Chapter 11 Constructions

**NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1**

**In each of the following, give the justification of the construction also:**

**Question 1.**

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

**Solution:**

**Steps of Construction:
** 1. Draw a line segment AB = 7.6 cm.

2. Draw an acute angle BAX on base AB. Mark the ray as AX.

3. Locate 13 points A

_{1}, A

_{2}, A

_{3}, …… , A

_{13}on the ray AX so that AA

_{1}= A

_{1}A

_{2}= ……… = A

_{12}A

_{13 }4. Join A

_{13}with B and at A

_{5}draw a line ∥ to BA

_{13}, i.e. A

_{5}C. The line intersects AB at C.

5. On measure AC = 2.9 cm and BC = 4.7 cm.

**Justification:**

In ∆AA

_{5}C and ∆AA

_{13}B,

∴ AC : BC = 5 : 8

**Question 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.

**Solution:**

**Steps of Construction:
** 1. Construct a ΔABC with AB = 4 cm, BC = 6 cm and AC = 5 cm.

2. Draw an acute angle CBX on the base BC at point B. Mark the ray as BX.

3. Mark the ray BX with B

_{1}, B

_{2}, B

_{3}such that

BB

_{1}= B

_{1}B

_{2}= B

_{2}B

_{3 }4. Join B

_{3 }to C.

5. Draw B

_{2}C’ ∥ B3C, where C’ is a point on BC.

6. Draw C’A’ ∥ AC, where A’ is a point on BA.

7. ΔA’BC’ is the required triangle.

**Justification:**In ∆A’BC and ∆ABC,

**Question 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.

**Solution:**

**Steps of Construction:
** 1. Draw a ΔABC with AB = 5 cm, BC = 7 cm and AC = 6 cm.

2. Draw an acute angle CBX below BC at point B.

3. Mark the ray BX as B

_{1}, B

_{2}, B

_{3}, B

_{4}, B

_{5}, B

_{6}and B

_{7}such that BB

_{1}= B

_{1}B

_{2}= B

_{2}B

_{3}= B

_{3}B

_{4}= B

_{4}B

_{5}= B

_{5}B

_{6}= B

_{6}B

_{7}.

4. Join B

_{5}to C.

5. Draw B

_{7}C’ parallel to B

_{5}C, where C’ is a point on extended line BC.

6. Draw A’C’ ∥ AC, where A’ is a point on extended line BA.

A’BC’ is the required triangle.

**Justification:** In ∆ABC and ∆A’BC’,

**Question 4.**

Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 112 times the corresponding sides of the isosceles triangle.

**Solution:**

**Steps of Construction:
** 1. Draw base AB = 8 cm.

2. Draw perpendicular bisector of AB. Mark CD = 4 cm, on ⊥ bisector where D is mid-point on AB.

3. Draw an acute angle BAX, below AB at point A.

4. Mark the ray AX with A

_{1}, A

_{2}, A

_{3}such that AA

_{1}=A

_{1}A

_{2}= A

_{2}A

_{3 }5. Join A

_{2}to B. Draw A

_{3}B’ ∥ A

_{2}B, where B’ is a point on extended line AB.

6. At B’, draw B’C’ 11 BC, where C’ is a point on extended line AC.

7. ∆AB’C’ is the required triangle.

**Justification:** In ∆ABC and ∆A’BC’,

**Question 5.**

Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 34 of the corresponding sides of the triangle ABC.

**Solution:**

**Steps of Construction:
** 1. Draw a line segment BC = 6 cm and at point B draw an ∠ABC = 60°.

2. Cut AB = 5 cm. Join AC. We obtain a ΔABC.

3. Draw a ray BX making an acute angle with BC on the side opposite to the vertex A.

4. Locate 4 points A

_{1}, A

_{2}, A

_{3}and A

_{4}on the ray BX so that BA

_{1}= A

_{1}A

_{2}= A

_{2}A

_{3}= A

_{3}A

_{4}.

5. Join A

_{4}to C.

6. At A

_{3}, draw A

_{3}C’ ∥ A

_{4}C, where C’ is a point on the line segment BC.

7. At C’, draw C’A’ ∥ CA, where A’ is a point on the line segment BA.

∴ ∆A’BC’ is the required triangle.

**Question 6.**

Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 43 times the corresponding sides of ∆ABC.

**Solution:**

In ∆ABC, ∠A + ∠B + ∠C = 180°

⇒ 105° + 45° + ∠C = 180°

⇒ 150° + ∠C = 180°

⇒ ∠C = 30°

**Steps of Construction:
** 1. Draw a line segment BC = 7 cm. At point B, draw an ∠B = 45° and at point C, draw an ∠C = 30° and get ΔABC.

2. Draw an acute ∠CBX on the base BC at point B. Mark the ray BX with B

_{1}, B

_{2}, B

_{3}, B

_{4}, such that BB

_{1}= B

_{1}B

_{2}= B

_{2}B

_{3}= B

_{3}B

_{4 }3. Join B

_{3}to C.

4. Draw B

_{4}C’ ∥ B

_{3}C, where C’ is point on extended line segment BC.

5. At C’, draw C’A’ ∥ AC, where A’ is a point on extended line segment BA.

6. ∆A’BC’ is the required triangle.

**Justification:** In ∆ABC and ∆A’BC’,

**Question 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 53F times the corresponding sides of the given triangle.

**Solution:**

**Steps of Construction:
** 1. Draw a right angled ∆ABC with AB = 4 cm, AC = 3 cm and ∠A = 90°.

2. Make an acute angle BAX on the base AB at point A.

3. Mark the ray AX with A

_{1}, A

_{2}, A

_{3}, A

_{4}, A

_{5}such that AA

_{1}= A

_{1}A

_{2}= A

_{2}A

_{3}= A

_{3}A

_{4}= A

_{4}A

_{5}.

4. Join A

_{3}B. At A

_{5}, draw A

_{5}B’ ∥ A

_{3}B, where B’ is a point on extended line segment AB.

5. At B’, draw B’C’ ∥ BC, where C’ is a point on extended line segment AC.

6. ∆AB’C’ is the required triangle.

**Justification:**

In ∆ABC and ∆AB’C’,

**NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2**

**In each of the following, give the justification of the construction also:**

**Question 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 of Construction:
** 1. Draw a circle of radius 6 cm.

2. Mark a point P at a distance 10 cm from the centre O.

3. Here OP = 10 cm, draw perpendicular bisector of OP, which intersects OP at O’.

4. Take O’ as centre, draw a circle of radius O’O, which passes through O and P and intersects the previous circle at points T and Q.

5. Join PT and PQ, measure them, these are the required tangents.

**Justification:**

Join OT and OQ.

A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.

∴ PT = PQ.

**Question 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 of Construction:**

1. Draw concentric circles of radius OA = 4 cm and OP = 6 cm having same centre O.

2. Mark these circles as C and C’.

3. Points O, A and P lie on the same line.

4. Draw perpendicular bisector of OP, which intersects OP at O’.

5. Take O’ as centre, draw a circle of radius OO’ which intersects the circle C at points T and Q.

6. Join PT and PQ, these are the required tangents.

7. Length of these tangents are approx. 4.5 cm.

**Justification:**

Join OT and OQ.

A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.

∴ PT = PQ.

**Question 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 of Construction:
**

1. Draw a circle of radius 3 cm, having centre O. Mark this circle as C.

2. Mark points P and Q along its extended diameter such that OP = 7 cm and OQ = 7 cm.

3. Draw perpendicular bisector of OQ, intersecting OQ at O’.

4. Draw perpendicular bisector of OP intersecting OP at O”.

5. Take O’ as centre and draw a circle of radius OO’ which passes through points O and Q, intersecting circle C at points R and T.

6. Take O” as centre and draw a circle of radius O”P which passes through points O and P, intersecting the, circle C at points S and U.

7. Join QR and QT; PS and PU, these are the required tangents.

**Justification:**

Join OR.

In ∆OQR,

OR ⊥ QR [Radius ± to tangent]

OQ

^{2}= OR

^{2}+ QR

^{2}

⇒ (7)

^{2}= (3)

^{2}+ QR

^{2}⇒ 49 = 9 + OR

^{2}

⇒ 40 = QR

^{2}⇒ QR = 2 10−−√ cm

Similary,

QT = 2 10−−√ cm

PS = 2 10−−√ cm

PU = 2 10−−√ cm

A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.

∴ QR = QT and PS = PU

**Question 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 of Construction:**

1. Draw a circle of radius 5 cm.

2. As tangents are inclined to each other at an angle of 60°.

∴ Angle between the radii of circle is 120°. (Use quadrilateral property)

3. Draw radii OA and OB inclined to each other at an angle 120°.

4. At points A and B, draw 90° angles. The arms of these angles intersect at point P.

5. PA and PB are the required tangents.

**Justification:**

In quadrilateral AOBP,

AP and BP are the tangents to the circle.

Join OP.

In right angled AOAP, OA ⊥ PA [Radius is ⊥ to tangent]

A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.

∴ PA = PB

**Question 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 of Construction:**

1. Draw a line segment AB = 8 cm.

2. Taking A as centre draw a circle C of radius 4 cm and taking B as centre draw a circle C’ of radius 3 cm.

3. Draw perpendicular bisector of AB, which intersects AB at point O.

4. Taking point O as centre draw a circle of radius 4 cm passing through points A and B which intersect circle C at P and S and circle C’ at points R and Q.

5. Join AQ, AR, BP and BS. These are the required tangents.

**Justification:**

Join AP.

A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.

∴ AQ = AR and BP = BS.

**Question 6.**

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

**Solution:**

**Steps of Construction:**

1. Draw a right triangle ABC with AB = 6 cm, BC = 8 cm and ZB = 90°.

2. From B, draw BD perpendicular to AC.

3. Draw perpendicular bisector of BC which intersect BC at point O’.

4. Take O’ as centre and O’B as radius, draw a circle C’ passes through points B, C and D.

5. Join O’A and draw perpendicular bisector of O’A which intersect O’A at point K.

6. Take K as centre, draw an arc of radius KO’ intersect the previous circle C’ at T.

7. Join AT, AT is required tangent.

**Justification:**

∠BDC = 90°

∴ BC acts as diameter.

AB is tangent to circle having centre O’

A pair of tangents can be drawn to a circle from an external point outside the circle. These two tangents are equal in lengths.

∴ AB = AT.

**Question 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 of Construction:**

1. Draw a circle C’ with the help of a bangle, for finding the centre, take three non-collinear points A, B and C, lying on the circle. Join AB and BC and draw perpendicular bisector of AB and BC, both intersect at a point O,

‘O’ is centre of the circle.

2. Take a point P outside the circle. Join OP.

3. Draw perpendicular bisector of OP, which intersects OP at point O’.

4. Take O’ as the centre with OO’ as radius draw a circle which passes through O and P, intersecting previous circle C’ at points R and Q.

5. Join PQ and PR.

6. PQ and PR are the required pair of tangents.

**Justification:**

Join OQ and OR.

In AOQP and AOPR, OQ = OR [Radii of the circle]

OP = OP [Common]

∠Q = ∠R = 90° [Radius is ⊥ to tangent]

∆OQP ≅ ∆ORP [by RHS]

∴ PQ = PR [By C.P.C.T]

A pair of tangents can be drawn to a circle from an external point lying outside the circle. These two tangents are equal in lengths.

∴ PQ = PR