NCERT Textual Exercise (Solved)

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Constructions

EXERCISE – 4.1 Instructions: In each of the following, give the justification of the construction

also. 1. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure

the two parts. 2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar

to it whose sides are 23

of the corresponding sides of the first triangle.

3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle

whose sides are 75

of the corresponding sides of the first triangle.

4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and

then another triangle whose sides are 1 12

times the corresponding sides of isosceles triangle.

5. Draw a ∆ABC with BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3

4 of the corresponding sides of ∆ABC.

6. Draw a ∆ABC with BC = 7 cm, ∠B = 45°, ∠A = 105°. Then construct a

triangle whose sides are 34

times the corresponding sides of ∆ABC.

7. Draw a right triangle in which the sides (other than hypotenuse) are of length 4 cm and 3 cm. Then construct another triangle whose sides are 5

3 times the

corresponding sides of the given triangle.

Test Yourself – CNST 1 1. Draw a line segment of length 6 cm. Divide it into seven equal parts. 2. Draw a line segment of length 4 cm. Divide it in the ratio 2 : 3 internally. 3. Construct a triangle ABC in which BC = 6 cm, ∠B = 70° and ∠C = 50°.

Construct a triangle DBE similar to the given ∆ABC such that BE is 34

of BC.

4. Divide a line segment of 8 cm into the ratio 4 : 3 internally.

5. Construct a triangle similar to a given ∆ABC with its sides equal to 45 of

the corresponding sides of the ∆ABC.

NCERT Textual Exercise (Solved)

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Constructions

6. Construct a triangle with sides 5 cm, 7 cm and 8.5 cm and then construct

another triangle similar to it whose sides are 53

times the corresponding sides of the first triangle.

7. Draw ∆ABC with AB = 7 cm, ∠CAB = 60° and ∠ABC = 105° and then

construct another triangle whose sides are 34

th of the corresponding sides of the ∆ABC.

8. Construct ∆ABC right angled at B, BC = 7 cm and AB = 6 cm. Then construct another triangle similar to ∆ABC so that the corresponding sides of the

second triangle be 57

times than those of the ∆ABC.

9. Draw a line segment AB and locate a point C on AB such that AC = 27

AB.

10. Draw a line segment PQ = 7.8 cm and divide it in the ratio 7 : 10. Mark the point of division as R and measure the lengths of PR and QR.

EXERCISE – 4.2 In each of the following, give also the justification of the construction. 1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre,

construct the pair of tangents to the circle and measure their lengths. 2. Construct a tangent to a circle of radius 4 cm from a point on the concentric

circle of radius 6 cm and measure its length. Also verify the measurement by actual calculations.

3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameters each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.

4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.

5. Draw a line seg AB of length 8 cm. Taking A as centre, draw a circle of radius 4cm & taking B as centre draw another circle of radius 3 cm. Construct tangents to each circle from the centre of other circle.

6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C and D is drawn. Construct tangents from A to this circle.

7. Draw a circle with the help of a triangle. Take a point outside the circle. Construct the pair of tangent from this point to the circle.

NCERT Textual Exercise (Solved)

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Constructions

Test Yourself – CNST 2 1. Draw a circle of radius 4.5 cm. Mark a point P outside the circle. Through

P, draw a tangent to the circle, using the centre of the circle. 2. Draw a circle of radius 3.5 cm. Take a point M outside it and draw a tangent

at the point M to the circle. 3. Draw a circle of diameter 7 cm. Mark a point M at a distance of 9 cm from

its centre. Through M, draw tangents to the circle. Measure the lengths of the tangent segments. Are the two lengths equal ?

4. Draw a circle with radius = 4.2 cm. Mark a point A such that OA = 10 cm. Draw a circle with OA as diameter. From A, draw tangents to the circle.

5. Draw a circle of diameter 10 cm. Take a chord PQ of length of 6.7 cm. Draw tangents at P and Q.

6. Draw a pair of tangents to a circle of radius 4 cm which are inclined to each other at 120°.

7. Draw a line segment PQ = 9 cm. Taking P as centre, draw circle of radius 5cm and taking Q as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

8. Construct a tangent to a circle of radius 3 cm from a point on the concentric circle of radius 5 cm and measure its length.

9. Draw a circle of radius 4 cm. Take two points A and B on one of its extended diameter each at a distance of 9 cm from its centre. Draw tangents to the circle from these two points A and B.

10. Draw a line segment AB of length 9 cm. Taking A as centre, draw a circle of radius 5 cm. Construct pair of tangents to the circle from the point B.

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NCERTTextualExercisesandAssignments

Exercise–4.1 In each of the following, give the justification of the construction also. 1.

Stepsofconstruction: 1. Draw any ray AX making an acute with AB. 2. Locate 13 (5 + 8) points A1, A2, A3, A4, A5, A6, A7, A8, A9, A10, A11, A12, A13 on AX such that AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7 = A7A8 = A8A9 = A9A10 = A10A11 = A11A12 = A12A13.

3. Join A13 (thirteenth point) to B. 4. Through the point A5 , draw a line parallel to A13 B intersecting AB at C. 5. Then AC : CB = 5 : 8. Justification: In ∆AA13B A5C || to A13 B .... (by construction)

∴ AAA A

ACBC

5

5 13

= .... (by Basic proportionality theorem)

but AAA A

5

5 13

58

= ,

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∴ ACBC= 58

2.

Stepsofconstruction: 1. Draw the given ∆ABC. 2. Draw any ray BX making an acute angle with BC on the opposite side of the vertex A.

3. Mark off 3 points (3 being greater than 2 in 23

) B1, B2 and B3 on BX such that, BB1 = B1B2 = B2B3.

4. Join B3 (third point) to C. 5. Draw a line through B2 (second point) parallel to B3C to intersect BC at C′. 6. Draw a line through C′ parallel to line CA to intersect BA at A′. 7. ∆A′BC′ is the required triangle. Justification: A′C′ || AC .... (by construction) ∆A′BC′ ∼ ∆ABC .... (AA similarity)

A B AB

BCBC

A CAC

.... (corresponding sides of similar triangles)

but BCBC

BBBB

23

23

∴ BCBC

23

∴ A B AB

BCBC

A CAC

23

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3. mn= 75

m > n ⇒ 7 > 5 Construct triangle larger than the given one.

Stepsofconstruction: 1. Draw the given ∆ABC. 2. Draw any ray BX making an acute angle with BC on the opposite side of the vertex A.

3. Mark 7 points (7 being greater than 5 in 75

)

B1, B2, B3, B4, B5, B6 and B7 on BX such that BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7. 4. Join B5 (fifth point) to C. 5. Draw a line through B7 (seventh point) parallel to B5C intersecting the extended line segment

BC at C′. 6. Draw a line through C′ parallel to line CA intersecting the extended line segment BA at A′. 7. ∆A′BC′ is the required triangle. Justification: A′C′ || AC .... (by construction)

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∆A′BC′ ∼ ∆ABC .... (AA similarity)

A B AB

BCBC

A CAC

.... (corresponding sides of similar triangles)

but BCBC

BBBB

′= =7

5

75

∴ BCBC

75

∴ A B AB

BCBC

A CAC

75

4.

Stepsofconstruction: 1. Draw seg BC = 8 cm 2. Draw line PQ the ⊥ bisector of BC intersecting BC at M. 3. Taking M as centre and radius = 4 cm, cut an arc on PQ at A such that MA = 4 cm. 4. Join A to B and C. 5. Draw ray BD making an acute angle with BC on the opposite side of vertex A.

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6. Mark 3 points (3 being greater than 2 in 32

) B1,B2 and B3 on ray BD such that BB1 = B1B2 = B2B3. 7. Join B2 (second point) to ‘C’. 8. Draw a line through B3 (third point) parallel to B2C intersecting the extended line segment

BC at C′. 9. Draw a line through C′ parallel to CA intersecting the extended line segment BA at A′. Then ∆A′BC′ is the required triangle. Justification:

A′C′ || AC .... (by construction) ∆A′BC′ ∼ ∆ABC .... (AA similarity)

A B AB

BCBC

A CAC

.... (corresponding sides of similar triangles)

But BCBC

BBBB

32

32

∴ BCBC

32

∴ A BAB

BCBC

A CAC

32

5.

C

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Stepsofconstruction: 1. Draw the given ∆ABC. 2. Draw any ray BX making an acute angle with BC on the opposite side of the vertex A.

3. Mark off 4 points (4 being greater than 3 in 34

) B1, B2, B3 and B4 on BX such that BB1 = B1B2 = B2B3 = B3B4. 4. Join B4 (fourth point) to C. 5. Draw a line through B3 (third point) parallel to B4C to intersect BC at C′. 6. Draw a line through C′ parallel to line CA to intersect BA at A′. 7. ∆A′BC′ is the required triangle. Justification: A′C′ || AC .... (by construction) ∆A′BC′ ∼ ∆ABC .... (AA similarity)

A B AB

BCBC

A CAC

.... (corresponding sides of similar triangles)

but BCBC

BBBB

34

34

∴ BCBC

34

∴ A B AB

BCBC

A CAC

34

6. Analysis In ∆ABC, ∠B = 45°, ∠A = 105° ∠C = 180° – (∠A + ∠B) = 180° – (105° + 45°) = 180° – 150° ∠C = 30° Stepsofconstruction: 1. Draw the given ∆ABC. 2. Draw ray BQ making an

acute angle with BC on the opposite side of vertex A.

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3. Mark 4 points (4 being greater than 3 in 43 ) B1, B2 ,B3 and B4 such that

BB1= B1B2 = B2B3 = B3B4. 4. Join B3 (third point) to C. 5. Draw a line through B4 (fourth point) parallel to B3C intersecting the extended line segment BC at C′. 6. Draw a line through C′ parallel to CA intersecting the extended line segment BA at A′. 7. ∆A′BC′ is the required triangle. Justification: A′C′ || AC .... (by construction) ∆A′BC′ ∼ ∆ABC .... (AA similarity)

∴ A B AB

BCBC

A CAC

.... (corresponding sides of similar triangles)

But BCBC

BBBB

43

43

∴ BCBC

43

∴ A BAB

BCBC

A CAC

43

7. Stepsofconstruction: 1. Draw the given ∆ABC. 2. Draw any ray BX making an acute angle with

BC on the opposite side of the vertex A.

3. Mark off 5 points (5 being greater than 3 in 53 )

B1, B2 ,B3 ,B4 and B5 on BX such that, BB1 = B1B2 = B2B3 = B3B4 = B4B5. 4. Join B3 (third point) to C. 5. Draw a line through B5 (fifth point) parallel to

B3C intersecting the extended line segment BC at C′.

6. Draw a line through C′ parallel to line CA intersecting the extended line segment

BA at A′. 7. ∆A′BC′ is the required triangle.

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Justification: A′C′ || AC .... (by construction) ∆A′BC′ ∼ ∆ABC .... (AA similarity)

A B AB

BCBC

A CAC

.... (corresponding sides of similar triangles)

but BCBC

BBBB

53

53

∴ BCBC

53

∴ A B AB

BCBC

A CAC

53

Exercise–4.2 In each of the following, give also the justification of the construction. 1.

∴ PQ = PR = 7.7 cm. Stepsofconstruction: We are given a circle with centre ‘O’ and point P outside it such that OP = 10 cm 1. Join PO and bisect it. Let M be the midpoint of PO. 2. Taking M as centre and MO as radius, draw a circle. Let it intersect the given circle at point

Q and R. 3. Join PQ and PR. PQ and PR are the required two tangents.

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Justification: Join OQ; then ∠PQO is an angle in a semicircle. ∴ ∠PQO = 90° ∴ Line PQ ⊥ radius OQ. ∴ PQ has to be a tangent to circle (A line ⊥to radius of a circle at its outer end is a tangent) Similarly, PR is also a tangent to the circle. 2.

PQ = PR = 4.5 cm (by measurement) Now PQ = OP OQ2 2− .... (by Pythagoras theorem)

= 6 42 2-

= 36 16−

= 20 = 4.47 ≈ 4.5 cm .... (similarly PR ≈ 4.5 cm) Stepsofconstruction: Draw the two concentric circles with centre ‘O’ of radii 4 cm and 6 cm. Take any point P on the

larger circle. 1. Join OP and bisect it. Let M be the midpoint of OP. 2. Taking M as centre and MO as radius, draw a circle. Let it intersect the smaller circle at point Q and R. 3. Join PQ and PR. PQ and PR are the required tangents.

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Justification: Join OQ then ∠PQO is an angle in a semicircle. ∴ ∠PQO = 90° ∴ Line PQ ⊥ radius OQ. ∴ PQ has to be a tangent to circle (A line ⊥ to radius of a circle at its outer end is a tangent.) Similarly, PR is also a tangent to the circle. 3.

Stepsofconstruction: Draw the circle of radius 3 cm. Take the points P and Q on one of its extended diameter each at

a distance of 7 cm from centre ‘O’. 1. Draw ⊥ bisector of seg OP and let its midpoint be M. 2. Taking M as centre and MO as radius, draw a circle to cut the given circle at A and B. 3. Join PA and PB. 4. PA and PB are the required two tangents. 5. Taking N as centre and radius = NO, draw a circle to cut the given circle at C and D. 6. Join QC and QD. 7. QC and QD are the required two tangents.

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Justification Join OA, then ∠OAP is an angle in a semicircle. ∴ ∠OAP = 90° ∴ Line PA ⊥ radius OA. ∴ PA has to be a tangent to circle (A line ⊥ to radius of a circle at its outer end is a tangent) Similarly, we can show that PB, QC and QD are the tangents. 4.

Analysis Let ‘O’ be the centre of the circle with radius 5 cm. Suppose PA and PB are tangents to the circle at A and B such that ∠APB = 60° ∠OAP = ∠OBP = 90° ...(tangent at any point on the circle is ⊥ to radius through the point of contact) In OAPB, ∠OAP + ∠APB + ∠OBP + ∠AOB = 360° ...(II) ...(sum of the measures of all angles of quadrilateral is 360°) ∴ 90° + 60° + 90° + ∠AOB = 360° ... from (I) and (II) ∴ ∠AOB = 360° – 240° ∴ ∠AOB = 120° StepsofConstruction: 1. Draw ∠AOB = 120°. 2. Draw perpendiculars to radii OA and OB at A and B to intersect at P. 3. Join PA and PB. 4. PA and PB are the required tangents.

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Justification: PA and PB are ⊥ to radii PA and PB Hence PA and PB are the tangents to the circle. ..... (A line ⊥ to radius of a circle at its outer end is a tangent.) 5.

Stepsofconstruction: 1. Draw AB of length 8 cm. Taking A and B as centres draw circles of radii 4 cm and 3 cm

respectively. 2. Draw ⊥ bisector of seg AB and let M be the midpoint of AB. 3. Taking M as centre and radius as MA, draw a circle to intersect the two circles at P, Q, R

and S. 4. Draw AP, AQ, BR and BS. Justification: Draw AR ∠ARB is an angle in a semicircle. ∴ ∠ARB = 90° ∴ BR ⊥ radius AR. ∴ BR is a tangent to the circle ..... (A line ⊥ to radius of a circle at its outer end is a tangent) Similarly, we can show that BS, AP and AQ are the tangents.

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6.

Stepsofconstruction: ABC is right triangle with ∠B = 90°, AB = 6 cm and BC = 8 cm ⊥ is drawn from B to AC at D. 1. Draw the given ∆ABC. 2. Draw ⊥ bisector of BC and let X be midpoint of BC. 3. Taking X as centre and radius = XB, draw a circle passing through B and C and intersecting

AC at D. 4. Taking A as centre and radius = AB, cut an arc on the circle intersecting it at P. 5. Join AP. 6. AP and AB are the required tangents. Justification: ∠ABX = 90° and hence AB ⊥ radius XB ∴ AB is a tangent to circle ....... (A line ⊥ to radius of a circle at its outer end is a tangent) Length of the tangents drawn to a circle from an external point is always equal. Hence, AP is the other tangent to the circle.

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7.

Stepsofconstruction: A circle and a point P outside it are given. 1. Draw ∆ABC and draw ⊥ bisectors of AB and BC to intersect at ‘O’. O is the centre of the

given circle. 2. Join OP, draw ⊥ bisector of OP and let M be its midpoint. 3. Taking M as centre and radius = MO, draw a circle to intersect the given circle at Q and R. 4. Draw PQ and PR. 5. PQ and PR are the required tangents. Justification: Join OQ, then ∠OQP is an angle in a semicircle. ∴ ∠OQP = 90° ∴ PQ ⊥ radius OQ. ∴ PQ is a tangent to the circle .......... (A line ⊥ to radius of a circle at its outer end is a tangent.) Similarly, PR is the other tangent.