BSEB Class 10 Maths Chapter 11 Constructions Ex 11.1 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 11 Constructions Ex 11.1 Book Answers ~ HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

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BSEB Class 10 Maths Chapter 11 Constructions Ex 11.1 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 11 Constructions Ex 11.1 Book Answers

BSEB Class 10th Maths Chapter 11 Constructions Ex 11.1 Textbooks Solutions and answers for students are now available in pdf format. Bihar Board Class 10th Maths Chapter 11 Constructions Ex 11.1 Book answers and solutions are one of the most important study materials for any student. The Bihar Board Class 10th Maths Chapter 11 Constructions Ex 11.1 books are published by the Bihar Board Publishers. These Bihar Board Class 10th Maths Chapter 11 Constructions Ex 11.1 textbooks are prepared by a group of expert faculty members. Students can download these BSEB STD 10th Maths Chapter 11 Constructions Ex 11.1 book solutions pdf online from this page.

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

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

Board	BSEB
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	10th
Subject	Maths Chapter 11 Constructions Ex 11.1
Chapters	All
Provider	Hsslive

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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 :

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, 𝐴𝑃𝑃𝐵 = AA5 A5 A13 = 58 [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 23 of the corresponding sides of it.
Solution:
Steps of Construction :

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 23 of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore
∆ ABC ~ ∆ EBD
and 𝐸𝐵𝐴𝐵 = 𝐷𝐸𝐶𝐴 = 𝐵𝐷𝐵𝐶 = 23 [Since 𝐵𝐷𝐵𝐶 = 23 ,by construction]
Hence, we get the new triangle similar to the given triangle whose sides are equal to 23 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 75 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.

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 75 of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore
∆ ABC ~ ∆ EBD and 𝐸𝐵𝐴𝐵 = 𝐷𝐸𝐶𝐴 = 𝐵𝐷𝐵𝐶 = 75
Hence, we get the new triangle similar to the given triangle whose sides are equal to 75 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 112 times the corresponding sides of the isosceles triangle.
Solution:
Steps of Construction :

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 𝐸𝐵𝐴𝐵 = 𝐷𝐸𝐶𝐴 = 𝐵𝐷𝐵𝐶 = 32
Hence, we get the new triangle similar to the given 32 triangle whose sides are 32, i.e., 1 12 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 34 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.

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 34 th of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore ∆ ABC ~ ∆ EBD
and 𝐸𝐵𝐴𝐵 = 𝐷𝐸𝐶𝐴 = 𝐵𝐷𝐵𝐶 = 34
Hence, we get the new triangle similar to the given triangle whose sides are equal to 34 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 43 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).

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 43 times of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore ∆ ABC ~ ∆ EBD
and 𝐸𝐵𝐴𝐵 = 𝐷𝐸𝐶𝐴 = 𝐵𝐷𝐵𝐶 = 43
Hence, we get the new triangle similar to the given triangle whose sides are equal to 43 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 53 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.

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 53 times of the corresponding sides of ∆ ABC.

Justification :
Since DE || CA, therefore ∆ ABC ~ ∆ EBD
and 𝐸𝐵𝐴𝐵 = 𝐷𝐸𝐶𝐴 = 𝐵𝐷𝐵𝐶 = 53
Hence, we get the new triangle similar to the given triangle whose sides are equal to 53 times of the corresponding sides of ∆ ABC.

BSEB Textbook Solutions PDF for Class 10th

Bihar Board Class 10th Maths Chapter 11 Constructions Ex 11.1 Textbooks for Exam Preparations

Bihar Board Class 10th Maths Chapter 11 Constructions Ex 11.1 Textbook Solutions can be of great help in your Bihar Board Class 10th Maths Chapter 11 Constructions Ex 11.1 exam preparation. The BSEB STD 10th Maths Chapter 11 Constructions Ex 11.1 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 10th Maths Chapter 11 Constructions Ex 11.1 Books State Board syllabus with maximum efficiency.

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