NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1.

• Constructions Class 10 Ex 11.1
• Constructions Class 10 Ex 11.2

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 11
Chapter Name	Constructions
Exercise	Ex 11.1
Number of Questions Solved	7
Category	NCERT Solutions

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 another line segment (or ray) AX at any convenient acute angle with AB.
3. Along AX, locate 13 (5 + 8) points A₁, A₂, A₃, …… , A₁₃ such that AA₁ = A₁A₂ = ……… = A₁₂A₁₃
4. Join BA₁₃.
5. At point, A₅ draw a straight line  A₅C parallel to A₁₃B making an angle equal to ∠AA₁₃B.
Thus, the point C divides the line segment AB in the ratio 5: 8.
Justification:

Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are of the corresponding sides of the first triangle.
Solution:
Steps of Construction:
1. Draw AC – 6 cm (ii) With A and Cas centers and radii 4 cm.
2. 5 cm respectively draw two ares intersecting each other at B. Join BA and BC.
3. Draw a ray AY making an acute angle with AC.
4. Locate three points and Ron AY, such that AP – POQR.
5. Join CR.
6. Through o. draw a line QC’ parallel to RC (by making an angle equal to ∠ARC) meeting the line segment AC at C’.
7. Similarly, through C, draw a line B’C parallel to CB
Thus, ABC is the required triangle, which is similar to ∆ABC with scale factor
Justification:
By construction, we have:

Question 3.
Construct a triangle with sides 5 cm, 6 cm, and 7 cm and then another triangle whose sides are of the corresponding sides of the first triangle.
Solution:
Steps of Construction:
1. Construct AABC such that AB = 5 cm, BC = 6 cm and CA = 7 cm.
2. Draw a ray BX making an acute angle with BC.
3. Locate 7 points on BX such that BD = DE = EF = FG – GH – IJ.
4. Join HC.
5. Through J, draw a line JC parallel to HC, meeting produced line BC at C’.
6. Through C’, draw a line 7 parallel to CA meeting the produced line BA at A’.
Thus A∆ABC is a required triangle
Justification:
By construction, we have:

Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1 times the corresponding sides of the isosceles triangle.
Solution:
Steps of Construction:
1. Construct an isosceles triangle ABC in which BC – 8 cm and altitude AD is 4 cm.
2. Draw a ray BX, making an acute angle with BC.
3. Locate 3 points on BX, such that BP – PQ = QR.
4. Join QC.
5. Through R, draw a line RC parallel to QC, meeting produced line BC at C’.
6. Through C, draw a line CA parallel to CA, meeting the produced line BA at A’.
Thus, ∆A’BC’ is the required isosceles triangle
Justification:

Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are of the corresponding sides of the triangle ABC.
Solution:
Steps of Construction:
1. Construct a ABC in which BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
2. Draw a ray BX making any acute angle with BC.
3. Locate 4 points B₁, B2, B₃ and B₄ on BX, such that BB₁ = B₁B₂ = B₂B₃ =B₃B₄
4. Join B₄C.
5. Through B₃ draw a line B₃C’ parallel to B₄C, intersecting the line segment BC at C’.
6.  Through C, draw a line parallel to CA, intersecting the line segment BA at A’.
Thus, ∆A’BC’ is the required triangle.
Justification:

Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are times the corresponding sides of ∆ABC.
Solution:
Steps of Construction:
1. Draw a line segment BC – 7 cm.
2. Draw ∠ABC = 45° and ∠ACB = 30°, i.e., ∠BAC = 105°.
3. We get ∆ABC
4. Draw a ray BX making an acute angle with BC
5. Mark four points B₁, B₂, B₃ and B₄ on BX, such that BB₁ = B₁B₂ = B₂B₃ = B₃B_4.
6.  Join B₃C.
7. Through B₄ draw a line B₄C’ parallel to B₃C, intersecting the extended line segment BC at C’.
8. Through C’, draw a line A’C’ parallel to CA, intersecting the extended line segment BA at A’.
Thus, A’BC’ is the required triangle.
Justification:

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 times the corresponding sides of the given triangle.
Solution:
Steps of Construction:
1. Construct a ∆ABC, such that BC = 4 cm, CA = 3 cm and ∠BCA = 90°
2. Draw a ray BX making an acute angle with BC.
3. Mark five points B₁, B₂, B₃, B₄ and B₅ on BX, such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B_5.
4. Join B₃C.
5. Through B₅, draw B₅C’ parallel to B₃C intersecting BC produced at C’.
6. Through C’, draw C’A’ parallel to CA intersecting AB produced at A’.
Thus, ∆A’BC’ is the required right triangle.
Justification:

 

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