $=\frac13\pi r^2h$ .

(b) $r=7$ cm, cone height $=24$ cm, cylinder height $=54-24=30$ cm.

$V=\pi r^2H+\frac13\pi r^2h=\frac{22}{7}\times49\times30+\frac13\times\frac{22}{7}\times49\times24=4620+1232=5852$ cm $^3$ .

(c) $l=\sqrt{24^2+7^2}=25$ cm.

Total surface area = base + curved surface of cylinder + curved surface of cone

$=\pi r^2+2\pi rH+\pi rl=154+2\times\frac{22}{7}\times7\times30+\frac{22}{7}\times7\times25$

$=154+1320+550=2024$ cm $^2$ .

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Koshi · 1+1+2 marks

A group of students constructed a square based pyramid shaped tent having length of base side 24 meter and vertical height 5 meter.
(a) How many triangular surfaces are there in the square based pyramid? Write it. (1)
(b) Find the slant height of the above square based tent. (1)
(c) What is the total cost of cloths required to make triangular surfaces at the rate of Rs.125 per square metre? Find it. (2)

Answer

(a) There are 4 triangular surfaces.

(b) $a=24$ m, $h=5$ m. $l=\sqrt{h^2+(a/2)^2}=\sqrt{25+144}=13$ m.

(c) Area of triangular surfaces $=2al=2\times24\times13=624$ m $^2$ .

Cost $=624\times125=$ Rs 78,000.

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Koshi · 1+2+2 marks

In the given figure, wooden cylinder and cone having equal base are shown.
(a) Write the formula to find the volume of a cone. (1)
(b) Find the volume of the cone in the given objects. (2)
(c) If given wooden cylinder is drilled out in the given conical shape, what will be the volume of remaining wood in cylinder? Find it. (2)
Figure: A cylinder with length (height) 24 m and base diameter 24 m, shown side by side with a cone whose slant side/radius is marked 16 m (the cone has the same base as the cylinder).

Answer

The cylinder and the cone share the same base, so $r=\dfrac{24}{2}=12$ m. The cylinder's height is $H=24$ m and the cone's height is $h=16$ m.

(a) Volume of a cone $=\dfrac13\pi r^2h$ .

(b) $V_{\text{cone}}=\frac13\times\frac{22}{7}\times12^2\times16=\frac{16896}{7}\approx2413.71\text{ m}^3$

(c) $V_{\text{cylinder}}=\pi r^2H=\frac{22}{7}\times12^2\times24=\frac{76032}{7}\approx10861.71\text{ m}^3$

Remaining wood $=V_{\text{cylinder}}-V_{\text{cone}}=\dfrac{76032-16896}{7}=\dfrac{59136}{7}=$ $8448$ m $^3$ .

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Madhesh · 1+1+3 marks

The volume of square based pyramid given in the figure is 512 cubic cm and length of side of base is 16 cm.
(a) How many plane surfaces area are counted to find the total surface area of a square based pyramid? Write it. (1)
(b) Find the vertical height of the pyramid. (1)
(c) Find the total surface area of the pyramid. (3)
Figure: A square-based pyramid drawn in perspective with apex above the square base; base side labelled 16 cm, vertical height labelled h, slant height marked on a lateral face.

Answer

(a) 5 plane surfaces (1 square base and 4 triangular faces).

(b) $V=\frac13a^2h\Rightarrow512=\frac13\times256\times h\Rightarrow h=\dfrac{1536}{256}=6$ cm.

(c) $l=\sqrt{h^2+(a/2)^2}=\sqrt{36+64}=10$ cm.

Total surface area $=a^2+2al=256+2\times16\times10=256+320=576$ cm $^2$ .

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Madhesh · 1+2+1 marks

Jyoti bought a tank made up of a cylinder and a hemisphere from the local market. The total height of the tank is 3.5 meter and radius of base is 1.05 meter.
(a) How many curved surfaces are there in a combined solid made of a cylinder and a hemisphere? Write it (1)
(b) Find the volume of the tank. (2)
(c) How much maximum liters of water is contained in the tank? Find it. (1)

Answer

(a) 2 curved surfaces (one of the cylinder and one of the hemisphere).

(b) $r=1.05$ m. Cylinder height $=3.5-1.05=2.45$ m.

$V=\pi r^2h+\frac23\pi r^3=\frac{22}{7}\times1.05^2\times2.45+\frac23\times\frac{22}{7}\times1.05^3$

$=8.48925+2.4255=10.91475$ m $^3$ .

(c) Maximum water $=10.91475\times1000=$ 10,914.75 litres.

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Bagmati · 1+1+2 marks

The diagram alongside is of a square based pyramid. Its each side of base is 6 meter and vertical height is 4 meter.
(a) Write the formula for finding the area of all triangular surfaces of pyramid. (1)
(b) Find the volume of the pyramid. (1)
(c) Find the total surface area of the pyramid. (2)
Figure: A square-based pyramid with base side labelled 6 m and vertical height (dotted line from apex to centre of base) labelled 4 m.

Answer

(a) Area of all triangular surfaces $=2al$ (that is $4\times\frac12al$ ).

(b) $V=\frac13a^2h=\frac13\times36\times4=48$ m $^3$ .

(c) $l=\sqrt{4^2+3^2}=5$ m.

Total surface area $=a^2+2al=36+2\times6\times5=36+60=96$ m $^2$ .

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Bagmati · 1+3+1 marks

Given solid object is made up of a cone and hemisphere. The diameters of the base of a cone and hemisphere are equal and the total height of object is 17 cm and diameter of base is 10 cm.
(a) Write the relation between height and radius of hemisphere. (1)
(b) Find the total surface area of solid object. (3)
(c) Will the amount of Rs. 150 be sufficient to color the surface of the solid object at the rate of 40 paisa per square cm? Give reason with calculation. (1)
Figure: A solid made of a cone placed on top of a hemisphere with a common base; diameter of the common base is 10 cm and total height of the solid (cone apex to bottom of hemisphere) is 17 cm.

Answer

(a) The height of a hemisphere equals its radius ( $h=r$ ).

(b) $r=5$ cm; hemisphere height $=5$ cm; cone height $=17-5=12$ cm.

$l=\sqrt{12^2+5^2}=13$ cm.

Total surface area = curved surface of cone + curved surface of hemisphere

$=\pi rl+2\pi r^2=\frac{22}{7}\times5\times13+2\times\frac{22}{7}\times25=\frac{1430+1100}{7}=\frac{2530}{7}\approx361.43$ cm $^2$ .

(c) Cost $=361.43\times0.40\approx$ Rs 144.57.

Since Rs 144.57 < Rs 150, yes, Rs 150 is sufficient (about Rs 5.43 is left).

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Bagmati · 2+2 marks

A rectangular room has length 12 ft, breadth 8 ft and height 9 ft. There are two windows of size 2.5 ft $\times$ 3 ft and two doors of size 6 ft $\times$ 2 ft.
(a) Find the total area of four walls, floor and ceiling. (2)
(b) How much less or more is the cost of coloring the four walls excluding the doors and windows at the rate of Rs. 175 per square feet than Rs. 50,000? Find it. (2)

Answer

(a) Four walls $=2(12+8)\times9=360$ sq. ft; floor $=12\times8=96$ sq. ft; ceiling $=96$ sq. ft.

Total $=360+96+96=$ 552 sq. ft.

(b) Windows $=2\times(2.5\times3)=15$ sq. ft; doors $=2\times(6\times2)=24$ sq. ft; total $=39$ sq. ft.

Area to colour $=360-39=321$ sq. ft.

Cost $=321\times175=$ Rs 56,175.

This is Rs $56175-50000=$ Rs 6,175 more than Rs 50,000.

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Gandaki · 1+2+2 marks

The vertical height of a square based pyramid is 12 cm and its base side is 10 cm.
(a) How many triangular surfaces are there in a square based pyramid? Write it.
(b) Find the volume of the pyramid.
(c) Find the total surface area of the pyramid.
Figure: A square-based pyramid with vertical height marked 12 cm (from apex to centre of the base) and base side marked 10 cm.

Answer

(a) There are 4 triangular surfaces.

(b) $V=\frac13a^2h=\frac13\times10^2\times12=400$ cm $^3$ .

(c) $l=\sqrt{12^2+5^2}=13$ cm.

Total surface area $=a^2+2al=100+2\times10\times13=100+260=360$ cm $^2$ .

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Gandaki · 1+2+1 marks

In the given figure, a combined solid object is formed with the combination of cylinder and cone having same radius. In the solid object, the length of cylindrical part is 28 cm and the slant height of conical part is 17 cm. The volume of the cylindrical part is 5632 cubic cm.
(a) What type is the shape of the base of solid object? Write it.
(b) Compare the height of conical part and the length of cylindrical part.
(c) Is the volume of cylindrical part five times the volume of conical part? Justify with calculation.
Figure: A solid made of a cylinder (length 28 cm) with a cone of the same radius attached at its right end; the slant height of the cone is marked 17 cm.

Answer

(a) The base of the solid is a circle.

(b) Let radius be $r$ . Cylinder: $\pi r^2\times28=5632\Rightarrow r^2=\dfrac{5632\times7}{22\times28}=64\Rightarrow r=8$ cm.

Cone height $h=\sqrt{17^2-8^2}=\sqrt{225}=15$ cm.

Height of cone $=15$ cm and length of cylinder $=28$ cm, so the conical height is less than the cylindrical length by 13 cm.

(c) Volume of cone $=\frac13\pi r^2h=\frac13\times\frac{22}{7}\times64\times15=\frac{7040}{7}\approx1005.71$ cm $^3$ .

Five times the cone $=5028.57$ cm $^3\ne5632$ cm $^3$ . In fact $5632\div1005.71=5.6$ .

So the cylinder volume is not five times the cone volume (it is 5.6 times).

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Gandaki · 1+3 marks

The length, breadth and height of a rectangular room are 12 m, 8 m and 3 m respectively. There are two square windows with edges 2 m and a door of size 1.5 m $\times$ 1 m in the room.
(a) Find the area of the floor of the room.
(b) How much does it cost to coloring the four walls and ceiling of the room excluding the area occupied by the windows and door at the rate of Rs. 15 per square meter? Calculate it.

Answer

(a) Floor area $=12\times8=96$ m $^2$ .

(b) Four walls $=2(12+8)\times3=120$ m $^2$ ; ceiling $=96$ m $^2$ ; total $=216$ m $^2$ .

Windows $=2\times(2\times2)=8$ m $^2$ ; door $=1.5\times1=1.5$ m $^2$ ; total $=9.5$ m $^2$ .

Area to colour $=216-9.5=206.5$ m $^2$ .

Cost $=206.5\times15=$ Rs 3,097.50.

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Lumbini · 1+1+2 marks

A solid object made up of a cone and a cylinder is given in the figure.
(a) How many curved surfaces are there in the given solid object? Write it. (1)
(b) Find the height of cone. (1)
(c) Compare the volume of cone and cylinder. (2)
Figure: A solid made of a cone placed on top of a cylinder (a dashed circle marks the joint). The slant height of the cone is labelled 25 cm, the height of the cylinder is 32 cm and the diameter of the base of the cylinder is 14 cm.

Answer

(a) 2 curved surfaces (curved surface of the cone and curved surface of the cylinder).

(b) $r=\frac{14}{2}=7$ cm, $l=25$ cm. Height of cone $h=\sqrt{25^2-7^2}=\sqrt{576}=24$ cm.

(c) Volume of cone $=\frac13\times\frac{22}{7}\times49\times24=1232$ cm $^3$ .

Volume of cylinder $=\frac{22}{7}\times49\times32=4928$ cm $^3$ .

Ratio cone : cylinder $=1232:4928=1:4$ . The cylinder's volume is 4 times the volume of the cone (greater by 3696 cm $^3$ ).

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Lumbini · 3+1 marks

The length, breadth and height of a rectangular classroom are 18 ft, 14 ft and 10 ft respectively. In the classroom, there are two windows with size 6 ft $\times$ 4 ft and two doors with size 6 ft $\times$ 3 ft.
(a) How much does it cost to paint four walls and ceiling of the classroom excluding doors and windows at the rate of Rs. 40 per square feet? Find it. (3)
(b) If a painter paints 202 square feet in a day, how many days will two painters take to paint the classroom? Find it. (1)

Answer

(a) Four walls $=2(18+14)\times10=640$ sq. ft; ceiling $=18\times14=252$ sq. ft; total $=892$ sq. ft.

Windows $=2\times(6\times4)=48$ sq. ft; doors $=2\times(6\times3)=36$ sq. ft; total $=84$ sq. ft.

Area to paint $=892-84=808$ sq. ft.

Cost $=808\times40=$ Rs 32,320.

(b) One painter takes $\dfrac{808}{202}=4$ days. Two painters take $4\div2=$ 2 days.

Answer checked by Claude Sonnet · Oct 21, 2026

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2081 Chaitra · Karnali · 1+1+2 marks

In the square based pyramid given in the figure, AH = 26 cm and AD = 24 cm.
(a) Write the relation of HD and EF. (1)
(b) Find the value of EF. (1)
(c) Find the total surface area of the pyramid. (2)
Figure: A square-based pyramid with apex A and square base BCFE (B, C at the back, E, F at the front). D is the centre of the base, with AD drawn perpendicular to the base (right-angle mark at D). H is the midpoint of side CF, and AH is the slant height drawn from A to H. Dashed line from D to H is marked equal to half of EF (equal-tick marks on DH and the sides).

Answer

(a) $D$ is the centre of the base and $H$ the midpoint of side $CF$ , so $HD=\dfrac12EF$ .

(b) In right triangle $ADH$ : $HD=\sqrt{AH^2-AD^2}=\sqrt{26^2-24^2}=\sqrt{100}=10$ cm.

$EF=2\times HD=20$ cm.

(c) Total surface area $=a^2+2al=20^2+2\times20\times26=400+1040=1440$ cm $^2$ .

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2081 Chaitra · Karnali · 1+2+2 marks

The given figure is of a combined solid object made with the combination of cylinder and hemisphere. The total height of the solid is 17 cm and circumference of the base is 44 cm.
(a) Write the formula to calculate volume of solid object. (1)
(b) What is the volume of hemispherical part? Find it. (2)
(c) Compare between the volume of cylinder and hemisphere. (2)
Figure: A solid made of a cylinder with a hemisphere on top (dome shaped, dotted shading); an arrow along the side marks the total height 17 cm.

Answer

(a) $V=\pi r^2h+\dfrac23\pi r^3$ (cylinder + hemisphere).

(b) $2\pi r=44\Rightarrow r=\dfrac{44\times7}{2\times22}=7$ cm.

Volume of hemisphere $=\frac23\pi r^3=\frac23\times\frac{22}{7}\times343=\dfrac{2156}{3}\approx718.67$ cm $^3$ .

(c) Cylinder height $=17-7=10$ cm.

Volume of cylinder $=\frac{22}{7}\times49\times10=1540$ cm $^3$ .

Ratio cylinder : hemisphere $=1540:\frac{2156}{3}=4620:2156=15:7$ .

The cylinder's volume is greater by $1540-718.67=821.33$ cm $^3$ .

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2081 Chaitra · Karnali · 3+1 marks

A rectangular tank of 5 m $\times$ 1 m $\times$ 4 m is filled with water at the rate of 50 paisa per litre.
(a) The water containing in full tank is enough for 20 families distributed equally for one month. How much cost of water should one family have to pay in one year? Find it. (3)
(b) If length, breadth and height each of the tank is increased by 1 m, by how many times the capacity of tank is increased? Find it. (1)

Answer

(a) Volume $=5\times1\times4=20$ m $^3=20{,}000$ litres.

Cost of a full tank $=20000\times0.50=$ Rs 10,000 (enough for one month).

Cost per family per month $=\dfrac{10000}{20}=$ Rs 500.

Cost per family for one year $=500\times12=$ Rs 6,000.

(b) New dimensions: $6\times2\times5=60$ m $^3$ .

$\dfrac{60}{20}=$ 3 times the capacity.

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2081 Chaitra · Sudurpashchim · 1+2+2 marks

In the figure, a metallic solid made of hemisphere and cone is given, where the height of cone is 24 cm and diameter of base is 14 cm.
(a) Write the formula to find the slant height of cone when vertical height and radius of base are given. (1)
(b) Find the volume of the solid object. (2)
(c) If the solid object is melted and turned into a cylindrical object of radius 7 cm, what is the height of cylinder? Calculate it. (2)
Figure: A solid made of a cone below and a hemisphere on top, sharing the same circular base. The base diameter is marked 14 cm and the vertical height of the cone (apex to base) is marked 24 cm.

Similar questions: Cone and hemisphere melted to cylinder

Answer

(a) $l=\sqrt{h^2+r^2}$ .

(b) $r=7$ cm, $h=24$ cm.

$V=\frac13\pi r^2h+\frac23\pi r^3=\frac13\times\frac{22}{7}\times49\times24+\frac23\times\frac{22}{7}\times343$

$=1232+\frac{2156}{3}=\frac{3696+2156}{3}=\frac{5852}{3}\approx1950.67$ cm $^3$ .

(c) Volume of cylinder $=$ volume of solid:

$\frac{22}{7}\times49\times H=\frac{5852}{3}\Rightarrow154H=\frac{5852}{3}\Rightarrow H=\dfrac{5852}{462}=\dfrac{38}{3}\approx12.67$ cm.

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2080 Chaitra · Koshi · 1+2+1 marks

The volume of a square based pyramid is 1728 cu. cm and its vertical height is 9 cm.
(a) Write the formula to find the volume of the square based pyramid. (1)
(b) Find the area of triangular surfaces of the pyramid. (2)
(c) Compare the area of the base and the area of the triangular surfaces of the pyramid. (1)

Answer

(a) $V=\dfrac13a^2h$ .

(b) $1728=\frac13a^2\times9\Rightarrow a^2=576\Rightarrow a=24$ cm.

$l=\sqrt{9^2+12^2}=15$ cm.

Area of triangular surfaces $=2al=2\times24\times15=720$ cm $^2$ .

(c) Area of base $=576$ cm $^2$ ; area of triangular surfaces $=720$ cm $^2$ .

The triangular surfaces' area is greater by $144$ cm $^2$ ; ratio base : triangular surfaces $=576:720=4:5$ .

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2080 Chaitra · Koshi · 1+1+3 marks

In the figure, the metallic object made up of a cylinder and a cone is designed for decoration.
(a) How many external surfaces are there in the given combined solid? Write it. (1)
(b) Find the height of the conical part. (1)
(c) If the cost of 1 cubic cm metal is Rs. 10, find the total cost of the metallic solid. (3)
Figure: A cone sitting on top of a cylinder of the same base. Diameter of the base is 14 cm; the height of the cylinder is 20 cm; the total height of the combined solid (cylinder plus cone) is 26 cm.

Answer

(a) 3 external surfaces (circular base, curved surface of the cylinder, curved surface of the cone).

(b) Height of cone $=26-20=6$ cm.

(c) $r=7$ cm.

Volume of cylinder $=\frac{22}{7}\times49\times20=3080$ cm $^3$ .

Volume of cone $=\frac13\times\frac{22}{7}\times49\times6=308$ cm $^3$ .

Total volume $=3388$ cm $^3$ .

Total cost $=3388\times10=$ Rs 33,880.

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Koshi · 2+2 marks

The length, breadth and height of a room are 7 meter, 6 meter and 4 meter respectively. In that room there are two windows having area (1.5 × 1) square meter each and a door with area 1.5 square meter.
(a) How much will it cost to plaster on the four walls of the room excluding windows and door at the rate of Rs. 50 per square meter? Find it. (2)
(b) When carpet is laid on the floor of the room, Rs. 900 has been spent. Does the wage of Rs. 25 per square meter of the carpet is suitable? Give reason. (2)

Answer

(a) Four walls $=2(7+6)\times4=104$ m $^2$ .

Windows $=2\times(1.5\times1)=3$ m $^2$ ; door $=1.5$ m $^2$ ; total $=4.5$ m $^2$ .

Area to plaster $=104-4.5=99.5$ m $^2$ .

Cost $=99.5\times50=$ Rs 4,975.

(b) Floor area $=7\times6=42$ m $^2$ .

At Rs 25 per m $^2$ the cost would be $42\times25=$ Rs 1,050, but only Rs 900 was spent.

The actual rate is $\dfrac{900}{42}\approx$ Rs 21.43 per m $^2$ . So the wage of Rs 25 per m $^2$ is not suitable (it is higher than the rate actually paid).

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Madhesh · 6 marks

The area of triangular surfaces of a square based solid pyramid is 260 square cm. The length of a side of its base is 10 cm.
(a) Write the formula for finding the area of triangular surfaces of the pyramid. (1)
(b) Find the volume of the pyramid. (3)
(c) Does Rs. 200 sufficient to paint on the total surface of the pyramid at the rate of 50 paisa per square centimeter? Give reason. (2)

Answer

(a) Area of triangular surfaces $=2al$ (or $4\times\frac12al$ ).

(b) $260=2\times10\times l\Rightarrow l=13$ cm.

$h=\sqrt{l^2-(a/2)^2}=\sqrt{169-25}=12$ cm.

$V=\frac13a^2h=\frac13\times100\times12=400$ cm $^3$ .

(c) Total surface area $=a^2+2al=100+260=360$ cm $^2$ .

Cost $=360\times0.50=$ Rs 180.

Since Rs 180 < Rs 200, yes, Rs 200 is sufficient.

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Madhesh · 4 marks

An event manager has built a cylindrical tent whose radius of the base is 7 m. The open upper part of the tent is covered by hemispherical tent. The total height of the tent so formed is 7 meter.
(a) Write the formula for calculating the base area of the combined solid object made up of cylinder and hemisphere. (1)
(b) Find the total cost of the required canvas cloth for making the tent at the rate of Rs. 135 per square meter. (2)
(c) How many persons can be accommodated on the floor of the tent where 2 persons can accommodate in a square meter of space? Find it. (1)

Answer

(a) Base area $=\pi r^2$ (the floor is a circle of radius $r$ ).

(b) $r=7$ m and total height $=7$ m. Height of hemisphere $=r=7$ m, so the cylindrical wall height is $7-7=0$ m; with the data given, the canvas covers only the hemispherical roof (in general canvas $=2\pi rh+2\pi r^2$ ).

Canvas area $=2\pi r^2=2\times\frac{22}{7}\times49=308$ m $^2$ .

Cost $=308\times135=$ Rs 41,580.

(c) Floor area $=\pi r^2=\frac{22}{7}\times49=154$ m $^2$ .

Persons $=154\times2=$ 308 persons.

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Madhesh · 3 marks

A farmer has a triangular field with sides of 60 meters, 60 meters and 80 meter. He wants to grow maize in the field. One kg maize is sufficient for 20 sq. meter land.
(a) How many kg of maize is required to grow in the triangular field? Find it. (2)
(b) What amount of total cost should the farmer bear paying as Rs. 2,000 for labourer charge and Rs. 100 per kg maize? Find it. (1)

Answer

(a) Sides $a=60,\ b=60,\ c=80$ m. $s=\dfrac{60+60+80}{2}=100$ m.

Area $=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{100\times40\times40\times20}=\sqrt{3200000}\approx1788.85$ m $^2$ .

Maize required $=\dfrac{1788.85}{20}\approx$ 89.44 kg.

(b) Cost of maize $=89.44\times100=$ Rs 8,944.

Total cost $=8944+2000=$ Rs 10,944 (approximately).

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Bagmati · 1+2+2 marks

In the figure the slant height of a square based pyramid is 24 cm and its length of slant edge is 25 cm.
a) Write the formula for finding lateral surface area of a square based pyramid. (1)
b) Find the area of the base of the pyramid. (2)
c) What is the total cost of coloring the total surfaces of the pyramid at the rate of 50 paisa per square centimeter? Find it. (2)
Figure: A square based pyramid drawn in perspective; the slant height (from the apex to the midpoint of a base edge, with a right-angle mark at the base edge) is labelled 24 cm and the slant edge (from the apex to a base corner) is labelled 25 cm.

Answer

(a) Lateral surface area $=2al$ .

(b) Slant height $l=24$ cm, slant edge $=25$ cm. Half of base side $=\sqrt{25^2-24^2}=\sqrt{49}=7$ cm, so $a=14$ cm.

Area of base $=14^2=$ 196 cm $^2$ .

(c) Total surface area $=a^2+2al=196+2\times14\times24=196+672=868$ cm $^2$ .

Cost $=868\times0.50=$ Rs 434.

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Bagmati · 1+3+1 marks

A cylindrical wooden log is drilled out in its upper part in a conical shape with equal radius of the cylinder and then it is filled with wax. Measurements of different parts of the log are labeled as in the given figure.
a) Find the vertical height of the cone. (1)
b) What is the total cost of painting the wooden log except conical part at the rate of 50 paisa per sq. cm? Calculate it. (3)
c) Is the volume of the conical portion one third of the volume of the cylindrical object? Give reason. (1)
Figure: A solid cylinder with a cone-shaped hollow (filled with wax) in its top face; the cone's base is the top circle of the cylinder. Total height of the cylinder is 67 cm, the slant height of the cone is 50 cm, and the diameter of the base is 28 cm.

Answer

(a) $r=\frac{28}{2}=14$ cm, $l=50$ cm.

$h=\sqrt{l^2-r^2}=\sqrt{2500-196}=\sqrt{2304}=48$ cm.

(b) The conical part is drilled in the top, so the surfaces to paint are the circular base and the curved surface of the cylinder (height 67 cm).

Area $=\pi r^2+2\pi rH=\frac{22}{7}\times196+2\times\frac{22}{7}\times14\times67=616+5896$

$=6512$ cm $^2$ (using base $\pi r^2=616$ ).

Cost $=6512\times0.50=$ Rs 3,256.

(c) Volume of cone $=\frac13\pi r^2\times48=16\pi r^2$ ; volume of cylinder $=\pi r^2\times67=67\pi r^2$ .

One third of the cylinder would be $22.33\pi r^2\ne16\pi r^2$ . So no: a cone has one third the volume of a cylinder only if they have the same height, but here the cone height (48 cm) is less than the cylinder height (67 cm).

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Bagmati · 1+2 marks

The length and breadth of the floor of a rectangular room are 20 feet and 16 feet respectively.
a) Write the formula for finding the area of a rectangle by letting length as ( $l$ ) and breadth as ( $b$ ). (1)
b) How much cost is required to pave the tiles of size 4 ft × 2 ft on the floor of the room at the rate of Rs. 200 per tile? Find it. (2)

Answer

(a) Area of rectangle $=l\times b$ .

(b) Area of floor $=20\times16=320$ sq. ft. Area of one tile $=4\times2=8$ sq. ft.

Number of tiles $=\dfrac{320}{8}=40$ .

Cost $=40\times200=$ Rs 8,000.

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Gandaki · 1+1+2 marks

A group of tourists planned to fix a pyramid shaped tent at the Everest Base Camp as shown in the figure. Length of a side of its base is 12 ft and area of triangular surfaces is 240 sq. ft.
a) Write the formula to find the area of triangular surfaces of a square based pyramid. (1)
b) Find the slant height of the tent. (1)
c) If 64 cu. ft. of air is required for a tourist, how many tourists can be accommodated easily in the tent? Find it. (2)
Figure: Photo of a pyramid-shaped tent with a square base; both base sides are labelled 12 ft.

Answer

(a) Area of triangular surfaces $=2al$ .

(b) $240=2\times12\times l\Rightarrow l=10$ ft.

(c) $h=\sqrt{l^2-(a/2)^2}=\sqrt{100-36}=8$ ft.

Volume $=\frac13a^2h=\frac13\times144\times8=384$ cu. ft.

Tourists $=\dfrac{384}{64}=$ 6 tourists.

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Gandaki · 1+2+3 marks

In the figure, a conical object with slant height 25 cm and a hemispherical object with radius 7 cm is composed to make a solid object.
a) What type of triangle is formed when the vertical height of the cone is drawn in the given figure? Write it. (1)
b) What will be the total cost to paint the solid at the rate of 25 paisa per square centimeter? Find it. (2)
c) What is the volume of the solid object? Find it. (3)
Figure: A cone sitting on top of a hemisphere sharing the same circular base; slant height of the cone is 25 cm and the radius of the hemisphere/base is 7 cm.

Answer

(a) A right-angled triangle is formed (the vertical height, the radius of the base and the slant height, with the right angle at the centre of the base).

(b) $r=7$ cm, $l=25$ cm.

Surface to paint = curved surface of cone + curved surface of hemisphere

$=\pi rl+2\pi r^2=\frac{22}{7}\times7\times25+2\times\frac{22}{7}\times49=550+308=858$ cm $^2$ .

Cost $=858\times0.25=$ Rs 214.50.

(c) $h=\sqrt{25^2-7^2}=24$ cm.

$V=\frac13\pi r^2h+\frac23\pi r^3=\frac13\times\frac{22}{7}\times49\times24+\frac23\times\frac{22}{7}\times343$

$=1232+\frac{2156}{3}=\frac{5852}{3}\approx1950.67$ cm $^3$ .

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Gandaki · 2+1 marks

The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Tiles of length 25 cm, breadth 20 cm and height 5 mm are paved on the floor of the room.
a) How much cost of carpeting is required in the room at the rate of Rs. 250 per sq.m? Find it. (2)
b) How many maximum numbers of tiles can be paved on the floor? Find it. (1)

Answer

(a) Floor area $=5\times4=20$ m $^2$ .

Cost of carpeting $=20\times250=$ Rs 5,000.

(b) Floor $=500\text{ cm}\times400\text{ cm}=200{,}000$ cm $^2$ . Area of one tile $=25\times20=500$ cm $^2$ .

Number of tiles $=\dfrac{200000}{500}=$ 400 tiles.

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Lumbini · 1+3 marks

The radius of the base and vertical height of a conical tent are 8 meter and 6 meter respectively. $[\pi=\frac{22}{7}]$
a) Write the formula for finding the curved surface area of a cone. (1)
b) Find the cost of the canvas used in the tent at the rate of Rs. 50 per square meter. (3)

Answer

(a) Curved surface area of a cone $=\pi rl$ .

(b) $r=8$ m, $h=6$ m. $l=\sqrt{r^2+h^2}=\sqrt{64+36}=10$ m.

Canvas area $=\pi rl=\frac{22}{7}\times8\times10=\frac{1760}{7}$ m $^2$ .

Cost $=\frac{1760}{7}\times50=\frac{88000}{7}\approx$ Rs 12,571.43.

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Lumbini · 1+1+1 marks

The solid object shown in the picture is made up of cuboid and pyramid with a square base. Where, the length of the base side is 12 cm, height of the cuboid is 9 cm and slant height of the pyramid is 10 cm.
a) How do you find the vertical height of the square based pyramid when slant height and length of the base are given? Write it. (1)
b) Find the volume of the pyramid. (1)
c) How much is the volume of the pyramid less or more than the volume of the cuboid? Compare it. (1)
Figure: A solid made of a cuboid with a square base of side 12 cm and height 9 cm, with a square-based pyramid of slant height 10 cm standing on its top face.

Answer

(a) Vertical height $h=\sqrt{l^2-\left(\dfrac a2\right)^2}$ , where $l$ is the slant height and $a$ the base side.

(b) $a=12$ cm, $l=10$ cm: $h=\sqrt{100-36}=8$ cm.

$V_{pyramid}=\frac13a^2h=\frac13\times144\times8=384$ cm $^3$ .

(c) Volume of cuboid $=12\times12\times9=1296$ cm $^3$ .

The pyramid's volume is less than the cuboid's by $1296-384=912$ cm $^3$ .

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Lumbini · 2+2+2 marks

As shown in the picture, 30 rings made of cement with internal diameter of 91 cm and height of 30 cm each are placed to make a well. 3 workers can build the well in 7 days. The cost of one ring is Rs. 900 and the daily wage per worker is Rs. 1500.
a) How much does it cost to build a well? Find it. (2)
b) How many maximum liters of water can be stored in the well? Find it. (2)
c) If there is 4000 liter of water in that well, how many rings are completely under the water level? Find it. (2)
Figure: Photograph looking down into a well lined with cement rings.

Answer

(a) Cost of rings $=30\times900=$ Rs 27,000.

Wages $=3\times7\times1500=$ Rs 31,500.

Total cost $=27000+31500=$ Rs 58,500.

(b) Inner radius $r=\frac{91}{2}=45.5$ cm. Total height $=30\times30=900$ cm.

$V=\pi r^2h=\frac{22}{7}\times45.5^2\times900=6506.5\times900=5{,}855{,}850$ cm $^3$ .

Capacity $=5{,}855{,}850\div1000\approx$ 5,855.85 litres.

(c) $4000$ L $=4{,}000{,}000$ cm $^3$ .

Height of water $=\dfrac{4000000}{\pi r^2}=\dfrac{4000000}{6506.5}\approx614.8$ cm.

Rings under water $=\dfrac{614.8}{30}\approx20.5$ , so 20 rings are completely under the water level.

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Karnali · 1+1+2+2 marks

A square based pyramid has its length of a side of base 10 meter and height 12 meter.
(a) Write the formula for finding the area of triangular surfaces of the pyramid. (1)
(b) Find the volume of the pyramid. (1)
(c) What will be the total cost of painting the total surface of the pyramid at the rate of Rs. 50 per square meter? Find it. (2)
(d) By how much the area of the base of the pyramid is less or more than the area of a triangular surface? Compare it. (2)

Answer

(a) Area of triangular surfaces $=2al$ .

(b) $V=\frac13a^2h=\frac13\times100\times12=400$ m $^3$ .

(c) $l=\sqrt{12^2+5^2}=13$ m.

Total surface area $=a^2+2al=100+2\times10\times13=360$ m $^2$ .

Cost $=360\times50=$ Rs 18,000.

(d) Area of base $=100$ m $^2$ . Area of one triangular surface $=\frac12\times10\times13=65$ m $^2$ .

The base area is greater than one triangular surface by $100-65=35$ m $^2$ .

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2080 Chaitra · Karnali · 1+3 marks

A water tank as shown in the figure has its lower part cylindrical and upper part hemispherical. The total height of the tank is 13 m and the diameter of its base is 14 m. $\left[\pi=\frac{22}{7}\right]$
(a) Which geometric shape is formed on the base of the tank? Write it. (1)
(b) How much does it cost to fill water in the tank at the rate of 50 paisa per liter? Find it. (3)
Figure: A water tank made of a cylinder with a hemisphere on top; total height 13 m marked on the right; base diameter 14 m marked at the bottom.

Answer

(a) A circle is formed on the base.

(b) $r=7$ m. Cylinder height $=13-7=6$ m.

$V=\pi r^2h+\frac23\pi r^3=\frac{22}{7}\times49\times6+\frac23\times\frac{22}{7}\times343$

$=924+\frac{2156}{3}=\frac{4928}{3}\approx1642.67$ m $^3$ .

Capacity $=\frac{4928}{3}\times1000\approx1{,}642{,}666.67$ litres.

Cost at Rs 0.50 per litre $=1642666.67\times0.5\approx$ Rs 8,21,333.33.

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Chaitra · Karnali · 2+1 marks

The radius of the base and vertical height of a conical object are 7 cm and 24 cm respectively.
(a) Find the curved surface area of the cone. (2)
(b) What can be done to increase the volume of the cone? Give a suitable reason. (1)

Answer

(a) $r=7$ cm, $h=24$ cm, $l=\sqrt{49+576}=25$ cm.

Curved surface area $=\pi rl=\frac{22}{7}\times7\times25=$ 550 cm $^2$ .

(b) Volume $=\frac13\pi r^2h$ . The volume can be increased by increasing the radius and/or the height of the cone, because volume is directly proportional to the height and to the square of the radius.

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2080 Chaitra · Sudurpashchim · 1+3+1 marks

The volume of the square based pyramid as shown in the figure is 384 cubic cm and the length of the side of the base is 12 cm.
(a) How many triangle's area are counted to find the lateral surface area of the square based pyramid? (1)
(b) Find the total surface area of the pyramid. (3)
(c) Compare the area of the triangular faces and the area of the base of the pyramid. (1)
Figure: A square-based pyramid with the base side marked 12 cm; the equal sides of the base are tick-marked.

Answer

(a) 4 triangles.

(b) $384=\frac13\times144\times h\Rightarrow h=8$ cm.

$l=\sqrt{8^2+6^2}=10$ cm.

Total surface area $=a^2+2al=144+2\times12\times10=144+240=384$ cm $^2$ .

(c) Area of triangular faces $=240$ cm $^2$ ; area of base $=144$ cm $^2$ .

The triangular faces' area is greater by $96$ cm $^2$ ; ratio $240:144=5:3$ .

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2080 Chaitra · Sudurpashchim · 1+2+1 marks

In the figure, the total height of the combined solid metal is 180 cm and radius of the base is 42 cm. $[\pi = \frac{22}{7}]$
(a) Write the formula for finding total surface area of the cone. (1)
(b) Find the volume of the combined solid object. (2)
(c) How much rupees is required to make the solid metal at the rate of 10 paisa per cubic centimeter? Find it. (1)
Figure: Two cones joined base to base (a double cone); common circular base of radius 42 cm; total height from apex to apex 180 cm.

Answer

(a) Total surface area of a cone $=\pi r(r+l)$ (i.e. $\pi rl+\pi r^2$ ).

(b) Each cone has height $=\frac{180}{2}=90$ cm and $r=42$ cm.

$V=2\times\frac13\pi r^2h=\frac23\times\frac{22}{7}\times42^2\times90=\frac23\times\frac{22}{7}\times1764\times90$

$=332{,}640$ cm $^3$ .

(c) Cost $=332640\times0.10=$ Rs 33,264.

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2080 Chaitra · Sudurpashchim · 2+2 marks

A water plant company built a reserve water tank whose lower part is cylindrical and upper part is hemi-spherical with equal diameter. The inner diameter and total height of the tank are 4.2 meter and 4.5 meter respectively. $[\pi = \frac{22}{7}]$
(a) Find the volume of the tank. (2)
(b) How many maximum number of water tanks of 1000 liter capacity can be filled by the full water of reserve tank? Find it. (2)

Answer

(a) $r=\frac{4.2}{2}=2.1$ m. Cylinder height $=4.5-2.1=2.4$ m.

$V=\pi r^2h+\frac23\pi r^3=\frac{22}{7}\times2.1^2\times2.4+\frac23\times\frac{22}{7}\times2.1^3$

$=33.264+19.404=52.668$ m $^3$ .

(b) Capacity $=52.668\times1000=52{,}668$ litres.

Number of tanks $=\dfrac{52668}{1000}=52.668$ , so a maximum of 52 tanks can be filled completely.

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2081 Shrawan · Grade increment · 1+2 marks

In the figure alongside, a combined solid made up of cone and hemisphere is given. The slant height of cone is 25 cm and radius of base is 7 cm.
(a) Write down the relationship among slant height (l), vertical height (h) and radius of base (r). (1)
(b) Find the total surface area of the combined solid. (2)
Figure: A cone with its base on top of a hemisphere (hemisphere below, curved side down). The slant height of the cone is labelled 25 cm and the radius of the common base is labelled 7 cm.

Answer

(a) $l^2=h^2+r^2$ .

(b) Total surface area = curved surface of cone + curved surface of hemisphere

$=\pi rl+2\pi r^2=\frac{22}{7}\times7\times25+2\times\frac{22}{7}\times49=550+308=$ 858 cm $^2$ .

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2080 Grade increment · 1+1+2 marks

A school is planning to construct a tent with shape of square based pyramid for a program in which the length of each side of the base is 12 m and volume is 384 cubic m.
(a) How many triangular surfaces are there in the pyramidic tent? Write it. (1)
(b) What is the height of tent? Find it. (1)
(c) How many square meter of the cloth is required to construct the tent? Find it. (2)

Answer

(a) 4 triangular surfaces.

(b) $384=\frac13\times12^2\times h\Rightarrow h=\dfrac{384\times3}{144}=$ 8 m.

(c) $l=\sqrt{8^2+6^2}=10$ m.

Cloth needed (the four triangular faces) $=2al=2\times12\times10=$ 240 m $^2$ .

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2080 Grade increment · 1+2+1 marks

A metallic solid made up of a cone and a cylinder is given in the figure. The radii of the base of the cone and cylinder are equal. The height of the cylinder is 50 cm, height of the cone is 30 cm and radius of the base of cone is 14 cm.
(a) Write the formula to find the volume of the cone. (1)
(b) Find the volume of the solid object. (2)
(c) How many times the volume of the cylindrical part is more than the volume of the conical part? Find it. (1)
Figure: A solid made of a cone standing on top of a cylinder with the same base radius. Cone: perpendicular height marked 30 cm, base radius marked 14 cm. Cylinder: height marked 50 cm.

Answer

(a) Volume of cone $=\frac13\pi r^2h$ .

(b) $r=14$ cm.

Volume of cone $=\frac13\times\frac{22}{7}\times196\times30=6160$ cm $^3$ .

Volume of cylinder $=\frac{22}{7}\times196\times50=30800$ cm $^3$ .

Total volume $=6160+30800=$ 36,960 cm $^3$ .

(c) $\dfrac{30800}{6160}=$ 5 times.

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2080 Grade increment · 3+2 marks

A garden is 12 m long and 6 m wide. It is planned to lay 60 cm long square turfs (chaparis) in the garden. The price of one turf (chapari) is Rs. 110.
(a) How many turfs (chaparis) are required to lay down in the garden? Find it. (3)
(b) To lay down turfs in the garden, 2 workers worked for 2 days. If the daily wages of a worker is Rs. 1500, comment on the wages and the cost of the turfs (Chapari). (2)

Answer

(a) Garden $=12$ m $\times$ 6 m. Turf side $=60$ cm $=0.6$ m.

Along length: $12\div0.6=20$ ; along width: $6\div0.6=10$ .

Number of turfs $=20\times10=$ 200.

(b) Cost of turfs $=200\times110=$ Rs 22,000.

Wages $=2\times2\times1500=$ Rs 6,000.

The cost of the turfs (Rs 22,000) is much higher than the wages (Rs 6,000): it is Rs 16,000 more, about 3.7 times the wages. So the turfs are the main expense.

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2075 Supplementary · 2 marks

The height of a triangular prism is $40$ cm. If the sides of the base of the prism are $8$ cm, $6$ cm and $10$ cm, find the area of the rectangular surfaces of the prism.

Answer

Area of the rectangular surfaces $=$ perimeter of the base $\times$ height.

Perimeter of the base $=8+6+10=24$ cm, height $=40$ cm.

$\text{Area}=24\times40=\textbf{960 cm}^2$

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2075 Supplementary · 5 marks

The given figure is a pillar made up of a square based prism and a pyramid. If the length of the side of the base is 6 ft, the height of the prism is 8 ft and the height of the pyramid is 4 ft, what will be the cost of painting the surfaces of the pillar at the rate of Rs. 90 per square feet? Find it.

Answer

The painted surface is the four side faces of the prism and the four slant faces of the pyramid. The ground base and the top of the prism (covered by the pyramid) are not painted.

Prism: lateral surface area $= 4 \times 6 \times 8 = 192$ sq. ft.

Pyramid: half of the base side is $3$ ft, so slant height

$l = \sqrt{4^2 + 3^2} = 5 \text{ ft}.$

Lateral surface area $= 4 \times \dfrac{1}{2} \times 6 \times 5 = 60$ sq. ft.

Total area to paint $= 192 + 60 = 252$ sq. ft.

$\text{Cost} = 252 \times 90 = \text{Rs. } 22680$

The cost of painting is Rs. 22,680.

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Grade increment · 1+1+2 marks

The given figure is a square based solid pyramid, where $BC = 18$ cm and $OP = 12$ cm.
(a) If the length of the base side of a pyramid is $a$ , the vertical height is $h$ and the volume is $V$ , then express $V$ in terms of $a$ and $h$ . (1)
(b) Find the volume of the pyramid. (1)
(c) Find the area of triangular surfaces of the pyramid. (2)
Figure: Square based pyramid $P\text{-}ABCD$ with $O$ the centre of the base, $PO$ the vertical height, $Q$ the mid-point of $CD$ and $PQ$ marked as the slant height; $BC = 18$ cm.

Answer

(a) $V = \dfrac{1}{3}a^2h$ .

(b) Base side $a = BC = 18$ cm and height $h = OP = 12$ cm.

$V = \frac{1}{3} \times 18^2 \times 12 = \frac{1}{3} \times 324 \times 12 = 1296 \text{ cm}^3$

(c) Let $Q$ be the mid-point of a base edge, so $OQ = \dfrac{18}{2} = 9$ cm. The slant height is

$PQ = \sqrt{OP^2 + OQ^2} = \sqrt{12^2 + 9^2} = \sqrt{225} = 15 \text{ cm}$

Area of the four triangular faces $= 4 \times \dfrac{1}{2} \times 18 \times 15 = 540 \text{ cm}^2$ .

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Grade increment · 1+3+2 marks

A toy is made with the combination of a cone and a hemi-sphere having the same base. The total height of the toy is 31 cm and the radius of the base is 7 cm.
(a) How many surfaces should there be in the toy? Write it. (1)
(b) Find the surface area of the conical part of the toy. (3)
(c) By how much is the surface area of the conical part more or less than the surface area of the hemi-spherical part? Compare with calculation. (2)

Answer

(a) The toy has 2 surfaces: the curved surface of the cone and the curved surface of the hemisphere (the common base is hidden inside the toy).

(b) Height of the hemisphere $= r = 7$ cm, so the height of the cone is $h = 31 - 7 = 24$ cm.

$l = \sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2} = \sqrt{625} = 25 \text{ cm}$

Curved surface area of the cone $= \pi r l = \dfrac{22}{7} \times 7 \times 25 = 550 \text{ cm}^2$ .

(c) Curved surface area of the hemisphere $= 2\pi r^2 = 2 \times \dfrac{22}{7} \times 7^2 = 308 \text{ cm}^2$ .

Difference $= 550 - 308 = 242 \text{ cm}^2$ .

So the surface area of the conical part is $242 \text{ cm}^2$ more than that of the hemispherical part.

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Grade increment · 2+1 marks

The given figure $ABCD$ is a quadrilateral shaped land in which $BF \perp AC$ , $DE \perp AC$ , $AC = 40$ m, $DE = 10$ m and $BF = 15$ m.
(a) Find the area of the land. (2)
(b) Does Rs. 34,000 suffice to level the land at the rate of Rs. 700 per 100 square meters? Calculate. (1)
Figure: Quadrilateral $ABCD$ with the diagonal $AC$ ; $E$ and $F$ are the feet of the perpendiculars from $D$ and $B$ on $AC$ .

Answer

(a) The diagonal $AC$ divides the land into $\triangle ABC$ (height $BF$ ) and $\triangle ACD$ (height $DE$ ).

$\text{Area} = \frac{1}{2} \times AC \times (BF + DE) = \frac{1}{2} \times 40 \times (15 + 10) = 500 \text{ m}^2$

(b) Cost of levelling $= \dfrac{500}{100} \times 700 = \text{Rs } 3{,}500$ .

Since $\text{Rs } 3{,}500 < \text{Rs } 34{,}000$ , Rs 34,000 is more than enough to level the land, with Rs 30,500 left over.

Answer checked by Claude Sonnet · Oct 21, 2026

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2079 Chaitra · Bagmati · 2 marks

Find the area of an equilateral triangle having perimeter 60 cm.

Answer

Side $a = \dfrac{60}{3} = 20$ cm.

$\text{Area} = \frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4} \times 20^2 = 100\sqrt{3} \approx 173.2 \text{ cm}^2$

Answer checked by Claude Sonnet · Oct 21, 2026

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2079 Chaitra · Bagmati · 2 marks

If the radius of a football is 7 cm, find its surface area.

Answer

A football is a sphere, so

$\text{Surface area} = 4\pi r^2 = 4 \times \frac{22}{7} \times 7^2 = 616 \text{ cm}^2$

Answer checked by Claude Sonnet · Oct 21, 2026

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2079 Chaitra · Bagmati · 2 marks

If the curved surface area of a cone is 550 sq. cm and the slant height is 25 cm, find the height of the cone.

Answer

Curved surface area $= \pi r l$ :

$550 = \frac{22}{7} \times r \times 25 \Rightarrow r = \frac{550 \times 7}{22 \times 25} = 7 \text{ cm}$

$h = \sqrt{l^2 - r^2} = \sqrt{25^2 - 7^2} = \sqrt{576} = 24 \text{ cm}$

Answer checked by Claude Sonnet · Oct 21, 2026

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2079 Chaitra · Bagmati · 2 marks

In the figure, $PQRS$ is a rectangle and $PQMN$ is a parallelogram. If $PQ = 8$ cm, $QR = 5$ cm, $MQ = 10$ cm and $MO \perp NP$ , find the length of $MO$ .
Figure: Rectangle $PQRS$ and parallelogram $PQMN$ stand on the same base $PQ$ and between the same parallel lines; $MO$ is drawn perpendicular to the side $NP$ (produced to $O$ ).

Answer

The rectangle $PQRS$ and the parallelogram $PQMN$ stand on the same base $PQ$ and lie between the same parallel lines, so their areas are equal.

Area of rectangle $= PQ \times QR = 8 \times 5 = 40\ \text{cm}^2$ .

In the parallelogram, the side $NP = MQ = 10$ cm and $MO \perp NP$ , so $MO$ is the height on the base $NP$ :

$NP \times MO = 40 \Rightarrow 10 \times MO = 40 \Rightarrow MO = 4\ \text{cm}$

Answer checked by Claude Sonnet · Oct 21, 2026

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2079 Chaitra · Bagmati · 4 marks

A person plans to make a combined solid of a cone and a hemisphere for decoration material from gold. If the diameter of the hemisphere and the base of the cone is 10 cm and the slant height of the cone is 13 cm, find the total volume of gold required to make the combined solid.

Answer

Radius of the hemisphere and of the cone base: $r = \frac{10}{2} = 5$ cm. Slant height $l = 13$ cm.

Height of the cone: $h=\sqrt{l^2-r^2}=\sqrt{169-25}=\sqrt{144}=12$ cm.

Volume of cone $=\frac13\pi r^2h=\frac13\pi(5)^2(12)=100\pi$ cm $^3$

Volume of hemisphere $=\frac23\pi r^3=\frac23\pi(5)^3=\frac{250\pi}{3}$ cm $^3$

$V = 100\pi+\frac{250\pi}{3}=\frac{550\pi}{3}\approx 575.96\ \text{cm}^3$

(Using $\pi = 3.1416$ .) The gold required is about 575.96 cm $^3$ .

Answer checked by Claude Sonnet · Oct 21, 2026

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2079 Chaitra · Bagmati · 5 marks

In the given square based pyramidic house, how much does it cost for metal plating the roof at the rate of Rs. 12,000 per square meter? Find it.
Figure: A square based pyramid-shaped house whose base side is 16 m and whose vertical height is 6 m.

Answer

The roof is the four slanting triangular faces, that is, the lateral surface of the pyramid.

Base side $b = 16$ m, vertical height $h = 6$ m, so the half-base is $8$ m.

Slant height:

$l=\sqrt{h^2+\left(\tfrac b2\right)^2}=\sqrt{6^2+8^2}=\sqrt{100}=10\ \text{m}$

Area of the roof $=$ lateral surface area $=2bl=2\times16\times10=320\ \text{m}^2$ .

Cost $=320\times12{,}000=$ Rs 38,40,000.

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Model · Set 1 · 1+1+2 marks

The height of a square based pyramid is 21 cm and the length of the base is 20 cm.
(a) How many triangular surfaces are there in a square based pyramid? Write it. (1)
(b) Find the slant height of the pyramid. (1)
(c) What is the total cost of painting the total surface area of the pyramid at the rate of Rs. 5 per square cm? Find it. (2)

Answer

(a) A square based pyramid has 4 triangular surfaces.

(b) Half of the base $=\frac{20}{2}=10$ cm and $h=21$ cm.

$l=\sqrt{h^2+10^2}=\sqrt{441+100}=\sqrt{541}\approx \textbf{23.26 cm}$

(c) Total surface area $=$ base $+$ lateral surface $=b^2+2bl$

$=20^2+2\times20\times23.26=400+930.38=1330.38\ \text{cm}^2$

Cost $=1330.38\times5\approx$ Rs 6,651.88 (about Rs 6,652).

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Model · Set 1 · 1+2+1 marks

A metallic solid made up of a cone and a cylinder is given in the figure. The radii of the base of the cone and the cylinder are equal. The height of the cylinder is 40 cm, the height of the cone is 24 cm and the radius of the base of the cone is 7 cm.
(a) If the radius of the base and the slant height are given, write the formula for finding the curved surface area of the cone. (1)
(b) Find the volume of the solid object. (2)
(c) Compare the volume of the cylindrical part and the volume of the conical part. (1)
Figure: A cone placed on top of a cylinder with the same base radius 7 cm; the height of the cone is 24 cm and the height of the cylinder is 40 cm.

Answer

(a) Curved surface area of a cone $=\pi r l$ , where $r$ is the base radius and $l$ the slant height.

(b) $r=7$ cm, cylinder height $H=40$ cm, cone height $h=24$ cm.

Volume of cylinder $=\pi r^2H=\frac{22}{7}\times49\times40=6160\ \text{cm}^3$

Volume of cone $=\frac13\pi r^2h=\frac13\times\frac{22}{7}\times49\times24=1232\ \text{cm}^3$

Volume of the solid $=6160+1232=$ 7,392 cm $^3$ .

(c) $\dfrac{\text{cylinder}}{\text{cone}}=\dfrac{6160}{1232}=5$

The volume of the cylindrical part is 5 times the volume of the conical part (ratio $5:1$ ).

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Model · Set 1 · 3+2 marks

The length, breadth and height of a rectangular room are 14 ft, 13 ft and 10 ft respectively. There are two square windows with 3 feet edges and two doors of size 6 ft $\times$ 3 ft in the room.
(a) How much does it cost to paint the four walls and the ceiling of the room excluding doors and windows at the rate of Rs. 36 per square feet? Find it. (3)
(b) How much will the total cost increase to paint the same part if the cost of painting per square meter is increased by one third of what it was before due to the increase in the market price? Find it. (2)

Answer

(a) Area of four walls $=2h(l+b)=2\times10\times(14+13)=540\ \text{ft}^2$

Area of ceiling $=l\times b=14\times13=182\ \text{ft}^2$

Area of 2 windows $=2\times(3\times3)=18\ \text{ft}^2$

Area of 2 doors $=2\times(6\times3)=36\ \text{ft}^2$

Area to be painted $=540+182-18-36=668\ \text{ft}^2$

Cost $=668\times36=$ Rs 24,048.

(b) The rate rises by one third: $\frac13\times36=\text{Rs } 12$ , so the new rate is Rs 48 per sq. ft. (the question says "square meter", but the rate is per square foot, as in part (a)).

New cost $=668\times48=\text{Rs } 32{,}064$

Increase in cost $=32064-24048=$ Rs 8,016.

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Model · Set 2 · 1+2+1 marks

A cone is filled with ice-cream whose upper part is hemispherical. The slant height of the cone is 25 cm and its radius is 7 cm.
(a) Write the relation among the height $(h)$ , radius $(r)$ and slant height $(l)$ of the cone. (1)
(b) Find the total volume of ice-cream in the conical and hemispherical parts. (2)
(c) Compare the quantities of ice-cream in the conical and hemispherical parts. (1)
Figure: An ice-cream cone with a hemispherical scoop on top of the same radius 7 cm; the slant height of the cone is 25 cm.

Answer

(a) $l^2=h^2+r^2$ .

(b) Height of the cone: $h=\sqrt{25^2-7^2}=\sqrt{576}=24$ cm.

Volume of the conical part:

$\frac13\pi r^2h=\frac13\times\frac{22}7\times7^2\times24=1232\text{ cm}^3.$

Volume of the hemispherical part:

$\frac23\pi r^3=\frac23\times\frac{22}7\times7^3=\frac{2156}{3}\approx718.67\text{ cm}^3.$

Total volume $=1232+\dfrac{2156}3=\dfrac{5852}3\approx1950.67$ cm $^3$ .

(c)

$\text{Cone}:\text{hemisphere}=1232:\frac{2156}3=3696:2156=12:7.$

The conical part holds more ice-cream (ratio $12:7$ ).

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Model · Set 2 · 3+2 marks

The length, breadth and height of a rectangular room are 13 ft, 12 ft and 10 ft respectively. There are 2 windows of size $3\text{ ft}\times 4\text{ ft}$ and a door of size $3\text{ ft}\times 6\text{ ft}$ .
(a) How much does it cost to paint the four walls and ceiling excluding the door and windows at the rate of Rs. 40 per square ft? Find it. (3)
(b) By how much will the total cost of painting the same parts of the room be increased if the rate of cost per square feet is increased by $25\%$ ? Find it. (2)

Answer

(a) Area of the four walls $=2(l+b)h=2(13+12)\times10=500$ sq. ft.

Area of the ceiling $=l\times b=13\times12=156$ sq. ft.

Area of the door and windows $=3\times6+2\times(3\times4)=18+24=42$ sq. ft.

Area to be painted $=500+156-42=614$ sq. ft.

Cost $=614\times40=$ Rs 24,560.

(b) New rate $=40+25\%\text{ of }40=40+10=$ Rs 50 per sq. ft.

New cost $=614\times50=$ Rs 30,700.

Increase in cost $=30{,}700-24{,}560=$ Rs 6,140.

Answer checked by Claude Sonnet · Oct 21, 2026

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2080 Model · Set 3 · 1+2+1 marks

A metallic solid made up of a cone and a cylinder is given in the figure. The radii of the base of the cone and the cylinder are equal. The height of the cylinder is 35 cm, the height of the cone is 21 cm and the radius of the base of the cone is 7 cm.
(a) When the radius of the base of a cone and the slant height are given, write the formula for finding the total surface area of the cone. (1)
(b) Find the volume of the solid object. (2)
(c) How many times is the volume of the cylindrical part more than the volume of the conical part? Find it. (1)
Figure: A cone of height 21 cm placed on top of a cylinder of height 35 cm; both have base radius 7 cm.

Answer

(a) Total surface area of a cone $=\pi rl+\pi r^2=\pi r(l+r)$ .

(b) Volume of the cylindrical part:

$\pi r^2h=\frac{22}7\times7^2\times35=5390\text{ cm}^3.$

Volume of the conical part:

$\frac13\pi r^2h=\frac13\times\frac{22}7\times7^2\times21=1078\text{ cm}^3.$

Volume of the solid $=5390+1078=6468$ cm $^3$ .

(c)

$\frac{5390}{1078}=5.$

The cylindrical part has 5 times the volume of the conical part.

Answer checked by Claude Sonnet · Oct 21, 2026

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2075 Chaitra · Set EP · 2 marks

The height of a triangular prism whose base is a right-angled triangle is 30 cm. If the sides of the triangle containing the right angle are 4 cm and 3 cm, find the area of the rectangular surfaces of the prism.

Answer

Hypotenuse of the base $= \sqrt{4^2+3^2} = 5$ cm.

Perimeter of the base $= 3+4+5 = 12$ cm.

Area of the rectangular surfaces $=$ perimeter of base $\times$ height $= 12 \times 30 =$ 360 sq. cm.

Answer checked by Claude Sonnet · Oct 21, 2026

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2078 Baisakh · Set A · 2 marks

Find the area of a triangle having lengths of sides 6 cm, 7 cm and 11 cm.

Answer

$a=6$ , $b=7$ , $c=11$ , so $s=\dfrac{6+7+11}{2}=12$ cm.

$A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{12\times6\times5\times1}=\sqrt{360}=6\sqrt{10}$

Area $=6\sqrt{10}\approx$ 18.97 sq. cm

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2078 Baisakh · Set A · 2 marks

If the volume of a sphere is $36\pi$ cubic cm, find its radius.

Answer

$\frac43\pi r^3=36\pi \Rightarrow r^3=\frac{36\times3}{4}=27$

$r=\sqrt[3]{27}=\textbf{3 cm}$

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2078 Baisakh · Set A · 2 marks

Find the area of the rectangular surfaces of the given prism where $DE = 6$ cm, $CE = 8$ cm, $AB = 10$ cm and $BC = 12$ cm.
Figure: A triangular prism with the triangular face $DCE$ (right-angled at $E$ , $DC = 10$ cm) on top and the rectangular face $ABCD$ at the front; the length of the prism is $BC = 12$ cm.

Answer

The triangle $DCE$ is right-angled at $E$ , with $DE=6$ cm, $CE=8$ cm and $DC=10$ cm (so $AB=DC=10$ cm). The prism has three rectangular surfaces, each of length $12$ cm.

Area of the rectangular surfaces $=$ (perimeter of the triangular face) $\times$ length

$=(6+8+10)\times12=24\times12=\textbf{288 sq. cm}$

(that is, $10\times12+6\times12+8\times12=120+72+96=288$ ).

Answer checked by Claude Sonnet · Oct 21, 2026

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2078 Baisakh · Set A · 4 marks

The area of the base and the vertical height of a square based solid pyramid are 144 sq. cm and 8 cm respectively. What will be the total surface area of the pyramid? Find it.

Answer

Base area $a^2=144\Rightarrow a=12$ cm; vertical height $h=8$ cm.

Slant height $l=\sqrt{h^2+\left(\frac a2\right)^2}=\sqrt{8^2+6^2}=10$ cm

$\text{Lateral area}=2al=2\times12\times10=240\text{ sq. cm}$

$\text{T.S.A.}=240+144=\textbf{384 sq. cm}$

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2078 Baisakh · Set A · 5 marks

A solid object is made with the combination of a cylinder and a cone having the same radii. The height of the cylinder and the slant height of the cone are 28 cm and 17 cm respectively. If the total cost of painting the total surface of the solid object at the rate of Rs. 140 per 100 sq. cm is Rs. 2851.20, find the height of the cone.

Answer

Let the common radius be $r$ cm. Use $\pi=\frac{22}{7}$ .

Rate $=\dfrac{140}{100}=$ Rs 1.40 per sq. cm, so

$\text{Total surface area}=\frac{2851.20}{1.40}=\frac{14256}{7}\text{ sq. cm}$

Surface $=$ curved surface of cylinder + base of cylinder + curved surface of cone:

$2\pi r(28)+\pi r^2+\pi r(17)=\pi r(73+r)$

$\frac{22}{7}r(73+r)=\frac{14256}{7}\Rightarrow r^2+73r-648=0$

$(r+81)(r-8)=0\Rightarrow r=8\ (\text{since } r>0)$

Height of the cone:

$h=\sqrt{l^2-r^2}=\sqrt{17^2-8^2}=\sqrt{225}=\textbf{15 cm}$

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2074 Chaitra · Set CC · 1 mark

The length of a side of the square base of a pyramid is $p$ cm and the slant height is $q$ cm. What is the total surface area of the pyramid? Write it.

Answer

Base area $= p^2$ and the four triangular faces have area $4\times\frac12\times p\times q = 2pq$ .

Total surface area $= p^2 + 2pq$ sq. cm

Answer checked by Claude Sonnet · Oct 21, 2026

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2074 Chaitra · Set CC · 2 marks

A cylindrical tank contains 172.48 litres of water. If its height is 0.7 m, then find the radius of the base.

Answer

Volume $= 172.48$ litres $= 172.48\ \text{dm}^3 = 172480\ \text{cm}^3$ , and height $h = 0.7\ \text{m} = 70$ cm.

$\pi r^2 h = V \implies \frac{22}{7}\times r^2\times 70 = 172480$

$220\,r^2 = 172480 \implies r^2 = 784 \implies r = 28$

The radius of the base is 28 cm.

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2074 Chaitra · Set CC · 2 marks

If the surface area of a sphere is 22176 sq. cm, then find its radius.

Answer

Surface area $= 4\pi r^2 = 22176$

$4\times\frac{22}{7}\times r^2 = 22176 \implies r^2 = \frac{22176\times 7}{88} = 1764$

$r = \sqrt{1764} = 42$

The radius is 42 cm.

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2074 Chaitra · Set CC · 2 marks

If the height of a cone is four times the radius of the base and its volume is $36\pi$ cu. cm, then find the radius of the base of the cone.

Answer

Let the radius be $r$ cm, so the height is $h = 4r$ cm.

$V = \frac13\pi r^2 h = \frac13\pi r^2(4r) = \frac{4}{3}\pi r^3$

$\frac43\pi r^3 = 36\pi \implies r^3 = 27 \implies r = 3$

The radius of the base is 3 cm.

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2074 Chaitra · Set CC · 4 marks

The sides of a triangular field are in the ratio $4:5:6$ . If its perimeter is 1500 m, what is the area of that field? Find it.

Answer

Let the sides be $4k$ , $5k$ and $6k$ metres.

$4k+5k+6k = 1500 \implies k = 100$

So the sides are $a=400$ m, $b=500$ m, $c=600$ m, and

$s = \frac{1500}{2} = 750\ \text{m}$

$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{750\times 350\times 250\times 150}$

Writing $750=2\cdot3\cdot5^3$ , $350=2\cdot5^2\cdot7$ , $250=2\cdot5^3$ , $150=2\cdot3\cdot5^2$ :

$\text{Area} = \sqrt{2^4\cdot 3^2\cdot 5^{10}\cdot 7} = 4\times 3\times 5^5\sqrt7 = 37500\sqrt7\ \text{m}^2 \approx 99{,}216\ \text{m}^2$

The area of the field is $37500\sqrt7$ m² (about 99,216 m²).

Answer checked by Claude Sonnet · Oct 21, 2026

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2074 Chaitra · Set CC · 5 marks

A water tank is formed with the combination of a cylinder and a hemisphere. The total height of the tank is 14 m and the base area is 38.5 sq. m. If the tank is filled with water at the rate of 24 paisa per litre, what is the total cost to fill the water? Find it.

Answer

Let the radius be $r$ m. The base area is $\pi r^2 = 38.5$ , so

$\frac{22}{7}r^2 = 38.5 \Rightarrow r^2 = 12.25 \Rightarrow r = 3.5\text{ m}.$

Height of the cylindrical part $h = 14 - r = 14 - 3.5 = 10.5$ m.

Volume of the tank = volume of cylinder + volume of hemisphere

$V = \pi r^2 h + \frac{2}{3}\pi r^3 = 38.5\times 10.5 + \frac{2}{3}\times 38.5\times 3.5$

$V = 404.25 + 89.83 = 494.083\text{ m}^3$

Since $1\text{ m}^3 = 1000$ litres, $V = 494{,}083.33$ litres.

Cost at 24 paisa $=$ Rs 0.24 per litre:

$\text{Cost} = 494083.33\times 0.24 = \text{Rs } 1{,}18{,}580$

The total cost of filling the tank is Rs 1,18,580.

Answer checked by Claude Sonnet · Oct 21, 2026

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2074 Model question · 2 marks

The area of an isosceles triangular land whose base side length is 10 meter is 60 square meter. Find the measure of its remaining sides.

Answer

Let the height on the base be $h$ .

$\text{Area}=\tfrac12\times 10\times h=60\Rightarrow h=12\text{ m}$

The height bisects the base in an isosceles triangle, giving a right triangle with legs $5$ m and $12$ m. Each equal side is

$\sqrt{5^2+12^2}=\sqrt{169}=13\text{ m}.$

The remaining two sides are 13 m each.

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2074 Model question · 2 marks

The volume of a cylinder with diameter of its base 14 cm is 2156 cubic cm. Find the height of the cylinder.

Answer

Radius $r=\dfrac{14}{2}=7$ cm. Using $V=\pi r^2h$ :

$2156=\frac{22}{7}\times 7^2\times h=154h\Rightarrow h=\frac{2156}{154}=14\text{ cm}$

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2074 Model question · 2 marks

In the given solid triangular prism, if $PQ \perp QR$ , $PQ = 8$ cm, $QR = 6$ cm and $RR_1 = 15$ cm, find the lateral surface area of the prism.
Figure: A right triangular prism $PQR\text{-}P_1Q_1R_1$ with the right angle at $Q$ and length $RR_1$ .

Answer

In right-angled $\triangle PQR$ ,

$PR=\sqrt{PQ^2+QR^2}=\sqrt{8^2+6^2}=10\text{ cm}.$

Perimeter of the base $=8+6+10=24$ cm, and the height of the prism is $RR_1=15$ cm.

$\text{Lateral surface area}=\text{perimeter of base}\times\text{height}=24\times15=360\text{ cm}^2$

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2074 Model question · 4 marks

Find the total surface area of a square based pyramid having the slant height 10 cm and the vertical height 8 cm.

Answer

Let the base side be $b$ , vertical height $h=8$ cm and slant height $l=10$ cm.

The vertical height, the slant height and half the base side form a right triangle:

$\left(\frac b2\right)^2=l^2-h^2=100-64=36\Rightarrow\frac b2=6\Rightarrow b=12\text{ cm}.$

$\text{Total surface area}=b^2+2bl=12^2+2\times12\times10=144+240=384\text{ cm}^2$

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2074 Model question · 5 marks

A person bought a water tank of circular base having the radius 1.05 meter and height 3.5 meter for the use of his own house from the shop. If the upper part of the tank is semi-spherical, how many litres of water will be contained in the tank? Find it.

Answer

Radius $r=1.05$ m and total height $3.5$ m. The top is a hemisphere of height $r$ , so the cylindrical part has height

$h=3.5-1.05=2.45\text{ m}.$

Volume of the tank = volume of cylinder + volume of hemisphere

$V=\pi r^2h+\frac23\pi r^3$

$\pi r^2=\frac{22}{7}\times1.05^2=3.465\text{ m}^2$

$V=3.465\times2.45+\frac23\times3.465\times1.05=8.48925+2.4255=10.91475\text{ m}^3$

Since $1\text{ m}^3=1000$ litres,

$V=10.91475\times1000=10{,}914.75\text{ litres}.$

The tank holds 10,914.75 litres of water.

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Pre-board · PABSON Set A · 1+2+1 marks

The given figure is a square based pyramid. The side of the base of the pyramid is 6 cm and the sloping edge of a triangular face is $\sqrt{34}$ cm.
(a) Write the formula to calculate the slant height. (1)
(b) Find the vertical height. (2)
(c) Find the volume of the pyramid. (1)

Answer

(a) $l=\sqrt{h^2+\left(\dfrac{a}{2}\right)^2}$ , where $h$ is the vertical height and $a$ is the side of the base.

(b) Half the diagonal of the base $=\dfrac{6\sqrt2}{2}=3\sqrt2$ cm. The vertical height, half the diagonal and the sloping edge form a right-angled triangle:

h^2=(\sqrt{34})^2-(3\sqrt2)^2=34-18=16\Rightarrow h=4\text{ cm}

(c) $V=\dfrac13a^2h=\dfrac13\times6^2\times4=\mathbf{48\text{ cm}^3}$ .

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2082 Pre-board · PABSON Set A · 1+2+1 marks

The circumference of the circular base of a cylindrical pillar is 13.2 meters and its height is 25 meters. A cone of height 2.8 meters with the same radius is attached on the top of the pillar.
(a) Write the formula to find the circumference of the base. (1)
(b) Find the surface area of the pillar. (2)
(c) Find the cost of painting the pillar at the rate of Rs. 40 per square meter. (1)

Answer

(a) Circumference of the base $=2\pi r$ .

(b) $2\pi r=13.2\Rightarrow r=\dfrac{13.2\times7}{2\times22}=2.1$ m.

Slant height of the cone: $l=\sqrt{r^2+h^2}=\sqrt{2.1^2+2.8^2}=\sqrt{12.25}=3.5$ m.

Surface area $=$ base $+$ curved surface of cylinder $+$ curved surface of cone

=\pi r^2+2\pi rh+\pi rl=\frac{22}{7}\times2.1^2+13.2\times25+\frac{22}{7}\times2.1\times3.5

=13.86+330+23.1=\mathbf{366.96\text{ m}^2}

(c) Cost $=366.96\times40=$ Rs 14,678.40.

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2082 Pre-board · PABSON Set A · 3+2 marks

Ramesh has a farm in the shape of a quadrilateral as shown in the figure, in which $BE \perp AC$ , $DF \perp AC$ , $AC = 48$ meter, $BE = 30$ meter and $DF = 25$ meter. He planted paddy in the farm on 15th Asar.
(a) Find the area of the farm. (3)
(b) Find the total cost of 3 male laborers at the rate of Rs. 1000 per person and 4 female laborers at the rate of Rs. 2.5 per square meter for sowing the paddy in the field. (2)
Figure: Quadrilateral $ABCD$ with the diagonal $AC$ ; $E$ and $F$ are the feet of the perpendiculars from $B$ and $D$ on $AC$ .

Answer

(a) The diagonal $AC$ splits the farm into $\triangle ABC$ (height $BE$ ) and $\triangle ACD$ (height $DF$ ).

\text{Area}=\tfrac12\times AC\times BE+\tfrac12\times AC\times DF=\tfrac12\times AC\,(BE+DF)

=\tfrac12\times48\times(30+25)=24\times55=\mathbf{1320\text{ m}^2}

(b) Cost for 3 male laborers $=3\times1000=$ Rs 3,000.

Cost for the female laborers (paid for the area sown) $=2.5\times1320=$ Rs 3,300.

Total cost $=3000+3300=$ Rs 6,300.

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2082 Pre-board · PABSON Set B · 1+3+1 marks

The combined solid object is given alongside with a cone and a hemisphere. The total height of the solid is 35 cm and the height of the cone is 20 cm.
(a) Write the relation among the height $(h)$ , the radius $(r)$ and the slant height $(l)$ of the cone. (1)
(b) Find the total surface area of the solid. (3)
(c) Are the total surface area and the curved surface area of the solid equal? Justify with a reason. (1)
Figure: A cone with a hemisphere of the same radius on its flat circular face (the hemisphere on top, the cone apex pointing down).

Answer

(a) $l^2=h^2+r^2$

(b) Radius of the hemisphere $=$ radius of the cone $=r=35-20=15$ cm.

Slant height of the cone: $l=\sqrt{20^2+15^2}=\sqrt{625}=25$ cm.

The circular faces are joined, so only the curved surfaces are exposed:

\text{TSA}=\pi rl+2\pi r^2=\pi\times15\times25+2\pi\times15^2=375\pi+450\pi=825\pi

=825\times\frac{22}{7}\approx\mathbf{2592.86\text{ cm}^2}

(c) Yes, they are equal. The flat circular face of the hemisphere and the base of the cone are joined together and are not on the outside, so the whole outer surface of the solid consists only of the two curved surfaces. Hence total surface area $=$ curved surface area $=825\pi\text{ cm}^2$ .

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2082 Pre-board · PABSON Set B · 2+2 marks

A rectangular hall is 46 ft long and 24 ft broad. It has four windows of area 24 sq ft each and two doors of size 7 ft $\times$ 4 ft each. The height of the hall is 10 ft.
(a) What is the cost of colouring its four walls excluding doors and windows at the rate of Rs. 65 per sq ft? (2)
(b) Is Rs. 42,000 sufficient to pave tiles of 2 ft $\times$ 2 ft on its floor at the rate of Rs. 150 per tile? Justify with your calculation. (2)

Answer

(a) Area of four walls $=2h(l+b)=2\times10\times(46+24)=1400$ sq ft.

Area of 4 windows $=4\times24=96$ sq ft; area of 2 doors $=2\times7\times4=56$ sq ft.

Area to be coloured $=1400-96-56=1248$ sq ft.

Cost $=1248\times65=$ Rs 81,120.

(b) Floor area $=46\times24=1104$ sq ft. Area of one tile $=2\times2=4$ sq ft.

Number of tiles $=\dfrac{1104}{4}=276$ .

Cost $=276\times150=$ Rs 41,400.

Since Rs 41,400 $<$ Rs 42,000, Rs 42,000 is sufficient (Rs 600 remains).

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Preparatory · PABSON Kaski · 1+2+1 marks

A square-based pyramid has vertical height 24 cm and the length of the base is 20 cm.
(a) Write the relationship among edge $(e)$ , slant height $(l)$ and length of base $(a)$ of a square-based pyramid in the form of a formula. (1)
(b) Find the total surface area of the given square-based pyramid. (2)
(c) Is Rs. 500 sufficient for colouring all its surfaces when the cost of painting is 25 paisa per square centimetre? Calculate it. (1)

Answer

(a) $e^2 = l^2 + \left(\dfrac{a}{2}\right)^2$ (the slant height is also given by $l^2 = h^2 + \left(\dfrac{a}{2}\right)^2$ ).

(b) $h = 24$ cm, $a = 20$ cm, so $\dfrac{a}{2} = 10$ cm.

$l = \sqrt{24^2 + 10^2} = \sqrt{676} = 26 \text{ cm}$

$\text{TSA} = a^2 + 2al = 20^2 + 2\times 20\times 26 = 400 + 1040 = 1440 \text{ cm}^2$

(c) Cost $= 1440\times 0.25 =$ Rs 360.

Since Rs 360 $<$ Rs 500, Rs 500 is sufficient (Rs 140 is left over).

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Preparatory · PABSON Kaski · 1+2+1 marks

A metallic solid is made by combining a cylinder and a cone having the same radius as shown in the diagram. The radius of the base is 3 m and the total height of the solid is 8 m.
(a) How many surfaces are there in the solid? (1)
(b) If the height of the cylinder and cone are equal, then find the volume of the solid. (2)
(c) What would be the difference in the volume of the cone and the cylinder if the height of the conical part is tripled and the cylindrical part is as it is? (1)
Figure: A cone on top of a cylinder with the same base radius 3 m; the total height of the solid is 8 m.

Answer

(a) There are 3 surfaces: the curved surface of the cone, the curved surface of the cylinder, and the flat circular base.

(b) Equal heights: $h = \dfrac{8}{2} = 4$ m, $r = 3$ m.

$V = \pi r^2 h + \dfrac{1}{3}\pi r^2 h = \pi\times 9\times 4 + \dfrac{1}{3}\pi\times 9\times 4 = 36\pi + 12\pi = 48\pi$

$V = 48\times\dfrac{22}{7} \approx 150.86 \text{ m}^3$

(c) The cone's height becomes $3\times 4 = 12$ m.

Cone: $\dfrac{1}{3}\pi\times 9\times 12 = 36\pi$ m $^3$ . Cylinder: $\pi\times 9\times 4 = 36\pi$ m $^3$ .

Difference $= 36\pi - 36\pi = 0$ . The two volumes would be equal.

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Preparatory · PABSON Kaski · 3+2 marks

A contractor company took a contract to prepare a program hall at Rs. 30,00,000. The hall has length 12 m, breadth 10 m and height 8 m. The cost of materials used in the hall is as follows:

Cement: 2000 packets at Rs. 400 per packet,

Rods: 6000 kg at Rs. 80 per kg,

and the cost of other materials such as aluminium, sand, wood etc.: Rs. 8,50,000.

The worker for plastering charged the cost of plastering its 4 walls and floor at the rate of Rs. 300 per sq. meter and the total wages of construction for other than plastering works was Rs. 3,56,000.
(a) Find the cost of construction of the hall including the cost of materials and labour charge. (3)
(b) The company put Zipson on the ceiling which costs Rs. 400 per sq. meter as the owner of the hall makes a special request to do it as a penalty for doing delay in completing the hall (the cost of Zipson was not included in the initial contract). Calculate whether the company got profit or loss from the construction of the hall. (2)

Answer

(a) Materials:

Cement: $2000\times 400 =$ Rs 8,00,000
Rods: $6000\times 80 =$ Rs 4,80,000
Other materials: Rs 8,50,000

Total materials $= 8,00,000 + 4,80,000 + 8,50,000 =$ Rs 21,30,000.

Plastering area = 4 walls + floor:

$2(12 + 10)\times 8 + 12\times 10 = 352 + 120 = 472 \text{ m}^2$

Plastering charge $= 472\times 300 =$ Rs 1,41,600.

Other wages $=$ Rs 3,56,000.

Total cost $= 21,30,000 + 1,41,600 + 3,56,000 =$ Rs 26,27,600.

(b) Ceiling area $= 12\times 10 = 120$ m $^2$ , so Zipson costs $120\times 400 =$ Rs 48,000.

Total expenditure $= 26,27,600 + 48,000 =$ Rs 26,75,600.

The company received only the contract amount of Rs 30,00,000 (Zipson was not included).

Profit $= 30,00,000 - 26,75,600 =$ Rs 3,24,400.

The company made a profit of Rs 3,24,400.

Answer checked by Claude Sonnet · Oct 21, 2026

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2082 Preparation · Bhaktapur · Bhaktapur Municipality · 1+1+2 marks

Ram thought of covering the roof of the house with zinc sheets. The roof of the house is in the shape of a pyramid with a square base. The height of the roof is 12 feet and the height of the rafters (slant height) is 13 feet.
(a) Find the length of one side of the square base of the roof. (1)
(b) Calculate the total area of zinc sheet required to cover the roof. (1)
(c) After covering the roof with zinc sheets, Ram decides to paint the inner surface of the roof. If the cost of painting is Rs. 60 per square foot and 15% of the surface area is not paintable due to wooden supports, calculate the total cost of painting the roof. (2)

Answer

(a) $l^2 = h^2 + \left(\dfrac{a}{2}\right)^2 \Rightarrow 13^2 = 12^2 + \left(\dfrac{a}{2}\right)^2 \Rightarrow \left(\dfrac{a}{2}\right)^2 = 169 - 144 = 25$ .

$\dfrac{a}{2} = 5 \Rightarrow a = 10$ ft.

(b) The roof covers only the four slanting faces:

$2al = 2\times 10\times 13 = 260 \text{ sq. ft}$

(c) The inner surface is also 260 sq. ft. Paintable part $= 85\%$ of 260 $= 221$ sq. ft.

Cost $= 221\times 60 =$ Rs 13,260.

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2082 Preparation · Bhaktapur · Bhaktapur Municipality · 1+2+2 marks

A water tank is formed with the combination of a cylinder and a hemisphere. The height of the tank is 24 m and the base area is 616 m $^2$ .
(a) Find the diameter of the base. (1)
(b) Find the volume of the tank. (2)
(c) Calculate the total outer surface area of the tank that needs plastering. (2)

Answer

(a) Base area $\pi r^2 = 616 \Rightarrow \dfrac{22}{7}r^2 = 616 \Rightarrow r^2 = 196 \Rightarrow r = 14$ m.

Diameter $= 2r = 28$ m.

(b) Height of the cylinder $= 24 - 14 = 10$ m.

$V = \pi r^2 h + \dfrac{2}{3}\pi r^3 = 616\times 10 + \dfrac{2}{3}\times\dfrac{22}{7}\times 14^3$

$= 6160 + \dfrac{2}{3}\times 8624 = 6160 + 5749.33 = 11,909.33 \text{ m}^3$

(c) Plastering is needed on the outside curved surfaces (the base rests on the ground):

Curved surface of the cylinder: $2\pi r h = 2\times\dfrac{22}{7}\times 14\times 10 = 880$ m $^2$ .

Curved surface of the hemisphere: $2\pi r^2 = 2\times 616 = 1232$ m $^2$ .

Total $= 880 + 1232 = 2112$ m $^2$ .

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2082 Preparation · Bhaktapur · Bhaktapur Municipality · 1+1+2 marks

The base of a hall is square-shaped with each side 14 m and the height of the hall is 6 m. The hall has

2 doors, each of size 2 m $\times$ 3 m, and

8 windows, each of size 2.5 m $\times$ 1.5 m.

On the basis of the given information, answer the following questions:
(a) Find the area of the base of the hall. (1)
(b) Find the total area of the four walls. (1)
(c) If only 90% of the wall area (excluding doors and windows) is to be painted, find the total cost of painting the walls at the rate of Rs. 120 per square meter. (2)

Answer

(a) Area of base $= 14\times 14 = 196$ m $^2$ .

(b) Area of four walls $= 4\times 14\times 6 = 336$ m $^2$ .

(c) Doors: $2\times(2\times 3) = 12$ m $^2$ . Windows: $8\times(2.5\times 1.5) = 30$ m $^2$ .

Wall area excluding doors and windows $= 336 - 12 - 30 = 294$ m $^2$ .

Area to be painted $= 90\%$ of $294 = 264.6$ m $^2$ .

Cost $= 264.6\times 120 =$ Rs 31,752.

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2082 Pre-qualifying · PABSON Kathmandu · 1+3+2 marks

The volume of a metallic square-based pyramid is 675 cm $^3$ and the length of the side of its base is 15 cm.
(a) What is the formula to find the lateral surface area of a square based pyramid? (1)
(b) Find the cost of colouring the surfaces of the pyramid at the rate of Rs. 3 per cm $^2$ . (3)
(c) If the pyramid is melted and recast into another square based pyramid of height 20 cm, what would be the length of its base? (2)

Answer

(a) Lateral surface area of a square-based pyramid $=2al$ (half the perimeter of the base times the slant height), where $a$ is the base side and $l$ the slant height.

(b) Volume $=\frac13a^2h$ : $675=\frac13\times15^2\times h\Rightarrow h=9\text{ cm}$ Slant height: $l=\sqrt{h^2+\left(\frac a2\right)^2}=\sqrt{81+56.25}=\sqrt{137.25}\approx11.72\text{ cm}$ Lateral surface area $=2\times15\times11.72\approx351.46\text{ cm}^2$ .

Total surface area (all surfaces, including the base) $=a^2+2al=225+351.46=576.46\text{ cm}^2$ .

Cost of colouring $=576.46\times3\approx\text{Rs }1{,}729.38$ .

(c) The volume stays $675\text{ cm}^3$ . For the new pyramid with height $20$ cm and base side $b$ : $\frac13b^2\times20=675\Rightarrow b^2=101.25\Rightarrow b\approx10.06\text{ cm}$ The base length of the new pyramid is about $10.06$ cm.

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2082 Pre-qualifying · PABSON Kathmandu · 1+2 marks

A solid object is made up of a hemisphere and a cylinder. The height of the cylinder is 70 cm and the total height of the solid is 91 cm.
(a) Which formula is used to calculate the volume of a hemisphere? (1)
(b) Find the total surface area of the solid. (2)

Answer

(a) Volume of a hemisphere $=\dfrac23\pi r^3$ .

(b) The hemisphere has the same radius as the cylinder, and the total height is the cylinder height plus the radius: $r=91-70=21\text{ cm}$ The surfaces of the solid are the curved surface of the cylinder, the circular base of the cylinder, and the curved surface of the hemisphere. $\text{TSA}=2\pi rh+\pi r^2+2\pi r^2=2\pi rh+3\pi r^2$ $=\pi r(2h+3r)=\frac{22}{7}\times21\times(140+63)=66\times203=13398\text{ cm}^2$ The total surface area is $13{,}398\text{ cm}^2$ .

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CDC Model Question 2080, Grade 10 Compulsory Mathematics · 1+2+1 marks

A metallic solid consists of a cone and a hemisphere. The height of the cone is 10 cm and the diameter of the base is 6 cm. [π = 3.14]
(a) How do you find the slanted height of the cone when vertical height and radius of the base is given? Write it. [1K]
(b) Find the volume of the solid. [2U]
(c) If the solid object was melted and turned into a cylindrical object of radius 3 cm, what is the height of the cylinder? Calculate it. [1HA]
Figure: A solid made of a cone placed on top of a hemisphere sharing the same circular base; the cone's vertical height is marked 10 cm and the diameter of the common base is marked 6 cm.

Similar questions: Cone and hemisphere melted to cylinder (2081 Chaitra · Sudurpashchim)

Answer

(a) Slant height $l=\sqrt{h^2+r^2}$ (square root of the sum of squares of vertical height and radius).

(b) $r=3$ cm, $h=10$ cm, $\pi=3.14$ .

$V=\frac13\pi r^2h+\frac23\pi r^3=\frac13\times3.14\times9\times10+\frac23\times3.14\times27$

$=94.2+56.52=$ 150.72 cm $^3$ .

(c) Volume of cylinder $=\pi r^2H=3.14\times9\times H=150.72$

$H=\dfrac{150.72}{28.26}\approx$ 5.33 cm.

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CDC Model Question 2080, Grade 10 Compulsory Mathematics · 1+1+2 marks

The height of the square based pyramid is 12 cm and the length of base is 10 cm.
(a) How many triangular faces are there in square base pyramid? [1K]
(b) Find the volume of the pyramid. [1U]
(c) Find the cost of coloring in the triangular parts of the pyramid when the cost of per square cm is 50 paisa. [2A]

Answer

(a) 4 triangular faces.

(b) $V=\frac13a^2h=\frac13\times100\times12=400$ cm $^3$ .

(c) $l=\sqrt{12^2+5^2}=13$ cm.

Area of triangular parts $=2al=2\times10\times13=260$ cm $^2$ .

Cost $=260\times0.50=$ Rs 130.

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CDC Model Question 2080, Grade 10 Compulsory Mathematics · 3+2 marks

The inner length, breadth and height of a rectangular tank made for drinking water by two families are 3 m, 1.5 m and 1.6 m respectively.
(a) What is the total cost of coloring the inner four walls of the tank at the rate of Rs 100 per 3 square meters? Calculate it. [3A]
(b) If 2 families pay equal amounts for consuming water from a full tank, how much will one family have to pay at the rate of Rs 0.50 per liter? [2HA]

Answer

(a) $l=3$ m, $b=1.5$ m, $h=1.6$ m.

Inner four walls $=2(l+b)h=2(3+1.5)\times1.6=14.4$ m $^2$ .

Rate $=\dfrac{100}{3}$ per m $^2$ . Cost $=14.4\times\dfrac{100}{3}=$ Rs 480.

(b) Volume $=3\times1.5\times1.6=7.2$ m $^3=7200$ litres.

Cost of full tank $=7200\times0.50=$ Rs 3,600.

Each family pays $3600\div2=$ Rs 1,800.

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Past questions and answers

1.Find the area of △MNO\triangle MNO△MNO in the given figure. Figure: Triangle MNOMNOMNO with MN=24MN = 24MN=24 cm, MO=7MO = 7MO=7 cm and NO=25NO = 25NO=25 cm.

2.Find the area of △XYZ\triangle XYZ△XYZ in the given figure. Figure: Triangle XYZXYZXYZ with XY=4XY = 4XY=4 cm, XZ=3XZ = 3XZ=3 cm and YZ=5YZ = 5YZ=5 cm.

3.Find the area of △XYZ\triangle XYZ△XYZ in the given figure. Figure: Triangle XYZXYZXYZ with XY=12XY = 12XY=12 cm, XZ=14XZ = 14XZ=14 cm and YZ=20YZ = 20YZ=20 cm.

4.If the volume of a hemisphere is 9216π9216\pi9216π cubic cm, then find its radius.

5.If the volume of a hemisphere is 144π144\pi144π cubic cm, then find its diameter.

6.The given figure is a combined solid made up of a cylinder and a cone. The diameter of the base of the object is 161616 cm, the length of the cylindrical part is 757575 cm and the total length of the solid object is 909090 cm. Find the total surface area of the combined solid.

7.The given figure is of a combined solid made up of a cylinder and a cone. The radius of the base of the object is 333 cm, the length of the cylindrical part is 202020 cm and the total length of the solid object is 242424 cm. Find the curved surface area of the solid object.

8.In the given figure, find the area of △ABC\triangle ABC△ABC. Figure: Triangle ABCABCABC with AB=5AB = 5AB=5 cm, AC=12AC = 12AC=12 cm and BC=13BC = 13BC=13 cm.

9.If the volume of a hemisphere is 19404 cubic cm, then find its radius.

10.The given figure is a combined solid made up of a cylinder and a cone. The diameter of the base of the object is 10 cm, the length of the cylindrical part is 128 cm and the total length of the solid object is 140 cm. Find the total surface area of the solid object.

11.If the volume of a hemisphere is 486π486\pi486π cubic cm, then find its radius.

12.The given figure is a combined solid made up of a cylinder and a cone. The diameter of the base of the solid object is 6 cm, the length of the cylindrical part is 116 cm and the total length of the solid object is 120 cm. Find the total surface area of the solid object.

13.If the volume of a hemisphere is 3888π3888\pi3888π cubic cm, then find its radius.

14.The given figure is a combined solid made up of a cylinder and a cone. The radius of the base of the solid object is 7 cm, the length of the cylindrical part is 35 cm and the total length of the solid object is 59 cm. Find the total surface area of the solid object.

15.Find the area of △ABC\triangle ABC△ABC in the given figure. Figure: Triangle ABCABCABC with AB=12AB = 12AB=12 cm, AC=8AC = 8AC=8 cm and BC=13BC = 13BC=13 cm.

16.If the volume of a hemisphere is 3888π3888\pi3888π cubic cm, then find its diameter.

17.The given figure is a combined solid made up of a cylinder and a cone. The radius of the base of the solid object is 6 cm, the length of the cylindrical part is 20 cm and the total length of the solid object is 28 cm. Find the total surface area of the solid object.

18.Find the area of △EFG\triangle EFG△EFG in the given figure. Figure: Triangle EFGEFGEFG with EF=15EF = 15EF=15 cm, EG=8EG = 8EG=8 cm and FG=17FG = 17FG=17 cm.

19.If the volume of a hemisphere is 144π144\pi144π cubic cm, then find its radius.

20.The given figure is a combined solid made up of a cylinder and a cone. The diameter of the base of the object is 14 cm, the length of the cylindrical part is 45 cm and the length of the conical part is 24 cm. Find the total surface area of that solid object.

21.Find the area of △MNO\triangle MNO△MNO in the given figure. Figure: Triangle MNOMNOMNO with MN=6MN = 6MN=6 cm, MO=7MO = 7MO=7 cm and NO=5NO = 5NO=5 cm.

22.If the volume of a hemisphere is 18π18\pi18π cubic cm, then find its radius.

23.The given figure is a combined solid object made up of a cylinder and a cone. The diameter of the base of the solid object is 32 cm, the length of the cylindrical part is 150 cm and the total length of the solid object is 180 cm. Find the total surface area of the solid object.

24.The area of the triangular surfaces of a square based pyramid is 320 cm2^22 and the length of the side of the base is 16 cm. (a) Write the formula to find the total surface area of a square based pyramid. (1) (b) Find the slant height of the pyramid. (1) (c) Find the volume of the pyramid. (2)

25.Write the area of an isosceles triangle having equal sides xxx cm and third side yyy cm.

40.A square based pyramid has slant height 10 cm and the length of a side of base is 12 cm. (a) How many triangular surfaces are there in a square based pyramid? Write it. (1) (b) What is the total surface area of the pyramid? Find it. (1) (c) What is the volume of the pyramid? Find it. (2)

41.The base area of a square based pyramid is 64 sq. cm and its height is 3 cm. (a) Write the formula to find the area of the triangular surfaces of a square based pyramid. (1) (b) Find the length of the side of the base of the pyramid. (1) (c) Find the total surface area of the pyramid. (2)

42.The length of the base side and slant height of a square based pyramid are 48 cm and 25 cm respectively. (a) How many surfaces are there in square based pyramid? Write it. (1) (b) Find the volume of the pyramid. (3) (c) Find the area of a triangular surface of the pyramid. (1)

43.The vertical height of a square based pyramid is 24 cm and length of the base side is 14 cm. (a) Write the formula to find the area of a triangular surface of the pyramid. (1) (b) Find the volume of the pyramid. (2) (c) Find the total surface area of the pyramid. (2)

44.The vertical height of the square based pyramid is 24 cm and the length of one side of base is 20 cm. (a) Write the formula to find the volume of the pyramid. (1) (b) Find the total surface area of the pyramid. (2)

45.The length of side of the base of a square based pyramid is 24 cm and height is 9 cm. (a) Write the formula for finding the volume of a square based pyramid. (1) (b) Find the volume of the given pyramid. (1) (c) Find the area of triangular surfaces of the pyramid. (2)

47.The area of triangular surfaces of a square based pyramid is 80 square cm and slant height is 5 cm. (a) How many surfaces are there in a square based pyramid? Write it. (1) (b) Find the total surface area of the pyramid. (3) (c) Find the volume of the pyramid. (2)

49.What is the area of an equilateral triangle having each side xxx cm? Write it.

50.The lengths of the sides of a triangle are xxx cm, yyy cm and zzz cm. What is the semi-perimeter of the triangle?

63.In the given figure, AW=20AW = 20AW=20 cm, WW′=10WW' = 10WW′=10 cm, WX=16WX = 16WX=16 cm and the area of the rectangular surfaces of the prism is 480 square cm. Find the length of AXAXAX. Figure: A triangular prism whose base is the triangle AWXAWXAWX and whose length is WW′WW'WW′.

76.The perimeter of a square base pyramid is 24 meter and its height is 4 meter. (a) Write the formula to find the volume of square based pyramid. (1) (b) What is the volume of pyramid? Find it. (1) (c) Find the total surface area of the pyramid. (2)

113.The radius of the base and vertical height of a conical object are 7 cm and 24 cm respectively. (a) Find the curved surface area of the cone. (2) (b) What can be done to increase the volume of the cone? Give a suitable reason. (1)

121.The height of a triangular prism is 404040 cm. If the sides of the base of the prism are 888 cm, 666 cm and 101010 cm, find the area of the rectangular surfaces of the prism.

126.Find the area of an equilateral triangle having perimeter 60 cm.

127.If the radius of a football is 7 cm, find its surface area.

128.If the curved surface area of a cone is 550 sq. cm and the slant height is 25 cm, find the height of the cone.

130.A person plans to make a combined solid of a cone and a hemisphere for decoration material from gold. If the diameter of the hemisphere and the base of the cone is 10 cm and the slant height of the cone is 13 cm, find the total volume of gold required to make the combined solid.

131.In the given square based pyramidic house, how much does it cost for metal plating the roof at the rate of Rs. 12,000 per square meter? Find it. Figure: A square based pyramid-shaped house whose base side is 16 m and whose vertical height is 6 m.

138.The height of a triangular prism whose base is a right-angled triangle is 30 cm. If the sides of the triangle containing the right angle are 4 cm and 3 cm, find the area of the rectangular surfaces of the prism.

139.Find the area of a triangle having lengths of sides 6 cm, 7 cm and 11 cm.

140.If the volume of a sphere is 36π36\pi36π cubic cm, find its radius.

142.The area of the base and the vertical height of a square based solid pyramid are 144 sq. cm and 8 cm respectively. What will be the total surface area of the pyramid? Find it.

144.The length of a side of the square base of a pyramid is ppp cm and the slant height is qqq cm. What is the total surface area of the pyramid? Write it.

145.A cylindrical tank contains 172.48 litres of water. If its height is 0.7 m, then find the radius of the base.

Find the area of $\triangle MNO$ in the given figure.
Figure: Triangle $MNO$ with $MN = 24$ cm, $MO = 7$ cm and $NO = 25$ cm.

Find the area of $\triangle XYZ$ in the given figure.
Figure: Triangle $XYZ$ with $XY = 4$ cm, $XZ = 3$ cm and $YZ = 5$ cm.

Find the area of $\triangle XYZ$ in the given figure.
Figure: Triangle $XYZ$ with $XY = 12$ cm, $XZ = 14$ cm and $YZ = 20$ cm.

If the volume of a hemisphere is $9216\pi$ cubic cm, then find its radius.

If the volume of a hemisphere is $144\pi$ cubic cm, then find its diameter.

The given figure is a combined solid made up of a cylinder and a cone. The diameter of the base of the object is $16$ cm, the length of the cylindrical part is $75$ cm and the total length of the solid object is $90$ cm. Find the total surface area of the combined solid.

The given figure is of a combined solid made up of a cylinder and a cone. The radius of the base of the object is $3$ cm, the length of the cylindrical part is $20$ cm and the total length of the solid object is $24$ cm. Find the curved surface area of the solid object.

In the given figure, find the area of $\triangle ABC$ .
Figure: Triangle $ABC$ with $AB = 5$ cm, $AC = 12$ cm and $BC = 13$ cm.

If the volume of a hemisphere is 19404 cubic cm, then find its radius.

The given figure is a combined solid made up of a cylinder and a cone. The diameter of the base of the object is 10 cm, the length of the cylindrical part is 128 cm and the total length of the solid object is 140 cm. Find the total surface area of the solid object.

If the volume of a hemisphere is $486\pi$ cubic cm, then find its radius.

The given figure is a combined solid made up of a cylinder and a cone. The diameter of the base of the solid object is 6 cm, the length of the cylindrical part is 116 cm and the total length of the solid object is 120 cm. Find the total surface area of the solid object.

If the volume of a hemisphere is $3888\pi$ cubic cm, then find its radius.

The given figure is a combined solid made up of a cylinder and a cone. The radius of the base of the solid object is 7 cm, the length of the cylindrical part is 35 cm and the total length of the solid object is 59 cm. Find the total surface area of the solid object.

Find the area of $\triangle ABC$ in the given figure.
Figure: Triangle $ABC$ with $AB = 12$ cm, $AC = 8$ cm and $BC = 13$ cm.

If the volume of a hemisphere is $3888\pi$ cubic cm, then find its diameter.

The given figure is a combined solid made up of a cylinder and a cone. The radius of the base of the solid object is 6 cm, the length of the cylindrical part is 20 cm and the total length of the solid object is 28 cm. Find the total surface area of the solid object.

Find the area of $\triangle EFG$ in the given figure.
Figure: Triangle $EFG$ with $EF = 15$ cm, $EG = 8$ cm and $FG = 17$ cm.

If the volume of a hemisphere is $144\pi$ cubic cm, then find its radius.

The given figure is a combined solid made up of a cylinder and a cone. The diameter of the base of the object is 14 cm, the length of the cylindrical part is 45 cm and the length of the conical part is 24 cm. Find the total surface area of that solid object.

Find the area of $\triangle MNO$ in the given figure.
Figure: Triangle $MNO$ with $MN = 6$ cm, $MO = 7$ cm and $NO = 5$ cm.

If the volume of a hemisphere is $18\pi$ cubic cm, then find its radius.

The given figure is a combined solid object made up of a cylinder and a cone. The diameter of the base of the solid object is 32 cm, the length of the cylindrical part is 150 cm and the total length of the solid object is 180 cm. Find the total surface area of the solid object.

The area of the triangular surfaces of a square based pyramid is 320 cm $^2$ and the length of the side of the base is 16 cm.
(a) Write the formula to find the total surface area of a square based pyramid. (1)
(b) Find the slant height of the pyramid. (1)
(c) Find the volume of the pyramid. (2)

Write the area of an isosceles triangle having equal sides $x$ cm and third side $y$ cm.

A square based pyramid has slant height 10 cm and the length of a side of base is 12 cm.
(a) How many triangular surfaces are there in a square based pyramid? Write it. (1)
(b) What is the total surface area of the pyramid? Find it. (1)
(c) What is the volume of the pyramid? Find it. (2)

The base area of a square based pyramid is 64 sq. cm and its height is 3 cm.
(a) Write the formula to find the area of the triangular surfaces of a square based pyramid. (1)
(b) Find the length of the side of the base of the pyramid. (1)
(c) Find the total surface area of the pyramid. (2)

The length of the base side and slant height of a square based pyramid are 48 cm and 25 cm respectively.
(a) How many surfaces are there in square based pyramid? Write it. (1)
(b) Find the volume of the pyramid. (3)
(c) Find the area of a triangular surface of the pyramid. (1)

The vertical height of a square based pyramid is 24 cm and length of the base side is 14 cm.
(a) Write the formula to find the area of a triangular surface of the pyramid. (1)
(b) Find the volume of the pyramid. (2)
(c) Find the total surface area of the pyramid. (2)

The vertical height of the square based pyramid is 24 cm and the length of one side of base is 20 cm.
(a) Write the formula to find the volume of the pyramid. (1)
(b) Find the total surface area of the pyramid. (2)

The length of side of the base of a square based pyramid is 24 cm and height is 9 cm.
(a) Write the formula for finding the volume of a square based pyramid. (1)
(b) Find the volume of the given pyramid. (1)
(c) Find the area of triangular surfaces of the pyramid. (2)

The area of triangular surfaces of a square based pyramid is 80 square cm and slant height is 5 cm.
(a) How many surfaces are there in a square based pyramid? Write it. (1)
(b) Find the total surface area of the pyramid. (3)
(c) Find the volume of the pyramid. (2)

What is the area of an equilateral triangle having each side $x$ cm? Write it.

The lengths of the sides of a triangle are $x$ cm, $y$ cm and $z$ cm. What is the semi-perimeter of the triangle?

In the given figure, $AW = 20$ cm, $WW' = 10$ cm, $WX = 16$ cm and the area of the rectangular surfaces of the prism is 480 square cm. Find the length of $AX$ .
Figure: A triangular prism whose base is the triangle $AWX$ and whose length is $WW'$ .

The perimeter of a square base pyramid is 24 meter and its height is 4 meter.
(a) Write the formula to find the volume of square based pyramid. (1)
(b) What is the volume of pyramid? Find it. (1)
(c) Find the total surface area of the pyramid. (2)

The radius of the base and vertical height of a conical object are 7 cm and 24 cm respectively.
(a) Find the curved surface area of the cone. (2)
(b) What can be done to increase the volume of the cone? Give a suitable reason. (1)

The height of a triangular prism is $40$ cm. If the sides of the base of the prism are $8$ cm, $6$ cm and $10$ cm, find the area of the rectangular surfaces of the prism.

Find the area of an equilateral triangle having perimeter 60 cm.

If the radius of a football is 7 cm, find its surface area.

If the curved surface area of a cone is 550 sq. cm and the slant height is 25 cm, find the height of the cone.

A person plans to make a combined solid of a cone and a hemisphere for decoration material from gold. If the diameter of the hemisphere and the base of the cone is 10 cm and the slant height of the cone is 13 cm, find the total volume of gold required to make the combined solid.

In the given square based pyramidic house, how much does it cost for metal plating the roof at the rate of Rs. 12,000 per square meter? Find it.
Figure: A square based pyramid-shaped house whose base side is 16 m and whose vertical height is 6 m.

The height of a triangular prism whose base is a right-angled triangle is 30 cm. If the sides of the triangle containing the right angle are 4 cm and 3 cm, find the area of the rectangular surfaces of the prism.

Find the area of a triangle having lengths of sides 6 cm, 7 cm and 11 cm.

If the volume of a sphere is $36\pi$ cubic cm, find its radius.

The area of the base and the vertical height of a square based solid pyramid are 144 sq. cm and 8 cm respectively. What will be the total surface area of the pyramid? Find it.

The length of a side of the square base of a pyramid is $p$ cm and the slant height is $q$ cm. What is the total surface area of the pyramid? Write it.

A cylindrical tank contains 172.48 litres of water. If its height is 0.7 m, then find the radius of the base.