Geometry

a) Teaching the area of a triangle with base $b$ and height $h$: $\frac{1}{2}bh$

Specific objective: Students will derive and use the formula for the area of a triangle.

Teaching materials: Rectangular paper, grid paper, scissors, chalkboard, chart, geoboard.

Previous knowledge: Area of a rectangle $= \text{length} \times \text{breadth}$.

Teaching steps:

1. Review: Ask students to find the area of a rectangle of length $12$ cm and breadth $8$ cm.
   Area $= 12 \times 8 = 96$ cm$^2$.

2. Activity: Give each student a rectangular paper of $12$ cm $\times$ $8$ cm. Ask them to draw one diagonal and cut along it. They get two triangles. By placing one triangle over the other, they see that the two triangles are congruent.
   So each triangle has area $= \frac{96}{2} = 48$ cm$^2$.

3. Derivation: In the rectangle, base $b = 12$ cm and height $h = 8$ cm.
   Area of each triangle $= \frac{1}{2} \times$ area of rectangle
   $= \frac{1}{2} \times b \times h$.
   Hence,
   [ \text{Area of triangle} = \frac{1}{2}bh ]

4. Clarify height: Show that height is the perpendicular distance from the base to the opposite vertex. Draw acute, right and obtuse triangles on the board and mark base and height.

5. Grid paper verification: Draw a triangle on grid paper. Count the full and half squares to estimate area. Compare with $\frac{1}{2}bh$.

6. Application: Solve examples:

   • $b = 10$ cm, $h = 6$ cm, area $= \frac{1}{2} \times 10 \times 6 = 30$ cm$^2$.
   • If area $= 24$ cm$^2$ and $b = 8$ cm, find $h$:
     $24 = \frac{1}{2} \times 8 \times h \Rightarrow h = 6$ cm.

7. Evaluation: Ask oral and written questions:

   • Why do we multiply by $\frac{1}{2}$?
   • What happens if height is not perpendicular?
   • Find area when $b = 7$ cm, $h = 4$ cm.

Homework: Draw two triangles of different shapes, measure base and height, and find their areas.

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b) Solution of the problem

Given:
$\frac{2}{3}$ bag of fertilizer sows $\frac{4}{5}$ part of the field.

To find:
How many parts of the field can be sown by $1$ bag?

Solution by unitary method:

[ \frac{2}{3} \text{ bag} \rightarrow \frac{4}{5} \text{ part} ]

[ 1 \text{ bag} \rightarrow \frac{4}{5} \div \frac{2}{3} ]

[ = \frac{4}{5} \times \frac{3}{2} ]

[ = \frac{12}{10} = \frac{6}{5} = 1\frac{1}{5} ]

Answer: One bag of fertilizer can sow $\frac{6}{5}$ parts, i.e. $1\frac{1}{5}$ parts, of the field.
So it can sow the whole field and $\frac{1}{5}$ part more.

Check:
[ \frac{2}{3} \times \frac{6}{5} = \frac{12}{15} = \frac{4}{5} ] which matches the given condition.

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How to solve this problem while teaching in the classroom

Objective: Students will solve unitary method problems involving fractions.

Materials: Fraction chart, bar model, chalkboard, fraction strips.

Steps:

1. Present the problem in a familiar context:
   “$\frac{2}{3}$ bag of fertilizer sows $\frac{4}{5}$ part of a farmer’s field. How much of the field will $1$ bag sow?”

2. Draw a bar model:
   Draw a rectangle and divide it into $5$ equal parts. Shade $4$ parts.
   This shaded part represents $\frac{4}{5}$ of the field.

3. Link with the fertilizer:
   Tell students that these $4$ shaded parts are sown by $\frac{2}{3}$ bag.
   Since $\frac{2}{3}$ means $2$ equal parts out of $3$,
   [ \frac{2}{3} \text{ bag} \rightarrow 4 \text{ parts} ] So,
   [ \frac{1}{3} \text{ bag} \rightarrow 2 \text{ parts} ]

4. Find for one full bag:
   One bag $= 3$ times $\frac{1}{3}$ bag.
   [ 1 \text{ bag} \rightarrow 3 \times 2 = 6 \text{ parts} ] The field has $5$ parts. Therefore,
   [ 6 \text{ parts} = \frac{6}{5} \text{ of the field} ]

5. Connect with symbolic method:
   [ \frac{4}{5} \div \frac{2}{3} = \frac{4}{5} \times \frac{3}{2} = \frac{6}{5} ] Explain that dividing by a fraction means multiplying by its reciprocal.

6. Practice: Give similar problems:

   • $\frac{3}{4}$ bag sows $\frac{2}{3}$ part of a field. How much will $1$ bag sow?
   • $\frac{1}{2}$ bag sows $\frac{3}{7}$ part. Find the part sown by $1$ bag.

7. Evaluation: Ask students to explain each step in their own words. Check by multiplying back: [ \frac{2}{3} \times \frac{6}{5} = \frac{4}{5} ]

8. Homework: Solve two similar problems using both the diagram method and the unitary method.

1. Congruency and Similarity of Triangles — Major Differences (4 marks)

Both relations compare two triangles by matching corresponding vertices, sides and angles, but they differ in what they demand and in what they preserve.

Key idea to stress in class: congruency is a special case of similarity in which the scale factor is exactly 1. So "similar" is the wider (more general) relation and "congruent" is the narrower one.

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2. Teaching Aid: Constructing Triangles that are Similar but Not Necessarily Equal (2 marks)

Purpose: to let pupils see that same shape does not mean same size.

Materials: ruler, protractor, pencil, two sheets of paper (or graph paper).

Procedure:

1. On the first sheet draw $\triangle ABC$ with $AB = 4$ cm, $BC = 5$ cm and $CA = 6$ cm. Measure and record $\angle A,\ \angle B,\ \angle C$.
2. On the second sheet draw $\triangle A'B'C'$ taking each side $1.5$ times as long: $A'B' = 6$ cm, $B'C' = 7.5$ cm, $C'A' = 9$ cm.
3. Measure the angles of $\triangle A'B'C'$ with the protractor. Pupils find $\angle A' = \angle A,\ \angle B' = \angle B,\ \angle C' = \angle C$.
4. Compute the ratios: $\frac{A'B'}{AB}=\frac{6}{4}=1.5,\qquad \frac{B'C'}{BC}=\frac{7.5}{5}=1.5,\qquad \frac{C'A'}{CA}=\frac{9}{6}=1.5$
5. Since the corresponding angles are equal and the corresponding sides are in a constant ratio $1.5:1$, the triangles are similar ($\triangle ABC \sim \triangle A'B'C'$ by AAA/SSS similarity).
6. Now try to place one triangle exactly over the other — they do not coincide, because the sides are unequal. Hence they are not congruent.

Conclusion drawn by pupils: equal angles with proportional (unequal) sides give similarity, not congruency. The scale factor here is $1.5$, so the area of the second triangle is $(1.5)^2 = 2.25$ times the area of the first.

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3. Theorem: The Sum of the Three Exterior Angles of a Triangle is $360^\circ$ (4 marks)

Statement

If the three sides of a triangle are produced in order, the sum of the three exterior angles so formed is $360^\circ$.

Teaching procedure (activity first, proof next)

Step 1 — Recall prior knowledge. Interior angle sum of a triangle $= 180^\circ$; angles on a straight line (linear pair) $= 180^\circ$.

Step 2 — Paper-cutting activity. Draw any triangle $ABC$ on a sheet. Produce $BC$ to $X$, $CA$ to $Y$ and $AB$ to $Z$, so that the exterior angles are $\angle ACX$, $\angle BAY$ and $\angle CBZ$. Colour the three exterior angles in three different colours, cut them out and paste them with their vertices at one common point $O$, side by side. Pupils observe that the three pieces exactly fill up a complete turn about $O$, i.e. a full circle. Hence the sum is $360^\circ$.

Step 3 — Verification by measurement. Ask pupils to measure the three exterior angles of their own triangle with a protractor and add. Every group gets a total very close to $360^\circ$ (small error due to measurement). This establishes the result experimentally for many different triangles.

Step 4 — Formal reasoning (deductive proof).

Given: $\triangle ABC$ whose sides $BC$, $CA$ and $AB$ are produced to $X$, $Y$ and $Z$ respectively. Let the interior angles be $\angle A = x$, $\angle B = y$, $\angle C = z$, and the exterior angles be $\angle ACX = a$, $\angle BAY = b$, $\angle CBZ = c$.

To prove: $a + b + c = 360^\circ$.

Construction: Produce the three sides as shown (no extra line needed).

Proof:

Conclusion: The sum of the three exterior angles of a triangle is $360^\circ$. $\blacksquare$

Note for the teacher

• The result is independent of the shape and size of the triangle — it holds for acute, right and obtuse triangles alike.
• The same reasoning gives the general result for any polygon: sum of exterior angles $= 360^\circ$ always, while the sum of interior angles $=(n-2)\times 180^\circ$.
• A common pupil error is to add only two exterior angles or to confuse the exterior angle with the interior angle at the same vertex; the linear-pair relation $x + a = 180^\circ$ must be drilled before the proof.

a) Five properties of a parallelogram:

1. Opposite sides are equal and parallel.
2. Opposite angles are equal.
3. Adjacent (consecutive) angles are supplementary, i.e. their sum is (180^\circ).
4. The diagonals bisect each other.
5. Each diagonal divides the parallelogram into two congruent triangles.

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b) Theorem: The diagonals of a parallelogram bisect each other.

Given:
Let (ABCD) be a parallelogram. Its diagonals (AC) and (BD) intersect at (O).

To prove:
(AO = OC) and (BO = OD), i.e. the diagonals bisect each other at (O).

Proof:

In (\triangle AOB) and (\triangle COD),

1. (\angle OAB = \angle OCD)
   [ [\because AB \parallel DC \text{ and } AC \text{ is a transversal, so alternate angles are equal}] ]

2. (\angle OBA = \angle ODC)
   [ [\because AB \parallel DC \text{ and } BD \text{ is a transversal, so alternate angles are equal}] ]

3. (AB = DC)
   [ [\because \text{opposite sides of a parallelogram are equal}] ]

Therefore, [ \triangle AOB \cong \triangle COD ] by the ASA congruence criterion.

Hence, by corresponding parts of congruent triangles,

[ AO = OC ] and [ BO = OD ]

So, the diagonals (AC) and (BD) bisect each other at (O).

Proved.

a) Teaching students to construct a parallelogram when two sides and the included angle are given

Let the two adjacent sides be $AB=a$, $AD=b$, and the angle between them be $\angle DAB=\theta$.

Construction steps to demonstrate on the blackboard:

1. Draw a line segment $AB=a$ using a ruler.
2. At $A$, construct $\angle BAX=\theta$ with the help of a protractor or compass.
3. On the ray $AX$, mark $AD=b$.
4. With $B$ as centre and radius $b$, draw an arc.
5. With $D$ as centre and radius $a$, draw another arc cutting the first arc at $C$.
6. Join $BC$ and $DC$.

Then $ABCD$ is the required parallelogram.

Justification:
$AB=DC=a$ and $AD=BC=b$ by construction. Also, opposite sides are parallel, so $ABCD$ is a parallelogram with the given sides and included angle.

Teaching method:

• First review the properties of a parallelogram and the use of ruler, compass and protractor.
• Demonstrate the construction step by step, and ask students to follow in their copies.
• Let students work in pairs or groups using different measurements.
• Ask them to verify by measuring opposite sides and opposite angles.
• Finally, give a short class exercise and provide remedial help where needed.

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b) Pythagoras theorem and its use in finding the distance between two points

Statement of Pythagoras theorem:
In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

If the legs are $a$ and $b$, and the hypotenuse is $c$, then

$c^2=a^2+b^2$

Converse:
If in a triangle the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Use in finding the distance between two points:
Let the two points be

$P(x_1,y_1)\quad \text{and}\quad Q(x_2,y_2).$

Draw a horizontal line through $P$ and a vertical line through $Q$, meeting at $R(x_2,y_1)$. Then $\triangle PRQ$ is right-angled at $R$.

Now,

$PR=|x_2-x_1|,\qquad QR=|y_2-y_1|.$

By Pythagoras theorem,

$PQ^2=PR^2+QR^2$

$PQ^2=(x_2-x_1)^2+(y_2-y_1)^2$

Therefore,

$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

This is the distance formula. It is used to find the distance between any two points in the coordinate plane, to find the lengths of sides of triangles and quadrilaterals, to prove whether a triangle is right-angled, and to solve many real-life problems involving distance.

a) Area of square $=484$ cm², so its side $=\sqrt{484}=22$ cm.

The side of the square equals the diameter of the circle, so $d=22$ cm and $r=11$ cm.

$\text{Area of circle}=\pi r^2=\frac{22}{7}\times11\times11=\frac{2662}{7}\approx380.29\ \text{cm}^2$

b) The wording of this part is incomplete, so the usual interpretation is taken: the largest circle has diameter $48$ cm, and each following circle has half the diameter of the one before it, continuing without end.

For the cylinder, the base circumference is $2\pi r=44$ cm, so $r=\dfrac{44\times7}{2\times22}=7$ cm (base diameter $14$ cm).

Largest circle: $r_1=24$ cm, so $A_1=\pi(24)^2=576\pi$ cm².

Halving the diameter halves the radius, so each area is $\left(\tfrac12\right)^2=\tfrac14$ of the previous one. The areas form an infinite geometric series $576\pi+144\pi+36\pi+\cdots,\qquad a=576\pi,\ r=\tfrac14.$

Since $|r|<1$, the sum is $S_\infty=\frac{a}{1-r}=\frac{576\pi}{1-\frac14}=768\pi=768\times\frac{22}{7}=\frac{16896}{7}\approx2413.7\ \text{cm}^2.$

1. Show that (ax^2+2hxy+by^2=0) represents a pair of straight lines through the origin. [2]

The equation is homogeneous of degree 2. Treat it as a quadratic in (y):

[ by^2+2hxy+ax^2=0 ]

If (b\neq 0),

[ y=\frac{-2hx\pm \sqrt{4h^2x^2-4abx^2}}{2b} ]

[ y=\frac{-h\pm \sqrt{h^2-ab}}{b}x ]

Let

[ m_1=\frac{-h+\sqrt{h^2-ab}}{b},\qquad m_2=\frac{-h-\sqrt{h^2-ab}}{b} ]

Then the equation gives

[ y=m_1x \quad \text{and} \quad y=m_2x ]

These are two straight lines passing through the origin. Hence

[ ax^2+2hxy+by^2=0 ]

always represents a pair of straight lines through the origin.

If (b=0), the equation becomes

[ x(ax+2hy)=0 ]

which gives (x=0) and (ax+2hy=0), again two straight lines through the origin. Similarly, the case (a=0) also gives a pair of lines through the origin.

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2. Factorise the given pair of lines. [2]

Given:

[ x^2-5xy+4y^2=0 ]

[ x^2-4xy-xy+4y^2=0 ]

[ x(x-4y)-y(x-4y)=0 ]

[ (x-y)(x-4y)=0 ]

So the given pair of straight lines is

[ x-y=0 \quad \text{and} \quad x-4y=0 ]

Their slopes are

[ m_1=1,\qquad m_2=\frac14 ]

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3. Find the equation of straight lines through the origin and at right angles to the given lines. [2]

The lines perpendicular to the given lines have slopes

[ -\frac1{m_1}=-1,\qquad -\frac1{m_2}=-4 ]

Hence the required lines through the origin are

[ y=-x \quad \text{and} \quad y=-4x ]

or,

[ x+y=0 \quad \text{and} \quad 4x+y=0 ]

Therefore, the combined equation of the required pair of straight lines is

[ (x+y)(4x+y)=0 ]

[ \boxed{4x^2+5xy+y^2=0} ]

So, the required equation is

[ \boxed{4x^2+5xy+y^2=0} ]

a) Euclid’s fifth postulate

If a straight line (EF) falls on two straight lines (AB) and (CD) such that the sum of the interior angles on the same side of (EF) is less than two right angles, then (AB) and (CD), if produced indefinitely, meet on that same side on which the sum of the angles is less than two right angles.

Figure:

        E
        |
A-------1-------B
        |
C-------2-------D
        |
        F

Here (\angle 1) and (\angle 2) are the interior angles on the same side of the transversal (EF).
If (\angle 1+\angle 2<180^\circ), then (AB) and (CD) meet on that side when produced.

Any three properties of a parallelogram:

1. Opposite sides are equal.
2. Opposite angles are equal.
3. Diagonals bisect each other.

(Also: consecutive angles are supplementary; each diagonal divides the parallelogram into two congruent triangles.)

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b) Theorem: The angle at the centre of a circle is double the angle at the circumference standing on the same arc.

Given:
A circle with centre (O). Arc (AB) subtends (\angle AOB) at the centre and (\angle ACB) at the circumference at point (C) on the remaining part of the circle.

To prove:
[ \angle AOB = 2\angle ACB ]

Construction:
Join (C) to (O) and produce it to (D).

Proof:

1. In (\triangle OAC),
   [ OA=OC \quad \text{(radii of the same circle)} ] So, [ \angle OAC=\angle OCA ] Let (\angle OCA=x).

2. The exterior angle at (O) of (\triangle OAC) is (\angle AOD).
   [ \angle AOD=\angle OAC+\angle OCA=x+x=2x ]

3. Similarly, in (\triangle OBC),
   [ OB=OC \quad \text{(radii)} ] So, [ \angle OBC=\angle OCB ] Let (\angle OCB=y).

4. The exterior angle at (O) of (\triangle OBC) is (\angle BOD).
   [ \angle BOD=\angle OBC+\angle OCB=y+y=2y ]

5. Since (OD) lies inside (\angle AOB),
   [ \angle AOB=\angle AOD+\angle BOD=2x+2y=2(x+y) ]

6. But, [ x+y=\angle OCA+\angle OCB=\angle ACB ]

Therefore, [ \angle AOB=2\angle ACB ]

Hence, the angle at the centre is double the angle at the circumference standing on the same arc.
Proved.