# Class 10 Maths Chapter 11 Constructions

Posted on: October 29, 2021 Posted by: user Comments: 0

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 A1, A2, A3, …… , A13 on the ray AX so that AA1 = A1A2 = ……… = A12A13
4. Join A13 with B and at A5 draw a line ∥ to BA13, i.e. A5C. The line intersects AB at C.
5. On measure AC = 2.9 cm and BC = 4.7 cm. Justification:
In ∆AA5C and ∆AA13B, ∴ 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 B1, B2, B3 such that
BB1 = B1B2 = B2B3
4. Join Bto C.
5. Draw B2C’ ∥ 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 B1, B2, B3, B4, B5, B6 and B7 such that BB1= B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7.
4. Join B5 to C.
5. Draw B7C’ parallel to B5C, 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 A1, A2, A3 such that AA1 =A1A2 = A2A3
5. Join A2 to B. Draw A3B’ ∥ A2 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 A1, A2, A3 and A4 on the ray BX so that BA1 = A1A2 = A2A3 = A3A4.
5. Join A4 to C.
6. At A3, draw A3C’ ∥ A4C, 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 B1, B2, B3, B4, such that BB1 = B1B2 = B2B3 = B3B4
3. Join B3 to C.
4. Draw B4C’ ∥ B3C, 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 A1, A2, A3, A4, A5 such that AA1 = A1A2 = A2A3 = A3A4 = A4A5.
4. Join A3B. At A5, draw A5B’ ∥ A3B, 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]
OQ2 = OR2 + QR2
⇒ (7)2 = (3)2 + QR2 ⇒ 49 = 9 + OR2
⇒ 40 = QR2 ⇒ 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:
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