These NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.1

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:

- Draw a line segment AB = 7.6 cm.
- Draw an acute angle BAX on base AB. Mark the ray as AX.
- 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} - 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.

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

Then AC : CB = 5 : 8

Since A_{3}C is parallel to A_{13}B therefore

This show that C divides AB in the ratio 5

Question 2.

Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac { 2 }{ 3 }\) of the corresponding sides of the first triangle.

Solution:

Steps of Construction:

First we draw line AB 6 cm and then cut an arc AC = 4 cm. After that cut an another are BC = 5 cm. Join A to C and B to C. We find a AABC.

- Draw ray AX making an acute angle with AB.
- Locate points A
_{1}, A_{2}, A_{3}, A_{4}, A_{5}on AX so that AA_{1}= A_{1}A_{3}, = A_{2}A_{3} - Join BA
_{3}. - Join A
_{2}B’ such that A_{2}B’ || A_{3}B and (∠AA_{3}B = ∠AA_{2}B’) - Through B draw a ray B’C || BC’ and ∠ABC – ∠AB’C.

Hence ABC is the required triangle.

Question 3.

Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac { 7 }{ 5 }\) of the corresponding sides of the first triangle.

Solution:

Steps of Construction:

- Draw a ΔABC with AB = 5 cm, BC = 7 cm and AC = 6 cm.
- Draw an acute angle CBA below BC at point B.
- 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}. - Join B
_{5}to C. - Draw B
_{7}C’ parallel to B_{5}C, where C’ is a point on extended line BC. - Draw A’C’ ∥ AC, where A’ is a point on extended line BA.

∴ A’BC’ is the required triangle.

Question 4.

Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1\(\frac { 1 }{ 2 }\) times the corresponding sides of the isosceles triangle.

Solution:

Steps of Construction:

- Draw base AB = 8 cm.
- Draw perpendicular bisector of AB. Mark CD = 4 cm, on ⊥ bisector where D is mid-point on AB.
- Draw an acute angle BAX, below AB at point A.
- Mark the ray AX with A
_{1}, A_{2}, A_{3}such that AA_{1}=A_{1}A_{2}= A_{2}A_{3 } - Join A
_{2}to B. Draw A_{3}B’ ∥ A_{2}B, where B’ is a point on extended line AB. - At B’, draw B’C’ 11 BC, where C’ is a point on extended line AC.

∴ ∆AB’C’ is the required triangle.

Question 5.

Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are \(\frac { 3 }{ 4 }\) of the corresponding sides of the triangle ABC.

Solution:

Steps of Construction:

- Draw a line segment BC = 6 cm and at point B draw an ∠ABC = 60°.
- Cut AB = 5 cm. Join AC. We obtain a ΔABC.
- Draw a ray BX making an acute angle with BC on the side opposite to the vertex A.
- 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}. - Join A
_{4}to C. - At A
_{3}, draw A_{3}C’ ∥ A_{4}C, where C’ is a point on the line segment BC. - 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 \(\frac { 4 }{ 3 }\) 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:

- 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.
- 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} - Join B
_{3}to C. - Draw B
_{4}C’ ∥ B_{3}C, where C’ is point on extended line segment BC. - At C’, draw C’A’ ∥ AC, where A’ is a point on extended line segment BA.

∴ ∆A’BC’ is the required triangle.

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 \(\frac { 5 }{ 3F }\) times the corresponding sides of the given triangle.

Solution:

Steps of Construction:

- Draw a right angled ∆ABC with AB = 4 cm, AC = 3 cm and ∠A = 90°.
- Make an acute angle BAX on the base AB at point A.
- 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}. - Join A
_{3}B. At A_{5}, draw A_{5}B’ ∥ A_{3}B, where B’ is a point on extended line segment AB. - At B’, draw B’C’ ∥ BC, where C’ is a point on extended line segment AC.

∴ ∆AB’C’ is the required triangle.