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NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

Complete solutions of Ex 11.1 Class 10 Maths Chapter 11 with additional questions and answers from new NCERT syllabus textbook Class 10 Maths.

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:

Ex 11.1 Class 10 Maths 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.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions 1
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:
NCERT Solutions for Class 10 Maths Chapter 11 Constructions 2

Ex 11.1 Class 10 Maths 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:
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’.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions 3
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 \(\frac { 2 }{ 3 } \)
Justification:
By construction, we have:
NCERT Solutions for Class 10 Maths Chapter 11 Constructions 4

Ex 11.1 Class 10 Maths 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. 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’.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions 5
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:
NCERT Solutions for Class 10 Maths Chapter 11 Constructions 6

Ex 11.1 Class 10 Maths 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:
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. T
hrough C, draw a line CA parallel to CA, meeting the produced line BA at A’.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions 7
Thus, ∆A’BC’ is the required isosceles triangle
Justification:
NCERT Solutions for Class 10 Maths Chapter 11 Constructions 8

Ex 11.1 Class 10 Maths 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. 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’.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions 9
Thus, ∆A’BC’ is the required triangle.
Justification:
NCERT Solutions for Class 10 Maths Chapter 11 Constructions 10

Ex 11.1 Class 10 Maths 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. 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 = B3B4.
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.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions 11
Justification:
NCERT Solutions for Class 10 Maths Chapter 11 Constructions 12

Ex 11.1 Class 10 Maths 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:
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 = B4B5.
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.
NCERT Solutions for Class 10 Maths Chapter 11 Constructions 13
Justification:

NCERT Solutions for Class 10 Maths Chapter 11 Constructions 14

We hope the NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1, drop a comment below and we will get back to you at the earliest.

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