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:

NCERT Solutions Class 8 Science	Important Maths Formulas	Basic Maths
Arithmetic	Geometry	Number system

CBSE X Related Questions

1.
If for a data, median is 5 and mode is 4, then mean is equal to :
- 7
- 11
- \(\frac{11}{2}\)
- \(\frac{14}{3}\)
View Solution
2.
The value of k for which sum of the zeroes of the polynomial \(p(x) = 3x^2 - kx + 6\) is 2, is
- 2
- \(-6\)
- \(-2\)
- 6
View Solution
3.
If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is
- \(\frac{7}{2}\)
- \(\frac{6}{7}\)
- \(\frac{-7}{6}\)
- \(\frac{7}{6}\)
View Solution
4.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is
- \(0\)
- \(\frac{80}{81}\)
- \(1\)
- \(\frac{82}{80}\)
View Solution
5.
The graph of a quadratic polynomial f(x) passes through (5,0), (0, -1) and (-2, 0). The two factors of the polynomial are
- \((x + 2), (x - 5)\)
- \((x + 5), (x - 2)\)
- \((x + 1), (x - 5)\)
- \((x - 1), (x + 2)\)
View Solution
6.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is
- \(4 \text{ cm}\)
- \(4\sqrt{2} \text{ cm}\)
- \(8 \text{ cm}\)
- \(2\sqrt{2} \text{ cm}\)
View Solution

View All

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NCERT Solutions for Class 8 Mathematics Chapter 8: Comparing Quantities

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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.
If for a data, median is 5 and mode is 4, then mean is equal to :

2.
The value of k for which sum of the zeroes of the polynomial \(p(x) = 3x^2 - kx + 6\) is 2, is

3.
If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is

4.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

5.
The graph of a quadratic polynomial f(x) passes through (5,0), (0, -1) and (-2, 0). The two factors of the polynomial are

6.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

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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.If for a data, median is 5 and mode is 4, then mean is equal to :

2.The value of k for which sum of the zeroes of the polynomial \(p(x) = 3x^2 - kx + 6\) is 2, is

3.If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is

4.For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

5.The graph of a quadratic polynomial f(x) passes through (5,0), (0, -1) and (-2, 0). The two factors of the polynomial are

6.If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
If for a data, median is 5 and mode is 4, then mean is equal to :

2.
The value of k for which sum of the zeroes of the polynomial \(p(x) = 3x^2 - kx + 6\) is 2, is

3.
If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is

4.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

5.
The graph of a quadratic polynomial f(x) passes through (5,0), (0, -1) and (-2, 0). The two factors of the polynomial are

6.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is