Q: Using ruler and compass, construct a triangle $ABC$ in which $AB = 6\text{ cm}, \angle BAC = 120^\circ$ and $AC = 5\text{ cm}$. Construct a circle passing through $A, B$ and $C$. Measure and write down the radius of the circle.

120° A B C O 6 cm 5 cm R ≈ 5.5 cm Figure: Obtuse $\triangle ABC$ with circumcentre $O$ outside the triangle and circumcircle Steps of Construction 1. Draw line segment $AB = 6\text{ cm}$. 2. At point $A$, construct an angle $\angle BAX = 120^\circ$ using a pair of compasses. 3. Cut off an arc of length $5\text{ cm}$ along the ray $AX$ to locate vertex $C$. Join $BC$ to complete $\triangle ABC$. 2 Marks 4. Draw the perpendicular bisectors of sides $AB$ and $AC$. Produce them to intersect at point $O$ (circumcentre lies outside because $\angle A$ is obtuse). 5. With $O$ as centre and radius $OA = OB = OC$, draw the circumscribing circle. 1 Mark Radius Measurement Measuring with ruler: $\mathbf{\text{Radius } R \approx 5.5\text{ cm}}$ (acceptable examination tolerance: $5.3\text{ to } 5.7\text{ cm}$). 1 Mark

Q: Using ruler and compass only, construct a $\triangle ABC$ such that $BC = 5\text{ cm}$ and $AB = 6.5\text{ cm}$ and $\angle ABC = 120^\circ$. (i) Construct a circum-circle of $\triangle ABC$ (ii) Construct a cyclic quadrilateral $ABCD$, such that $D$ is equidistant from $AB$ and $BC$.

O B C (5 cm) A (6.5 cm) D 120° Step 1 Construct $\triangle ABC$: • Draw base $BC = 5\text{ cm}$. • At $B$, construct $\angle CBX = 120^\circ$ using compass. • Cut an arc of radius $6.5\text{ cm}$ along ray $BX$ to mark $A$. Join $AC$. [1.5 Marks] Step 2: Circumcircle • Draw the perpendicular bisectors of sides $AB$ and $BC$. • Their point of intersection is the circumcentre $O$ (lying outside $\triangle ABC$ because $\angle B > 90^\circ$). • With centre $O$ and radius $OA = OB = OC$, draw the circumcircle. [1.5 Marks] Step 3: Point D and Quad ABCD • The locus of points equidistant from $AB$ and $BC$ is the internal angle bisector of $\angle ABC$. • Construct the bisector of $\angle ABC$ ($60^\circ$) and produce it to intersect the circumcircle at point $D$. • Join $AD$ and $CD$ to complete the required cyclic quadrilateral $ABCD$. [1 Mark] Result: $\triangle ABC$ constructed with circumcircle and cyclic quadrilateral $ABCD$.

Question 1

Using ruler and compasses only, construct a triangle ABC with AB = 6 cm, BC = 7 cm and ∠ABC = 60°. Construct its incircle and measure its radius.

Accepted Answer

Step 1: Triangle ABC Draw base BC = 7 cm. At B, construct ∠B = 60° using compass. Cut off AB = 6 cm. Join AC to complete ΔABC. [1 Mark]

Step 2: Incentre I Bisect ∠B and ∠C using compass arcs. The point of intersection of the angle bisectors is the incentre I. [1 Mark]

Step 3: Incircle From I, draw a perpendicular to BC meeting at D. With centre I and radius ID, draw the incircle touching all three sides. Measured inradius r = 1.8 cm (± 0.1 cm). [1 Mark]

Question 2

Construct ΔABC with AB = 6 cm, AC = 4.5 cm and ∠BAC = 120°. Construct circle circumscribing ΔABC. Measure and write down length of radius.

Accepted Answer

1. Draw line AB = 6 cm. Construct angle 120° at A using compass. Cut off AC = 4.5 cm. Join BC. [2 Marks]
2. Draw perpendicular bisectors of AB and AC intersecting at circumcentre O. [1 Mark]
3. With centre O, draw circle passing through A, B, C. Measured radius = 5.3 cm (±0.1 cm). [1 Mark]

Question 3

Use ruler and compass to answer this question. Construct ∠ABC = 90°, where AB = 6 cm, BC = 8 cm.
(a) Construct the locus of points equidistant from B and C.
(b) Construct the locus of points equidistant from A and B.
(c) Mark intersection as O. Construct locus of points at distance OA from O.
(d) Construct locus of points equidistant from BA and BC.

Accepted Answer

1. Draw BC = 8 cm, ∠ B = 90°, BA = 6 cm, join AC (10 cm). [1 Mark] 2. Perpendicular bisector of BC . [1 Mark] 3. Perpendicular bisector of AB meeting at O (midpoint of AC ). Draw circumcircle radius 5 cm. [1 Mark] 4. Angle bisector of ∠ ABC . [1 Mark]

Question 4

Using ruler and compass, construct a triangle $ABC$ in which $AB = 6	ext{ cm}, \angle BAC = 120^\circ$ and $AC = 5	ext{ cm}$. Construct a circle passing through $A, B$ and $C$. Measure and write down the radius of the circle.

Accepted Answer

120° A B C O 6 cm 5 cm R ≈ 5.5 cm Figure: Obtuse $	riangle ABC$ with circumcentre $O$ outside the triangle and circumcircle Steps of Construction 1. Draw line segment $AB = 6	ext{ cm}$. 2. At point $A$, construct an angle $\angle BAX = 120^\circ$ using a pair of compasses. 3. Cut off an arc of length $5	ext{ cm}$ along the ray $AX$ to locate vertex $C$. Join $BC$ to complete $	riangle ABC$. 2 Marks 4. Draw the perpendicular bisectors of sides $AB$ and $AC$. Produce them to intersect at point $O$ (circumcentre lies outside because $\angle A$ is obtuse). 5. With $O$ as centre and radius $OA = OB = OC$, draw the circumscribing circle. 1 Mark Radius Measurement Measuring with ruler: $\mathbf{	ext{Radius } R \approx 5.5	ext{ cm}}$ (acceptable examination tolerance: $5.3	ext{ to } 5.7	ext{ cm}$). 1 Mark

Question 5

Using ruler and compass construct a triangle $ABC$ where $AB = 3	ext{ cm}, BC = 4	ext{ cm}$ and $\angle ABC = 90^\circ$. Hence construct a circle circumscribing the triangle $ABC$. Measure and write down the radius of the circle.

Accepted Answer

O A B C AB = 3 cm BC = 4 cm AC = 5 cm (Hypotenuse) Figure: Right $	riangle ABC$ and circumcircle with centre $O$ at midpoint of hypotenuse $AC$ Step 1 Draw base segment $BC = 4	ext{ cm}$. At $B$, construct $\angle ABC = 90^\circ$ using compass. Cut off $AB = 3	ext{ cm}$. Join $AC$ to form $	riangle ABC$. [1.5 Marks] Step 2 Draw the perpendicular bisectors of $AB$ and $BC$. Their point of intersection is $O$, which lies exactly on the midpoint of hypotenuse $AC$. [1.5 Marks] Step 3 With $O$ as centre and radius $OA = OB = OC$, draw the circumcircle. Theoretical calculation: In right $	riangle ABC$, $AC = \sqrt{3^2 + 4^2} = 5	ext{ cm}$. Radius $R = \frac{AC}{2} = \frac{5}{2} = \mathbf{2.5	ext{ cm}}$. Measured radius: $2.5	ext{ cm}$ . [1 Mark]

Question 6

Use ruler and compass only for answering this question. Draw a circle of radius $4	ext{ cm}$. Mark the centre as $O$. Mark a point $P$ outside the circle at a distance of $7	ext{ cm}$ from the centre. Construct two tangents to the circle from the external point $P$. Measure and write down the leng

Accepted Answer

O M P T₁ T₂ OP = 7 cm r = 4 cm Tangent ≈ 5.7 cm Vector Diagram: Compass and straightedge construction of tangents PT₁ and PT₂ to a circle from external point P Step 1 Draw circle of radius $r = 4	ext{ cm}$ with centre $O$. [1 Mark] Step 2 Mark point $P$ such that $OP = 7	ext{ cm}$. Draw the perpendicular bisector of $OP$ to locate its midpoint $M$. [1 Mark] Step 3 With $M$ as centre and radius $MO = MP$, draw an auxiliary circle (or semi-circles) intersecting the given circle at points $T_1$ and $T_2$. [1 Mark] Step 4 Join $PT_1$ and $PT_2$. These are the required two tangents from $P$. Theoretical verification: In right $\Delta OT_1 P$: $PT_1 = \sqrt{OP^2 - OT_1^2} = \sqrt{7^2 - 4^2} = \sqrt{49 - 16} = \sqrt{33} \approx 5.74	ext{ cm}$. $\mathbf{	ext{Measured length of tangent: } 5.7	ext{ cm}}$ (acceptable range: $5.6	ext{ cm}$ to $5.8	ext{ cm}$). [1 Mark]

Question 7

Using a ruler and compass only, construct: $\Delta ABC$ such that $AB = 5	ext{ cm}$, $AC = 6.3	ext{ cm}$ and $\angle ABC = 60^\circ$. the circumcircle of $\Delta ABC$. the locus of points which are equidistant from $AB$ and $BC$.

Accepted Answer

O B A C 60° Angle Bisector Locus AB = 5 cm AC = 6.3 cm Vector Diagram: Circumcircle of ΔABC with angle bisector locus equidistant from AB and BC Step 1 (a) Construct $\Delta ABC$: Draw base $AB = 5	ext{ cm}$. At $B$, construct $\angle ABX = 60^\circ$ using compass arcs. With $A$ as centre and radius $6.3	ext{ cm}$, draw an arc cutting ray $BX$ at point $C$. Join $AC$ to complete $\Delta ABC$. [1.5 Marks] Step 2 (b) Construct Circumcircle of $\Delta ABC$: Draw the perpendicular bisectors of sides $AB$ and $AC$ (or $BC$). Their point of intersection is the circumcentre $O$. With $O$ as centre and radius $OA = OB = OC$, draw the circumcircle passing through all three vertices. [1.5 Marks] Step 3 (c) Construct Locus of points equidistant from $AB$ and $BC$: The locus of points equidistant from two intersecting lines $AB$ and $BC$ is the **internal angle bisector of $\angle ABC$**. Construct the bisector of $\angle ABC = 60^\circ$. [1 Mark]

Question 8

Using ruler and compass only, construct a $\triangle ABC$ such that $BC = 5\text{ cm}$ and $AB = 6.5\text{ cm}$ and $\angle ABC = 120^\circ$.
(i) Construct a circum-circle of $\triangle ABC$
(ii) Construct a cyclic quadrilateral $ABCD$, such that $D$ is equidistant from $AB$ and $BC$.

Accepted Answer

O B C (5 cm) A (6.5 cm) D 120° Step 1 Construct $	riangle ABC$: • Draw base $BC = 5	ext{ cm}$. • At $B$, construct $\angle CBX = 120^\circ$ using compass. • Cut an arc of radius $6.5	ext{ cm}$ along ray $BX$ to mark $A$. Join $AC$. [1.5 Marks] Step 2: Circumcircle • Draw the perpendicular bisectors of sides $AB$ and $BC$. • Their point of intersection is the circumcentre $O$ (lying outside $	riangle ABC$ because $\angle B > 90^\circ$). • With centre $O$ and radius $OA = OB = OC$, draw the circumcircle. [1.5 Marks] Step 3: Point D and Quad ABCD • The locus of points equidistant from $AB$ and $BC$ is the internal angle bisector of $\angle ABC$. • Construct the bisector of $\angle ABC$ ($60^\circ$) and produce it to intersect the circumcircle at point $D$. • Join $AD$ and $CD$ to complete the required cyclic quadrilateral $ABCD$. [1 Mark] Result: $	riangle ABC$ constructed with circumcircle and cyclic quadrilateral $ABCD$.

Question 9

Using a ruler and a compass construct a triangle $ABC$ in which $AB = 7\text{ cm}$, $\angle CAB = 60^\circ$ and $AC = 5\text{ cm}$. Construct the locus of:
(i) points equidistant from $AB$ and $AC$.
(ii) points equidistant from $BA$ and $BC$.
Hence construct a circle touching the three sides of the triangle internally.

Accepted Answer

I A B (7 cm) C 5 cm 60° Step 1: Triangle ABC Draw base $AB = 7	ext{ cm}$. At $A$, construct $\angle CAB = 60^\circ$. Cut arc $AC = 5	ext{ cm}$ on the ray. Join $BC$. [1 Mark] Step 2: Locus (i) Locus of points equidistant from $AB$ and $AC$ is the angle bisector of $\angle CAB$ ($30^\circ$). Construct this ray. [1 Mark] Step 3: Locus (ii) Locus of points equidistant from $BA$ and $BC$ is the angle bisector of $\angle ABC$. Construct this ray. [1 Mark] Step 4: Incircle Their intersection is the incentre $I$. Drop a perpendicular from $I$ to $AB$ to obtain the inradius $r$. Draw the incircle touching all three sides. [1 Mark] Result: $	riangle ABC$ and its incircle successfully constructed.

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