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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Arithmetic	Geometry	Number system

CBSE X Related Questions

1.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.
View Solution
2.
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
3.
In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.
View Solution
4.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.
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 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.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.

2.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

3.
In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.

4.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

5.
The HCF of 960 and 432 is :

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

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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.A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.

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

3.In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.

4.PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

5.The HCF of 960 and 432 is :

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

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.

2.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

3.
In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.

4.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

5.
The HCF of 960 and 432 is :

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