Bihar Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1

Bihar Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1 Textbook Questions and Answers.

BSEB Bihar Board Class 10th Maths Solutions Chapter 11 Constructions Ex 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 :
Bihar Board Class 10th Maths Solutions 11 Constructions Ex 11.1 1
1. Draw a line segment AB = 7.6 cm.

2. Draw a ray AC making any acute angle with AB, as shown in the figure.

3. On ray AC, starting from A, mark 5 + 8 = 13 equal line segments : AA₁, A₁A₂, A₂A_3, A₃A₄, A₄A₅, A₅A₆, A₆A₇, A₇A₈, A₈A₉, A₉A₁₀, A₁₀A₁₁, A₁₁A₁₂ and A₁₂A₁₃.

4. Join A₁₃B.

5. From A₅, draw A₅P || A₁₃B, meeting AB at P.

6. Thus, P divides AB in the ratio 5 : 8.
On measuring the two parts, we find that AP = 2.9 cm and PB = 4.7 cm (approx.).

Justification :
In ∆ ABA₁₃, PA₅ || BA₁₃.
So, \(\frac { AP }{ PB }\) = \(\frac{\mathrm{AA}_{5}}{\mathrm{~A}_{5} \mathrm{~A}_{13}}\) = \(\frac { 5 }{ 8 }\) [By BPT]
Thus, AP : PB = 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 \(\frac { 2 }{ 3 }\) of the corresponding sides of it.
Solution:
Steps of Construction :
Bihar Board Class 10th Maths Solutions 11 Constructions Ex 11.1 2
1. Draw a line segment BC = 6 cm.

2. With B as centre and radius equal to 5 cm, draw an arc.

3. With C as centre and radius equal to 4 cm, draw an arc intersecting the previously drawn arc at A.

4. Join AB and AC. Then ∆ ABC is the required triangle.

5. Below BC, make an acute angle CBX.

6. Along BX, mark off three (bigger of the numerator and denominator) points : B₁, B₁ and B₃ such that BB₁ = B₁B₂ = B₂B₃

7. Join B₃C. [Note that 3 is denominator]

8. From B₂ draw B₂D || B₃C, meeting BC at D. [Note that 2 is numerator]

9. From D, draw DE || CA, meeting BA at E. Then, ∆ EBD is the required triangle whose sides are \(\frac { 2 }{ 3 }\) of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore
∆ ABC ~ ∆ EBD
and \(\frac { EB }{ AB }\) = \(\frac { DE }{ CA }\) = \(\frac { BD }{ BC }\) = \(\frac { 2 }{ 3 }\) [Since \(\frac { BD }{ BC }\) = \(\frac { 2 }{ 3 }\) ,by construction]
Hence, we get the new triangle similar to the given triangle whose sides are equal to \(\frac { 2 }{ 3 }\) of the corresponding sides of ∆ ABC.

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 :
1. With the given data, construct ∆ ABC in which BC = 7 cm, CA = 4 cm and AB = 5 cm.

2. Below BC, make an acute angle CB₁.
Bihar Board Class 10th Maths Solutions 11 Constructions Ex 11.1 3
3. Along BX, mark off seven (bigger of the numerator and denominator) points : B₁, B₂, B₃, B₄, B₅, B₆ and B₇ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇

4. Join B₅C. [Note that 5 is denominator]

5. From B₇, draw B₇D || B₅C, meeting BC produced at D. [Note that 7 is numerator]

6. From D, draw DE || CA, meeting BA produced at E. Then, ∆ EBD is the required triangle whose sides are \(\frac { 7 }{ 5}\) of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore
∆ ABC ~ ∆ EBD and \(\frac { EB }{ AB }\) = \(\frac { DE }{ CA }\) = \(\frac { BD }{ BC }\) = \(\frac { 7 }{ 5 }\)
Hence, we get the new triangle similar to the given triangle whose sides are equal to \(\frac { 7 }{ 5 }\) of the corresponding sides of ∆ ABC.

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 :
Bihar Board Class 10th Maths Solutions 11 Constructions Ex 11.1 4
1. Draw BC = 8 cm.

2. Construct PQ, the perpendicular bisector of line segment BC meeting BC at M.

3. Along MP, cut off MA = 4 cm.

4. Join BA and CA. Then, ∆ ABC so obtained is the required triangle.

5. Make an acute angle CBX below BC.

6. Mark three points B₁, B₂ and B₃ on BX such that BB₁ = B₁B₂ = B₂B₃

7. Join BA₂C.

8. From B₃, draw B₃D || B₂C meeting BC produced at D.

9. From D, draw DE || CA meeting BA produced at E. Then, ∆ EBD is the required triangle.

Justification :
Since DE || CA, therefore ∆ ABC ~ ∆ EBD
and \(\frac { EB }{ AB }\) = \(\frac { DE }{ CA }\) = \(\frac { BD }{ BC }\) = \(\frac { 3 }{ 2 }\)
Hence, we get the new triangle similar to the given \(\frac { 3 }{ 2 }\) triangle whose sides are \(\frac { 3 }{ 2 }\), i.e., 1 \(\frac { 1 }{ 2 }\) times of the corresponding sides of the isosceles ∆ ABC.

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 :
1. With the given data, construct ∆ ABC in which BC 6 cm, ∠ABC = 60° and AB = 5 cm.

2. Below BC, make an acute angle CB₁.

3. Along BX, mark off 4 points : B₁, B₂, B₃ and B₄ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.

4. Join B₄C.
Bihar Board Class 10th Maths Solutions 11 Constructions Ex 11.1 5

5. From B₃, draw B₃D || B₄C to meet BC at D.

6. From D, draw ED || AC, meeting BA at E. Then, ∆ EBD is the required triangle whose sides are \(\frac { 3 }{ 4 }\) th of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore ∆ ABC ~ ∆ EBD
and \(\frac { EB }{ AB }\) = \(\frac { DE }{ CA }\) = \(\frac { BD }{ BC }\) = \(\frac { 3 }{ 4 }\)
Hence, we get the new triangle similar to the given triangle whose sides are equal to \(\frac { 3 }{ 4 }\) of the corresponding sides of ∆ ABC.

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:
Steps of Construction :
1. With the given data, construct ∆ ABC in which BC = 7 cm, ∠B = 45°, ∠C = 180° – (∠A + ∠B).
Bihar Board Class 10th Maths Solutions 11 Constructions Ex 11.1 6
i. e., ∠C = 180° – (105° + 45°)
= 180° – 150° = 30°.

2. Below BC, make an acute ∠CBX.

3. Along BX, mark off four points : B₁, B₂, B₃ and B₄ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.

4. Join B₃C.

5. From B₄, draw B₄D || B₃C, meeting BC produced at D.

6. From D, draw DE || CA, meeting BA produced at E. Then, ∆ EBD is the required triangle whose sides are \(\frac { 4 }{ 3 }\) times of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore ∆ ABC ~ ∆ EBD
and \(\frac { EB }{ AB }\) = \(\frac { DE }{ CA }\) = \(\frac { BD }{ BC }\) = \(\frac { 4 }{ 3 }\)
Hence, we get the new triangle similar to the given triangle whose sides are equal to \(\frac { 4 }{ 3 }\) times of the corresponding I sides of ∆ ABC.

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 \(\frac { 5 }{ 3 }\) times the corresponding sides of the given triangle.
Solution:
Steps of Construction :
1. With given data, construct ∆ ABC in which BC = 4 cm, ∠B = 90° and BA = 3 cm.

2. Below BC, make an acute angle CBX.

3. Along BX, mark off four points : B₁, B₂, B₃ and B₄ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.

4. Join B₃C.
Bihar Board Class 10th Maths Solutions 11 Constructions Ex 11.1 7
5. From B₅, draw B₅D || B₃C, meeting BC produced at D.

6. From D, draw DE || CA, meeting BA produced at E.
Then, EBD is the required triangle whose sides are \(\frac { 5 }{ 3 }\) times of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore ∆ ABC ~ ∆ EBD
and \(\frac { EB }{ AB }\) = \(\frac { DE }{ CA }\) = \(\frac { BD }{ BC }\) = \(\frac { 5 }{ 3 }\)
Hence, we get the new triangle similar to the given triangle whose sides are equal to \(\frac { 5 }{ 3 }\) times of the corresponding sides of ∆ ABC.