NCERT Solutions Class 10 Maths Chapter 11: Constructions

Exercise 11.1

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.

Solution

Steps for the construction are as follows:

1. Draw a line segment AB = 7.6 cm.
2. Draw a ray AX that makes an acute angle BAX.
3. Since, the line segment needs to be divided in 5: 8 ratio. Therefore, mark 5 + 8 = 13 points on the ray AX such that AA₁ = A₁A₂ = A₂A₃ = ... ... = A₁₂A₁₃.
4. Join A₁₃ with point B.
5. Draw a line from A₅ to AB such that it touches AB at a point C and is parallel to BA₁₃.
6. The obtained point C is the one that divides line segment AB into the ratio 5: 8.
7. Measure AC and BC.
8. Result is AC = 2.9 cm and BC = 4.7 cm.

Justification:

A₅C is parallel to A₁₃B by construction. Therefore, we can use basic proportionality theorem in triangle AA₁₃B.

AC/BC = AA₅/A₅A₁₃

By construction, AA₅ has 5 equal divisions of segment AA₁₃ and A₅A₁₃ has 8 equal divisions. Thus,

AA₅/A₅A₁₃= 5/8

Therefore, AC/BC = 5/8. Hence, the construction is justified.

2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.

Solution

Steps for the construction are as follows:

1. Draw a line segment AB = 4 cm.
2. Taking the point A as the centre, draw an arc of radius 5 cm.
3. Taking the point B as the centre, draw an arc of radius 6 cm.
4. Mark the point of intersection of the two arcs as C.
5. We obtained AC = 5 cm and BC = 6 cm with ABC as the required triangle.
6. Draw a ray AX that makes an acute angle BAX.
7. Mark 3 points on the ray AX such that AA₁ = A₁A₂ = A₂A₃.
8. Join A₃ with point B.
9. Draw a line A₂B' so that it is parallel to A₃B and the point B' lies on the segment AB.
10. Draw a line B'C' so that it is parallel to BC and the point C' lies on the segment AC.
11. Now, AB'C' is the required second triangle.

Justification:

B'C' is parallel to BC by construction. Therefore

∠ABC = ∠AB'C'

In ΔAB'C' and ΔABC,

∠ABC = ∠AB'C' (Already Proven)

∠BAC = ∠B'AC' (Common angle)

Therefore, by the AA similarity criterion ΔAB'C' ∼ ΔABC.

This implies that

AB'/AB = B'C'/BC = AC'/AC (Corresponding sides of similar triangles)

A₂B' is parallel to A₃B by construction. Therefore

∠AA₂B' = ∠AA₃B

In ΔAA₂B' and ΔAA₃B,

∠A₂AB' = ∠A₃AB (Common angle)

∠AA₂B' = ∠AA₃B (Already Proven)

Therefore, by the AA similarity criterion ΔAA₂B' ~ ΔAA₃B.

This implies that

AB'/AB = AA₂/AA₃ (Corresponding sides of similar triangles)

By construction, AA₂ has 2 equal divisions of segment AA₃ and AA₃ has 3 equal divisions. Thus,

AA₂/AA₃ = 2/3

Therefore,

AB'/AB = B'C'/BC = AC'/AC = 2/3

AB' = (2/3)AB

B'C' = (2/3)BC

AC'= (2/3)AC

Hence, the construction is justified.

3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle.

Solution

Steps for the construction are as follows:

1. Draw a line segment AB = 5 cm.
2. Taking the point A as the centre, draw an arc of radius 6 cm.
3. Taking the point B as the centre, draw an arc of radius 7 cm.
4. Mark the point of intersection of the two arcs as C.
5. We obtained AC = 6 cm and BC = 7 cm with ABC as the required triangle.
6. Draw a ray AX that makes an acute angle BAX.
7. Mark 7 points on the ray AX such that AA₁ = A₁A₂ = A₂A₃ = ... ... = A₆A₇.
8. Join A₅ with point B.
9. Draw a line A₇B' so that it is parallel to A₅B and the point B' lies on the extended segment AB.
10. Draw a line B'C' so that it is parallel to BC and the point C' lies on the extended segment AC.
11. Now, AB'C' is the required second triangle.

Justification:

B'C' is parallel to BC by construction. Therefore

∠ABC = ∠AB'C'

In ΔAB'C' and ΔABC,

∠ABC = ∠AB'C' (Already Proven)

∠BAC = ∠B'AC' (Common angle)

Therefore, by the AA similarity criterion ΔAB'C' ∼ ΔABC.

This implies that

AB'/AB = B'C'/BC = AC'/AC (Corresponding sides of similar triangles)

A₇B' is parallel to A₅B by construction. Therefore

∠AA₇B' = ∠AA₅B

In ΔAA₇B' and ΔAA₅B,

∠A₇AB' = ∠A₅AB (Common angle)

∠AA₇B' = ∠AA₅B (Already Proven)

Therefore, by the AA similarity criterion ΔAA₇B' ~ ΔAA₅B.

This implies that

AB'/AB = AA₇/AA₅ (Corresponding sides of similar triangles)

By construction, AA₇ has 7 equal divisions of segment AA₇ and AA₅ has 5 equal divisions. Thus,

AA₇/AA₅ = 7/5

Therefore,

AB'/AB = B'C'/BC = AC'/AC = 7/5

AB' = (7/5)AB

B'C' = (7/5)BC

AC'= (7/5)AC

Hence, the construction is justified.

4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1� times the corresponding sides of the isosceles triangle.

Solution

Steps for the construction are as follows:

1. Draw a line segment BC = 8 cm.
2. Draw the perpendicular bisector of line segment BC which bisects it at the point D.
3. Taking the point D as the centre, draw an arc of radius 4 cm.
4. Mark the point at which the arc cuts the perpendicular bisector of BC as A.
5. Join A with points B and C.
6. ABC is the required triangle.
7. Draw a ray BX that makes an acute angle CBX.
8. Mark 3 points on the ray BX such that BB₁ = B₁B₂ = B₂B₃.
9. Join B₂ with point C.
10. Draw a line B₃C' so that it is parallel to B₂C and the point C' lies on the extended segment BC.
11. Draw a line C'A' so that it is parallel to AC and the point A' lies on the extended segment BA.
12. Now, A'BC' is the required second triangle.

Justification:

A'C' is parallel to AC by construction. Therefore

∠A'C'B = ∠ACB

In ΔA'BC' and ΔABC,

∠A'BC' = ∠ABC (Common angle)

∠A'C'B = ∠ACB (Already proven)

Therefore, by the AA similarity criterion ΔA'BC' ∼ ΔABC.

This implies that

A'B/AB = BC'/BC= A'C'/AC (Corresponding sides of similar triangles)

B₂C is parallel to B₃C' by construction. Therefore

∠BB₂C = ∠BB₃C'

In ΔBB₂C and ΔBB₃C',

∠CBB₂ = ∠C'BB₃ (Common angle)

∠BB₂C = ∠BB₃C' (Already Proven)

Therefore, by the AA similarity criterion ΔBB₂C and ΔBB₃C'.

This implies that

BB₂/BB₃ = BC/BC' (Corresponding sides of similar triangles)

By construction, BB₃ has 3 equal divisions of segment BB₃ and BB₂ has 2 equal divisions. Thus,

BB₂/BB₃ = 2/3

Therefore,

A'B/AB = BC'/BC = A'C'/AC = 3/2

A'B= (3/2)AB

BC' = (3/2)BC

A'C'= (3/2)AC

Hence, the construction is justified.

5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are ¾ of the corresponding sides of the triangle ABC.

Solution

Steps for the construction are as follows:

1. Draw a line segment BC = 6 cm.
2. Construct ∠B = 60° and produce the line.
3. Taking B as the centre draw an arc of radius 5 cm that cuts the line in step 2 and mark this point as A.
4. Triangle ABC has been drawn.
5. Draw a ray BX that makes an acute angle CBX.
6. Mark 4 points on the ray BX such that BB₁ = B₁B₂ = ... ... = B₃B₄.
7. Join B₄ with point C.
8. Draw a line B₃C' so that it is parallel to B₄C and the point C' lies on the segment BC.
9. Draw a line C'A' so that it is parallel to AC and the point A' lies on the segment BA.
10. Now, A'BC' is the required second triangle.

Justification:

A'C' is parallel to AC by construction. Therefore

∠A'C'B = ∠ACB

In ΔA'BC' and ΔABC,

∠A'BC' = ∠ABC (Common angle)

∠A'C'B = ∠ACB (Already proven)

Therefore, by the AA similarity criterion ΔA'BC' ∼ ΔABC.

This implies that

A'B/AB = BC'/BC= A'C'/AC (Corresponding sides of similar triangles)

B₄C is parallel to B₃C' by construction. Therefore

∠BB₄C = ∠BB₃C'

In ΔBB₄C and ΔBB₃C',

∠CBB₄ = ∠C'BB₃ (Common angle)

∠BB₄C = ∠BB₃C' (Already Proven)

Therefore, by the AA similarity criterion ΔBB₄C and ΔBB₃C'.

This implies that

BB₄/BB₃ = BC/BC' (Corresponding sides of similar triangles)

By construction, BB₃ has 3 equal divisions of segment BB₄ and BB₄ has 4 equal divisions. Thus,

BB₄/BB₃ = 4/3

Therefore,

A'B/AB = BC'/BC = A'C'/AC = 3/4

A'B= (3/4)AB

BC' = (3/4)BC

A'C'= (3/4)AC

Hence, the construction is justified.

6. Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ∆ ABC.

Solution

Steps for the construction are as follows:

1. Draw a line segment BC = 7 cm.
2. Construct ∠B = 45° and produce the line.
3. We can find ∠C = 30° by using the Angle Sum property of a triangle (∠C = 180° - ∠A - ∠B). Produce the line.
4. Mark the point at which the lines in step 2 and step 3 intersect as A.
5. ABC is the required triangle.
6. Draw a ray BX that makes an acute angle CBX.
7. Mark 4 points on the ray BX such that BB₁ = B₁B₂ = ... ... = B₃B₄.
8. Join B₃ with point C.
9. Draw a line B₄C' so that it is parallel to B₃C and the point C' lies on the extended segment BC.
10. Draw a line C'A' so that it is parallel to AC and the point A' lies on the extended segment BA.
11. Now, A'BC' is the required second triangle.

Justification:

A'C' is parallel to AC by construction. Therefore

∠A'C'B = ∠ACB

In ΔA'BC' and ΔABC,

∠A'BC' = ∠ABC (Common angle)

∠A'C'B = ∠ACB (Already proven)

Therefore, by the AA similarity criterion ΔA'BC' ∼ ΔABC.

This implies that

A'B/AB = BC'/BC= A'C'/AC (Corresponding sides of similar triangles)

B₃C is parallel to B₄C' by construction. Therefore

∠BB₃C = ∠BB₄C'

In ΔBB₃C and ΔBB₄C',

∠CBB₃ = ∠C'BB₄ (Common angle)

∠BB₃C = ∠BB₄C' (Already Proven)

Therefore, by the AA similarity criterion ΔBB₃C and ΔBB₄C'.

This implies that

BB₃/BB₄ = BC/BC' (Corresponding sides of similar triangles)

By construction, BB₃ has 3 equal divisions of segment BB₄ and BB₄ has 4 equal divisions. Thus,

BB₃/BB₄ = 3/4

Therefore,

A'B/AB = BC'/BC = A'C'/AC = 4/3

A'B= (4/3)AB

BC' = (4/3)BC

A'C'= (4/3)AC

Hence, the construction is justified.

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

Solution

Steps for the construction are as follows:

1. Draw a line segment BC = 3 cm.
2. Construct a right angle at B and produce the perpendicular line.
3. Taking the point B as a centre, draw an arc of radius 4 cm.
4. Mark the point at which the arc cuts the perpendicular as A.
5. ABC is the required triangle.
6. Draw a ray BX that makes an acute angle CBX.
7. Mark 5 points on the ray BX such that BB₁ = B₁B₂ = ... ... = B₄B₅.
8. Join B₃ with point C.
9. Draw a line B₅C' so that it is parallel to B₃C and the point C' lies on the extended segment BC.
10. Draw a line C'A' so that it is parallel to AC and the point A' lies on the extended segment BA.
11. Now, A'BC' is the required second triangle.

Justification:

A'C' is parallel to AC by construction. Therefore

∠A'C'B = ∠ACB

In ΔA'BC' and ΔABC,

∠A'BC' = ∠ABC (Common angle)

∠A'C'B = ∠ACB (Already proven)

Therefore, by the AA similarity criterion ΔA'BC' ∼ ΔABC.

This implies that

A'B/AB = BC'/BC= A'C'/AC (Corresponding sides of similar triangles)

B₃C is parallel to B₅C' by construction. Therefore

∠BB₃C = ∠BB₅C'

In ΔBB₃C and ΔBB₅C',

∠CBB₃ = ∠C'BB₅ (Common angle)

∠BB₃C = ∠BB₅C' (Already Proven)

Therefore, by the AA similarity criterion ΔBB₃C and ΔBB₅C'.

This implies that

BB₃/BB₅ = BC/BC' (Corresponding sides of similar triangles)

By construction, BB₃ has 3 equal divisions of segment BB₅ and BB₅ has 5 equal divisions. Thus,

BB₃/BB₅ = 3/5

Therefore,

A'B/AB = BC'/BC = A'C'/AC = 5/3

A'B= (5/3)AB

BC' = (5/3)BC

A'C'= (5/3)AC

Hence, the construction is justified.

Exercise 11.2

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.

Solution

Steps for the construction are as follows:

1. Draw a circle of radius 6 cm with O as the centre.
2. Join O to an external point A such that OA = 10 cm.
3. Construct the perpendicular bisector of the line OA which bisects it at the point D.
4. Taking the point D as the centre, draw a circle of radius OD or AD.
5. The circle with centre D will intersect the circle with centre O at the points B and C respectively.
6. Join the point A to B and C.
7. AB and AC are the required tangents.
8. On measurement, AB = AC = 8 cm.

Justification:

We need to connect O with B and C.

∠B is an angle in a semi-circle subtended by a diameter. Therefore, it will be a right angle.

This means that OB will be perpendicular to AB.

Since, the radius OB of the circle is perpendicular to the line AB that touches it at one point. Therefore, AB is a tangent to this circle.

Similarly, AC is a tangent to this circle.

Hence, the construction is justified.

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 calculation.

Solution

Steps for the construction are as follows:

1. Draw a circle of radius 4 cm with the centre O.
2. Taking O as the centre, draw another circle with radius 6 cm.
3. Mark a point A on the bigger circle.
4. Join the point A to the point O.
5. Construct the perpendicular bisector of the line OA which bisects it at the point D.
6. Taking the point D as the centre, draw a circle of radius OD or AD.
7. The circle with centre D will intersect the circle with radius 4 cm and centre O at the points B and C respectively.
8. Join the point A to B and C.
9. AB and AC are the required tangents.
10. On measurement, AB = AC = 4.47 cm.

Verification:

ABO forms a right triangle. Therefore, by using Pythagoras Theorem in ABO:

AO² = AB² + OB²

6² = AB² + 4²

36 - 16 = AB²

20 = AB²

AB = √20 cm

AB = 4.47 cm

Tangents to a circle from an external point are equal, so

AB = AC = 4.47 cm

The result has been verified.

Justification:

We need to connect O with B and C.

∠B is an angle in a semi-circle subtended by a diameter. Therefore, it will be a right angle.

This means that OB will be perpendicular to AB.

Since, the radius OB of the circle is perpendicular to the line AB that touches it at one point. Therefore, AB is a tangent to this circle.

Similarly, AC is a tangent to this circle.

Hence, the construction is justified.

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

Solution

Steps for the construction are as follow:

1. Draw a circle with radius 3 cm and centre O.
2. Draw a diameter for the circle and extend it to 7 cm on both sides from the centre O.
3. Mark the two end points of the extended diameter as P and Q.
4. Construct the perpendicular bisector of the line OP which bisects it at the point M.
5. Taking the point M as the centre, draw a circle of radius OM or PM.
6. The circle with centre M will intersect the circle with centre O at the points A and B respectively.
7. Join the point P to A and B.
8. Construct the perpendicular bisector of the line OQ which bisects it at the point N.
9. Taking the point N as the centre, draw a circle of radius ON or QN.
10. The circle with centre N will intersect the circle with centre O at the points C and D respectively.
11. Join the point Q to C and D.
12. PA, PB, QC, and QD are the required tangents.

Justification:

We need to connect O with A, B, C and D.

∠A is an angle in a semi-circle subtended by a diameter. Therefore, it will be a right angle.

This means that OA will be perpendicular to PA.

Since, the radius OA of the circle is perpendicular to the line PA that touches it at one point. Therefore, PA is a tangent to this circle.

Similarly, PB, QC, and QD are tangents to this circle.

Hence, the construction is justified.

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

Solution

Steps for the construction are as follows:

1. Draw a circle with radius 5 cm and centre O.
2. Mark any point on the circle as B and join it with O to get radius OB.
3. Construct ∠B = 90° and produce the line outwards.
4. Construct ∠O = 120° and produce the line until it meets the circumference of the circle. Mark this point as C.
5. Construct ∠C = 90° and produce the line outwards until it intersects the line in drawn in step 3. Mark this external point as A.
6. AB and AC are the required tangents that are inclined at 60°.

Justification:

By construction we have,

∠B = 90°

∠C = 90°

And ∠O = 120°

OABC forms a quadrilateral, therefore

∠O + ∠A + ∠B + ∠C = 360°

120° + ∠A + 90° + 90° = 360°

∠A + 300° = 360°

∠A = 60°

Hence, the construction is justified.

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

Solution

Steps for the construction are as follows:

1. Draw a line segment AB = 8 cm.
2. Taking the point A as the centre draw a circle of radius = 4 cm.
3. Taking the point B as the centre draw a circle of radius = 3 cm.
4. Construct the perpendicular bisector of the line AB which bisects it at the point D.
5. Taking the point D as the centre, draw a circle of radius AD or BD.
6. Mark the points at which the circle with centre D intersects the circle with the centre A as P and Q respectively.
7. Mark the points at which the circle with centre D intersects the circle with the centre B as R and S respectively.
8. Join the points P and Q to B.
9. Join the point R and S to A.
10. PB, QB, AR and AS are the required tangents.

Justification:

We need to connect A with P and Q and connect B with R and S.

∠P is an angle in a semi-circle subtended by a diameter. Therefore, it will be a right angle.

This means that AP will be perpendicular to PB.

Since, the radius AP of the circle is perpendicular to the line PB that touches it at one point. Therefore, PB is a tangent to this circle.

Similarly, QB is also a tangent to the circle with centre A.

Similarly, we can prove that AR and AS are tangents to the circle with the centre B.

Hence, the construction is justified.

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, D is drawn. Construct the tangents from A to this circle.

Solution

Steps for the construction are as follows:

1. Draw a line segment BC = 8 cm.
2. Construct a right angle at B and produce the perpendicular line.
3. Taking the point B as a centre, draw an arc of radius 6 cm.
4. Mark the point at which the arc cuts the perpendicular as A and join the point A to C.
5. ABC is the required triangle.
6. Draw a perpendicular on hypotenuse AC from the point B.
7. Construct the perpendicular bisector of the line BC which bisects it at the point O.
8. Taking the point O as the centre, draw a circle of radius BO or CO.
9. Join the points A and O.
10. Construct the perpendicular bisector of the line AO which bisects it at the point M.
11. Taking M as the centre, draw a circle of radius AM or MO.
12. The circle with centre M will intersect the circle with centre O at the points B and E.
13. Join E to A.
14. AB and AE are the required tangents.

Justification:

We need to join O with E.

∠E is an angle in a semi-circle subtended by a diameter. Therefore, it will be a right angle.

This means that OE will be perpendicular to AE.

Since, the radius OE of the circle is perpendicular to the line AE that touches it at one point. Therefore, AE is a tangent to this circle.

Similarly, AB is a tangent to this circle.

Hence, the construction is justified.

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

Solution

Steps for the construction are as follows:

1. Draw a circle using a bangle.
2. Draw two non-parallel chords AB and CD in the circle.
3. Construct the perpendicular bisectors of AB and CD and mark the point where they intersect as O which will be the centre of this circle.
4. Take a point P outside the circle and join it to O.
5. Construct the perpendicular bisector of line OP which bisects it at the point M.
6. Taking the point M as the centre, draw a circle with radius PM or OM.
7. Marks the points at which the circle with centre M intersects the circle with centre O as Q and R.
8. Join P with Q and R.
9. PQ and PR are the required tangents.

Justification:

We need to join the points Q and R with centre O.

∠Q is an angle in a semi-circle subtended by a diameter. Therefore, it will be a right angle.

This means that OQ will be perpendicular to PQ.

Since, the radius OQ of the circle is perpendicular to the line PQ that touches it at one point. Therefore, PQ is a tangent to this circle.

Similarly, PR is a tangent to this circle.

Hence, the construction is justified.