Constructions — Chapter Test

Draw a circle of radius 3 cm. Mark its centre as C and mark a point P such that CP = 7 cm. Using ruler and compasses only, construct two tangents from P to the circle.

Answer

Steps of construction :

1. Construct a circle with centre as C and radius = 3 cm.

2. Mark a point P at a distance of 7 cm from the centre. Join CP. Draw its perpendicular bisector to meet CP at M.

3. With M as centre and CM (or MP) as radius, draw a circle. Let this circle intersect the circle with C as centre at points A and B.

4. Join PA and PB.

Hence, PA and PB are tangents to the circle with centre C.

Draw a line AQ = 7 cm. Mark a point P on AQ such that AP = 4 cm. Using ruler and compasses only, construct :

(i) a circle with AP as diameter

(ii) two tangents to the above circle from the point Q.

Answer

(i) Steps of construction :

1. Draw a line segment AQ = 7 cm. Mark a point P on AQ such that AP = 4 cm.

2. Draw the perpendicular bisector of AP and let it meet AP at O, so that AO = OP = radius of the circle with centre O.

3. With O as centre and radius AO draw a circle. This is the circle with AP as diameter.

(ii) Steps of construction :

1. Join OQ.

2. Draw perpendicular bisector of OQ to meet OQ at M.

3. With M as centre and OM (or MQ) as radius, draw a circle. Let this circle intersect the circle with O as centre at points C and D.

4. Join QC and QD.

Hence, QC and QD are tangents to the circle with centre O from the point Q.

Using ruler and compasses only, construct a triangle ABC having given c = 6 cm, b = 7 cm and ∠A = 30°. Measure side a. Draw the circumcircle of the triangle.

Answer

We know that a = BC, b = AC and c = AB.

Steps of construction :

1. Draw a line segment AB = 6 cm.

2. Construct ∠A = 30°. Extend segment from A to point C such that AC = 7 cm and ∠CAB = 30°.

3. Join C and B. Hence, △ABC is formed.

4. Draw perpendicular bisectors of AB and BC. Let these meet at O.

5. Now using O as centre and OA or OB as radius, draw a circle touching all the vertices of triangle ABC. This is the circumcircle of triangle ABC.

On measuring side BC we get, BC = 3.5 cm. Since a = BC,

∴ a = 3.5 cm.

Hence, the length of side a = 3.5 cm.

Using ruler and compasses only, construct an equilateral triangle of height 4 cm and draw its circumcircle.

Answer

Steps of construction :

1. Draw a line XY and take a point D on it.

2. At D, draw perpendicular and cut off DA = 4 cm.

3. From A, draw rays making an angle of 30° on each side of AD meeting the line XY at B and C.

4. Now draw perpendicular bisector of AB intersecting AD at O.

5. With centre O and radius OA or OB or OC, draw a circle which will pass through A, B and C.

This is the required circumcircle of △ABC.

Using ruler and compasses only :

(i) Construct a triangle ABC with the following data :

BC = 7 cm, AB = 5 cm and ∠ABC = 45°.

(ii) Draw the inscribed circle to △ABC drawn in part (i).

Answer

(i) Steps of construction :

1. Draw a line segment BC = 7 cm.

2. At B draw a ray BX making an angle of 45° and cut off BA = 5 cm.

3. Join AC.

Hence, the required triangle ABC is formed.

(ii) Steps of construction :

1. Draw the angle bisectors of ∠B and ∠C intersecting each other at I.

2. From I draw a perpendicular ID on BC.

3. With centre as I and radius ID, draw a circle which touches all the sides of △ABC.

This is the required inscribed circle to △ABC.

Draw a triangle ABC, given that BC = 4 cm, ∠C = 75° and that radius of circumcircle of △ABC is 3 cm.

Answer

Steps of construction :

1. Draw a line segment BC = 4 cm.

2. Draw the perpendicular bisector of BC.

3. From B, draw an arc of 3 cm which intersects the perpendicular bisector at O.

4. Draw a ray CX making an angle of 75°.

5. With centre O and radius 3 cm draw a circle which intersects the ray CX at A.

6. Join AB.

Hence, the required triangle ABC is formed.

Draw a regular hexagon of side 3.5 cm. Construct its circumcircle and measure its radius.

Answer

We know that each angle in a regular hexagon = 120°.

1. Draw a line segment AB = 3.5 cm.

2. At A and B draw rays making an angle of 120° each and cut off AF = BC = 3.5 cm.

3. At F and C, draw rays making angle of 120° each and cut off EF = CD = 3.5 cm.

4. Join ED. Hence, ABCDEF is the required hexagon.

5. Draw the perpendicular bisector of AB and BC. Let these bisectors meet at the point O.

6. With O as center and radius equal to OA or OB, draw a circle which passes through the vertices of the hexagon. This is the required circumcircle of hexagon ABCDEF.

On measuring we get OA = 3.5 cm.

Hence, the required hexagon with circumcircle of radius 3.5 cm. is formed.

Construct a triangle ABC with the following data :

AB = 5 cm, BC = 6 cm and ∠ABC = 90°.

(i) Find a point P which is equidistant from B and C and is 5 cm from A. How many such points are there ?

(ii) Construct the inscribed circle of △ABC drawn above.

Answer

(i) Steps of construction :

1. Draw a line segment BC = 6 cm.

2. At B, draw a ray BX making an angle of 90° and cut off BA = 5 cm.

3. Join AC.

4. Draw the perpendicular bisector of BC.

5. From A with 5 cm radius, draw arc which intersects the perpendicular bisector of BC at P and P'.

There are two points (P and P') equidistant from B and C and at a distance of 5 cm from A.

(ii) Steps of construction :

1. Draw the angle bisectors of ∠B and ∠C intersecting at O.

2. From O, draw OD ⊥ BC.

3. With centre O and radius OD, draw a circle which will touch all the sides of △ABC.

Hence, the required inscribed circle of △ABC is formed.

Use ruler and compasses for the following question taking a scale of 10 m = 1 cm.

A park in the city is bounded by straight fences AB, BC, CD and DA.

Given that AB = 50 m, BC = 63 m, ∠ABC = 75°. D is a point equidistant from the fences AB and BC. If ∠BAD = 90°, construct the outline of the park ABCD.

Also locate a point P on the line BD for the flag post which is equidistant from the corners of the park A and B.

Answer

We know that,

The locus of a point equidistant from two intersecting lines is pair of bisectors of the angles between the two lines.

The locus of a point which is equidistant from two given points is actually the perpendicular bisector of the segment that joins the two points.

Given,

Scale : 10 m = 1 cm

BC = 63 m = $\dfrac{63}{10}$ = 6.3 cm.

AB = 50 m = $\dfrac{50}{10}$ = 5 cm.

Steps of construction :

1. Draw a line BC = 6.3 cm.

2. Draw ∠ABC = 75° such that AB = 5 cm.

3. Draw BE, angle bisector of ∠ABC.

4. Construct ∠BAF = 90°, intersecting BE at D.

5. Join ABCD.

6. Construct XY, the perpendicular bisector of AB, intersecting BD at P.