NCERT Solutions for class 10 Mathematics Chapter 11: Constructions

NCERT Solutions for Class 10 Mathematics Chapter 11 Constructions are provided in this article. Some of the important topics in Constructions chapter include:

Expected no of questions: 1 to 2 questions of total 4 marks

Download PDF: NCERT Solutions for Class 10 Mathematics Chapter 11 pdf

NCERT Solutions for Class 10 Mathematics Chapter 11

NCERT Solutions for Class 10 Mathematics Chapter 11 Constructions is given below.

NCERT Solutions

Class 10 Mathematics Chapter 11 Constructions – Important Topics

Construction here in geometry means drawing geometrical figures such as shapes like circles, lines with the help of a compass or a ruler/scale. Note: You cannot measure angles using a protractor, or measure lengths with a ruler for constructions.

Some of the construction techniques covered in this chapter include:

Bisection of a Line Segment
Division of a Line Segment in the ratio m:n
Construction of Triangle with a scale factor m:n
Construction of Tangent to the Circle from a Point Outside the Circle
Construction of Tangent to the Circle from a Point on the Circle

NCERT Solutions for Class 10 Chapter 11 Exercises

NCERT Solutions for Class 10 Chapter 11 Constructions Exercises is given below.

Chapter Related Articles:

Line Segment	Line and ray	Equilateral Triangle
Scalene Triangle	Division Line Segments	Constructions MCQs
Formula of Perimeter Shapes	Area and Volume	Properties of Triangle

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NCERT Solutions for Class 10 Mathematics	Important Maths Formulas	Maths Study material
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Geometry	Number System	Mensuration

CBSE X Related Questions

1.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.
View Solution
2.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :
- \( 130^\circ \)
- \( 50^\circ \)
- \( 120^\circ \)
- \( 90^\circ \)
View Solution
3.
There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?
- 144
- 2
- 420
- 272
View Solution
4.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals
- \(2.5\) cm
- \(10.5\) cm
- \(3.5\) cm
- \(\frac{14}{3}\) cm
View Solution
5.
The HCF of 960 and 432 is :
- 48
- 54
- 72
- 36
View Solution
6.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)
View Solution

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NCERT Solutions for class 10 Mathematics Chapter 11: Constructions

Download PDF: NCERT Solutions for Class 10 Mathematics Chapter 11 pdf

NCERT Solutions for Class 10 Mathematics Chapter 11

Class 10 Mathematics Chapter 11 Constructions – Important Topics

NCERT Solutions for Class 10 Chapter 11 Exercises

CBSE X Related Questions

1.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

2.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

3.
There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?

4.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

5.
The HCF of 960 and 432 is :

6.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

Similar Mathematics Concepts

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NCERT Solutions for class 10 Mathematics Chapter 11: Constructions

Download PDF: NCERT Solutions for Class 10 Mathematics Chapter 11 pdf

NCERT Solutions for Class 10 Mathematics Chapter 11

Class 10 Mathematics Chapter 11 Constructions – Important Topics

​NCERT Solutions for Class 10 Chapter 11 Exercises

CBSE X Related Questions

1.D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

2.In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

3.There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?

4.In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

5.The HCF of 960 and 432 is :

6.Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

NCERT Solutions for Class 10 Chapter 11 Exercises

1.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

2.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

3.
There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?

4.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

5.
The HCF of 960 and 432 is :

6.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)