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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.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 1
Justification:
In ∆AA5C and ∆AA13B,
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 2
∴ 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.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 3
Justification: In ∆A’BC and ∆ABC,
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 4

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.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 5
Justification: In ∆ABC and ∆A’BC’,
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 6
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 7

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.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 8
Justification: In ∆ABC and ∆A’BC’,
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 9

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.

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

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.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 12
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.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 13
Justification:
In ∆ABC and ∆AB’C’,
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 14
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 15

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.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 1
Justification:
Join OT and OQ.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 2
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.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 3
Justification:
Join OT and OQ.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 4
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:
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 5

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:
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 6
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]
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 7
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:

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 8
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.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 9
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.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 10
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’
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 11
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.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 12
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

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