Geometry

(a) $\angle PAQ = \angle PBQ$ (inscribed angles standing on the same arc PQ are equal).

(b) Central angle $=2\times$ inscribed angle:

$12x + 4 = 2(3x+20)$

$12x+4 = 6x+40$

$6x = 36 \Rightarrow x = 6$

Check: $12(6)+4 = 76^\circ$ and $3(6)+20 = 38^\circ$.

(c) Experimental verification (central angle is twice the inscribed angle on the same arc)

1. Draw a circle of radius at least 3 cm with centre O. Mark two points on it and join them to O to form the central angle POQ, and mark a third point on the major arc and join it to the two points to form the inscribed angle PAQ on the same arc.
2. Measure central angle POQ and inscribed angle PAQ with a protractor.
3. Repeat with a second circle of a different radius.

Conclusion: In each case the central angle is double the inscribed angle standing on the same arc, i.e. central angle POQ $= 2\times$ inscribed angle PAQ.

A rectangle and a triangle on the same base and between the same parallel lines: the area of the rectangle is twice the area of the triangle, i.e. $\text{Area of rectangle}=2\times\text{Area of }\triangle$.

The opposite angles of a cyclic quadrilateral are supplementary, i.e. their sum is $180^\circ$.

A square and a triangle on the same base and between the same parallel lines: the area of the square is twice the area of the triangle, i.e. $\text{Area of square}=2\times\text{Area of }\triangle$.

A tangent is perpendicular to the radius at the point of contact, so $\angle OQP = \angle ORP = 90^\circ$.

In quadrilateral $OQPR$, the angle sum is $360^\circ$:

$\angle QPR = 360^\circ - (90^\circ + 90^\circ + 120^\circ) = 60^\circ$

So $\angle QPR = 60^\circ$.

A tangent is perpendicular to the radius at the point of contact, so $\angle OYX = \angle OZX = 90^\circ$.

In quadrilateral $OYXZ$:

$\angle YOZ = 360^\circ - (90^\circ + 90^\circ + 75^\circ) = 105^\circ$

A tangent is perpendicular to the radius at the point of contact, so $\angle XBA=\angle XCA=90^\circ$.

In the quadrilateral $ABXC$, the angle sum is $360^\circ$:

$\angle BAC+90^\circ+105^\circ+90^\circ=360^\circ \Rightarrow \angle BAC=360^\circ-285^\circ=75^\circ.$

So $\angle BAC=$ $75^\circ$.

Radii $OQ$ and $OR$ are perpendicular to the tangents at the points of contact, so $\angle OQP = \angle ORP = 90^\circ$.

In quadrilateral $OQPR$: $\angle QOR + 90^\circ + 85^\circ + 90^\circ = 360^\circ$

$\angle QOR = 360^\circ - 265^\circ = \textbf{95}^\circ$

A tangent is perpendicular to the radius at the point of contact, so $\angle OAP = \angle OBP = 90^\circ$.

In quadrilateral $OAPB$, the angle sum is $360^\circ$:

Answer: $\angle APB = 80^\circ$

A tangent is perpendicular to the radius at the point of contact, so $\angle OAP = \angle OBP = 90^\circ$.

In quadrilateral $OAPB$, the angle sum is $360^\circ$:

Answer: $\angle AOB = 100^\circ$

Take the points in the order $S, O, M, I$ on the circle, so $SM$ and $OI$ are the diagonals of the cyclic quadrilateral $SOMI$ and meet at $K$.

(i) Compare $\triangle SOK$ and $\triangle IMK$:

• $SO = IM$ (equal arcs have equal chords)
• $\angle OSK = \angle OSM = \angle OIM = \angle MIK$ (angles in the same segment on arc $OM$)
• $\angle SOK = \angle SOI = \angle SMI = \angle IMK$ (angles in the same segment on arc $SI$)

So $\triangle SOK \cong \triangle IMK$ (A.S.A.), and therefore

$\text{Area}(\triangle SOK) = \text{Area}(\triangle IMK).$

(ii) From the congruence, the corresponding sides are equal: $SK = IK$ and $OK = MK$.

$SM = SK + KM = IK + KO = IO$

Hence $SM = IO$. Proved.

The radius $PE$ meets the tangent $RE$ at the point of contact, so $\angle PER = 90^\circ$.

In $\triangle PER$:

$\angle EPR = 180^\circ - (90^\circ + 30^\circ) = 60^\circ$

So $\angle EPR = 60^\circ$.

Since $NA = NE$ and $AM = ME$, the quadrilateral $NAME$ is a kite, symmetrical about the diagonal $NM$.

Part 1: Construct the quadrilateral $NAME$

1. Draw $AM = 5.2$ cm.
2. At $A$ construct $\angle MAX = 75^\circ$ and cut $AN = 4.2$ cm along $AX$. Join $N$ to $M$.
3. With centre $N$ and radius $4.2$ cm, and then with centre $M$ and radius $5.2$ cm, draw arcs cutting each other at $E$ on the opposite side of $NM$ from $A$.
4. Join $NE$ and $ME$. $NAME$ is the required quadrilateral.

Part 2: Construct $\triangle AMT$ equal in area to $NAME$

5. Join the diagonal $AE$.
6. Through $N$ draw a line parallel to $AE$ (copy an angle with compass, or use set squares).
7. Produce $ME$ beyond $E$ so that it meets this parallel line at $T$.
8. Join $AT$. $\triangle AMT$ is the required triangle.

Justification: $\triangle AEN$ and $\triangle AET$ have the same base $AE$ and lie between the same parallels $AE$ and $NT$, so they are equal in area. Therefore

The radius is perpendicular to the tangent at the point of contact, so $\angle OAT = \angle OBT = 90^\circ$.

In quadrilateral $OATB$, the angle sum is $360^\circ$: $\angle ATB = 360^\circ - (90^\circ + 90^\circ + 60^\circ) = 120^\circ$

The question as printed gives $NO = OP = 5.5$ cm, $PM = 4.5$ cm and $\angle MNO = 75^\circ$, but a quadrilateral needs one more side. Taking $MN = 4.5$ cm (so that $PM = MN = 4.5$ cm), construct as follows.

Steps of construction

1. Draw $NO = 5.5$ cm.
2. At $N$, construct $\angle ONX = 75^\circ$ with a protractor. With centre $N$ and radius $4.5$ cm, cut $NX$ at $M$.
3. With centre $O$ and radius $5.5$ cm, and centre $M$ and radius $4.5$ cm, draw two arcs that intersect at $P$. Join $MP$ and $OP$. $MNOP$ is the required quadrilateral.
4. Join the diagonal $MO$.
5. Through $P$, draw a line parallel to $MO$ (copying the angle with compasses). Extend $NO$ to meet this line at $Q$.
6. Join $MQ$. $\triangle MNQ$ is the required triangle.

Proof

$\triangle MOP$ and $\triangle MOQ$ stand on the same base $MO$ and lie between the same parallels $MO$ and $PQ$, so their areas are equal.

$\text{Area}(MNOP) = \text{Area}(\triangle MNO) + \text{Area}(\triangle MOP) = \text{Area}(\triangle MNO) + \text{Area}(\triangle MOQ) = \text{Area}(\triangle MNQ)$

Hence $\triangle MNQ$ is equal in area to the quadrilateral $MNOP$.

Steps of construction

1. Draw $ON = 5.3$ cm.
2. At $O$, construct $\angle NOX = 75^\circ$. With centre $O$ and radius $4.3$ cm, cut $OX$ at $M$.
3. With centre $N$ and radius $5.3$ cm, and centre $M$ and radius $4.3$ cm, draw two arcs that intersect at $K$. Join $MK$ and $NK$. $MONK$ is the required quadrilateral.
4. Join the diagonal $MN$.
5. Through $K$, draw a line parallel to $MN$. Extend $ON$ to meet this line at $L$.
6. Join $ML$. $\triangle MOL$ is the required triangle.

Proof

$\triangle MNK$ and $\triangle MNL$ stand on the same base $MN$ and lie between the same parallels $MN$ and $KL$, so their areas are equal.

$\text{Area}(MONK) = \text{Area}(\triangle MON) + \text{Area}(\triangle MNK) = \text{Area}(\triangle MON) + \text{Area}(\triangle MNL) = \text{Area}(\triangle MOL)$

Hence $\triangle MOL$ is equal in area to the quadrilateral $MONK$.

Steps of construction

Quadrilateral $PQRS$

1. Draw $QR = 5.5$ cm.
2. At $Q$, construct $\angle PQR = 75^\circ$ and cut $QP = 5$ cm.
3. With centre $R$ and radius $5.5$ cm, draw an arc. With centre $P$ and radius $6.5$ cm, draw another arc cutting the first at $S$.
4. Join $PS$ and $RS$. $PQRS$ is the required quadrilateral.

Triangle $QRM$ equal in area to $PQRS$

5. Join the diagonal $QS$.
6. Through $P$, draw a line parallel to $QS$ (copy the angle at $P$ equal to $\angle PQS$ using alternate angles, or use set squares).
7. Produce $RS$ to meet this line at $M$.
8. Join $QM$. $\triangle QRM$ is the required triangle.

Reason

$\triangle QSP$ and $\triangle QSM$ stand on the same base $QS$ and lie between the same parallels $QS$ and $PM$, so they have equal area.

$\text{Area of } PQRS = \triangle QRS + \triangle QSP = \triangle QRS + \triangle QSM = \triangle QRM$

Part 1: construct quadrilateral $MNOP$.

1. Draw $NO=4.2$ cm.
2. At $O$ construct $\angle NOP=75^\circ$ and cut $OP=5.2$ cm.
3. With centre $N$ and radius $4.2$ cm, draw an arc. With centre $P$ and radius $5.2$ cm, draw another arc. Let them meet at $M$ (on the side opposite to $O$ with respect to $NP$).
4. Join $NM$ and $PM$. $MNOP$ is the required quadrilateral.

Part 2: construct $\triangle PQO$ equal in area.

5. Join the diagonal $PN$.
6. Through $M$, draw a line parallel to $PN$ (by copying the angle at $M$ equal to $\angle MNP$ using alternate angles, or with a set square). Extend $ON$ to meet this line at $Q$.
7. Join $PQ$. $\triangle PQO$ is the required triangle.

Justification. $\triangle PNM$ and $\triangle PNQ$ have the same base $PN$ and lie between the same parallels $PN$ and $MQ$, so

$\text{Area}(\triangle PNM)=\text{Area}(\triangle PNQ).$

Adding $\triangle PNO$ to both sides:

$\text{Area}(MNOP)=\text{Area}(\triangle PNO)+\text{Area}(\triangle PNM)=\text{Area}(\triangle PNO)+\text{Area}(\triangle PNQ)=\text{Area}(\triangle PQO).$

(Q lies on $ON$ produced, so $\triangle PNO$ and $\triangle PNQ$ together make $\triangle PQO$.)

Steps of construction

1. Draw $QR = 5.2$ cm.
2. At $Q$ construct $\angle RQX = 75^\circ$. On $QX$ mark $P$ with $QP = 4.2$ cm.
3. With centre $P$ and radius $4.2$ cm, and with centre $R$ and radius $5.2$ cm, draw arcs that cut each other at $S$ (on the side of $PR$ away from $Q$).
4. Join $PS$ and $RS$. $PQRS$ is the required quadrilateral.
5. Join the diagonal $QS$.
6. Through $P$ draw a line parallel to $QS$ (by copying the angle $\angle PQS$ at $P$ as an alternate angle) and extend $RS$ to meet it at $T$.
7. Join $QT$. Then $\triangle QRT$ is the required triangle.

Justification: $\triangle PQS$ and $\triangle TQS$ stand on the same base $QS$ and lie between the same parallels $QS$ and $PT$, so they have equal area.

$\text{Area of } \triangle QRT = \triangle QRS + \triangle QST = \triangle QRS + \triangle PQS = \text{Area of } \square PQRS$

Steps of construction

1. Draw $BE = 4.2$ cm.
2. At $E$, construct $\angle BES = 75^\circ$ and cut $ES = 5.2$ cm.
3. With centre $B$ and radius $4.2$ cm, draw an arc. With centre $S$ and radius $5.2$ cm, draw another arc. Let the arcs meet at $T$.
4. Join $BT$ and $ST$. $BEST$ is the required quadrilateral.
5. Join the diagonal $BS$.
6. Through $T$, draw a line parallel to $BS$ (copy the angle at $T$ equal to $\angle TBS$ using alternate angles, or use a set square).
7. Produce $ES$ beyond $S$ so that it meets this parallel line at $R$.
8. Join $BR$. $\triangle BER$ is the required triangle.

Justification

$\triangle BST$ and $\triangle BSR$ are on the same base $BS$ and between the same parallels $BS$ and $TR$, so

$\text{Area}(\triangle BST) = \text{Area}(\triangle BSR).$

Adding $\triangle BES$ to both:

$\text{Area}(BEST) = \text{Area}(\triangle BES) + \text{Area}(\triangle BST) = \text{Area}(\triangle BES) + \text{Area}(\triangle BSR) = \text{Area}(\triangle BER),$

because $E, S, R$ lie on one line. Hence $\triangle BER$ has the same area as quadrilateral $BEST$.

Given: $EF\parallel GH$, $EI\parallel FG$ and $FJ\parallel GI$, with $E,H,J,I$ on one line and $F,G$ on the line parallel to it.

Since $EF\parallel GH$ and $EH\parallel FG$, $EFGH$ is a parallelogram. Since $FJ\parallel GI$ and $JI\parallel FG$, $FGIJ$ is a parallelogram.

(i) To prove: $\triangle EFJ\cong\triangle HGI$.

(ii) To prove: Area of $EFGH$ = Area of $FGIJ$.

The trapezium $EFGI$ can be divided in two ways.

• By the line $FJ$: $\text{Area}(EFGI)=\text{Area}(\triangle EFJ)+\text{Area}(FGIJ)$.
• By the line $GH$: $\text{Area}(EFGI)=\text{Area}(EFGH)+\text{Area}(\triangle HGI)$.

Hence

$\text{Area}(\triangle EFJ)+\text{Area}(FGIJ)=\text{Area}(EFGH)+\text{Area}(\triangle HGI).$

From (i), $\text{Area}(\triangle EFJ)=\text{Area}(\triangle HGI)$ (congruent triangles). Cancelling them,

$\text{Area}(EFGH)=\text{Area}(FGIJ).$

Hence proved. (This agrees with the theorem that parallelograms on the same base and between the same parallels are equal in area.)

$MN$ is a diameter, so $\angle MPN = 90^\circ$ (angle in a semicircle).

In $\triangle PMN$: $6x + 3x + 90^\circ = 180^\circ \Rightarrow 9x = 90^\circ \Rightarrow x = 10^\circ$.

So $\angle MNP = 3x = 30^\circ$.

$\angle PQM$ and $\angle PNM$ are angles on the same arc $PM$, so they are equal:

$\angle PQM = \angle MNP = \textbf{30}^\circ$

(i) $PQRS$ is a parallelogram, so $PQ = SR$ and $PQ \parallel SR$.

$QRTU$ is a parallelogram, so $QU = RT$ and $QU \parallel RT$.

Since $QP \parallel RS$ and $QU \parallel RT$ (with the same directions), $\angle PQU = \angle SRT$.

In $\triangle PQU$ and $\triangle SRT$:

• $PQ = SR$
• $\angle PQU = \angle SRT$
• $QU = RT$

So $\triangle PQU \cong \triangle SRT$ (SAS).

(ii) $PT \parallel QR$, so $QRTP$ is a trapezium. It can be divided in two ways:

$\text{Trapezium } QRTP = \square PQRS + \triangle SRT$

$\text{Trapezium } QRTP = \square QRTU + \triangle PQU$

Hence $\square PQRS + \triangle SRT = \square QRTU + \triangle PQU$.

From (i), $\triangle PQU \cong \triangle SRT$, so their areas are equal. Cancelling them,

$\text{Area of } \square PQRS = \text{Area of } \square QRTU.$

Hence proved.

Given: $BEND$ and $BEST$ are parallelograms, with $D, N, T, S$ on one line parallel to $BE$ (in this order).

(i) To prove: $\triangle DBT \cong \triangle NES$

(ii) To prove: area of parallelogram $BEND$ = area of parallelogram $BEST$

Both parallelograms stand on the same base $BE$ and lie between the same parallel lines $BE$ and $DS$.

Hence area of $\square BEND$ = area of $\square BEST$. (Parallelograms on the same base and between the same parallels are equal in area.)

Proved.

Given: $WZ \parallel XU$, $WV \parallel XY$, $WX \parallel ZY$.

Since $WZ \parallel XU$ and $WX \parallel ZU$ (part of $ZY$), $ZWXU$ is a parallelogram. Since $WV \parallel XY$ and $WX \parallel VY$, $WXYV$ is a parallelogram.

(i) In $\triangle WZV$ and $\triangle XUY$:

• $WZ = XU$ (opposite sides of parallelogram $ZWXU$)
• $WV = XY$ (opposite sides of parallelogram $WXYV$)
• $ZU = WX$ (opposite sides of $ZWXU$) and $VY = WX$ (opposite sides of $WXYV$), so $ZU = VY$. Adding $UV$ to both: $ZU + UV = UV + VY$, i.e. $ZV = UY$.

Hence $\triangle WZV \cong \triangle XUY$ (SSS).

(ii) Parallelograms $ZWXU$ and $WXYV$ stand on the same base $WX$ and lie between the same parallel lines $WX$ and $ZY$. Therefore

$\text{Area of } ZWXU = \text{Area of } WXYV.$

(Alternatively, each equals the common part plus a congruent triangle: $\triangle WZV \cong \triangle XUY$ from part (i).)

Given: $NY \parallel GA$, $GN \parallel AU$, $GX \parallel AY$.

Since $GN \parallel AU$ and $NU \parallel GA$ (part of $NY$), $GAUN$ is a parallelogram. Since $GX \parallel AY$ and $XY \parallel GA$, $GAYX$ is a parallelogram.

(i) In $\triangle NGX$ and $\triangle UAY$:

• $GN = AU$ (opposite sides of parallelogram $GAUN$)
• $GX = AY$ (opposite sides of parallelogram $GAYX$)
• $NU = GA$ (opposite sides of $GAUN$) and $XY = GA$ (opposite sides of $GAYX$), so $NU = XY$. Adding $UX$ to both: $NU + UX = UX + XY$, i.e. $NX = UY$.