NCERT Solutions for Class 8 Mathematics Chapter 3: Understanding Quadrilaterals

NCERT Solutions for class 8 Mathematics Chapter 3 Understanding Quadrilaterals are provided in the article below. Some of the important topics in the Understanding Quadrilaterals chapter include:

Download PDF: NCERT Solutions for Class 8 Mathematics Chapter 3 pdf

NCERT Solutions for Class 8 Mathematics Chapter 3

NCERT Solutions for Class 8 Mathematics Chapter 3 Understanding Quadrilaterals is given below.

NCERT Solutions for Class 8 Mathematics

Quadrilateral is any figure that has 4 sides and whose interior angles are 360 degrees. Quadrilaterals are divided into several categories such as:

Square: A square is a quadrilateral, or a special type of parallelogram, with all of its sides equal.
Rectangle: A rectangle is a quadrilateral with parallel and equal opposite sides.
Parallelogram: A quadrilateral with opposite parallel sides is known as a parallelogram (and therefore opposite angles equal).
Rhombus: A quadrilateral with four equal-length sides is known as a rhombus. Because of its characteristic of equality of length, it is also known as an equilateral quadrilateral.
Trapezoid: A trapezium is a quadrilateral having at least one pair of parallel sides that is convex in shape.

Area of Square = a²

Area of Rectabgle = Length x Breadth

Area of Parallelogram = Base x height

Area of Rhombus = ½ x diagonal 1 x diagonal 2

Area of Trapezium = ½ x (a + b) x h

NCERT Solutions for Class 8 Maths Chapter 3 Exercises

NCERT Solutions for Class 8 Maths Chapter 3 Exercises are given below.

Read More:

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CBSE X Related Questions

1.
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
2.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).
View Solution
3.
The HCF of 960 and 432 is :
- 48
- 54
- 72
- 36
View Solution
4.
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
5.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.
View Solution
6.
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

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NCERT Solutions for Class 8 Mathematics Chapter 3: Understanding Quadrilaterals

Download PDF: NCERT Solutions for Class 8 Mathematics Chapter 3 pdf

NCERT Solutions for Class 8 Mathematics Chapter 3

NCERT Solutions for Class 8 Mathematics

NCERT Solutions for Class 8 Maths Chapter 3 Exercises

CBSE X Related Questions

1.
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 ?

2.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

3.
The HCF of 960 and 432 is :

4.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

5.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

6.
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 :

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NCERT Solutions for Class 8 Mathematics Chapter 3: Understanding Quadrilaterals

Download PDF: NCERT Solutions for Class 8 Mathematics Chapter 3 pdf

NCERT Solutions for Class 8 Mathematics Chapter 3

NCERT Solutions for Class 8 Mathematics

NCERT Solutions for Class 8 Maths Chapter 3 Exercises

CBSE X Related Questions

1.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 ?

2.In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

3.The HCF of 960 and 432 is :

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

5.In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

6.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 :

Comments

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1.
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 ?

2.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

3.
The HCF of 960 and 432 is :

4.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

5.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

6.
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 :