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.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals
- \(\frac{21}{4} \text{ cm}\)
- \(\frac{28}{3} \text{ cm}\)
- \(\frac{12}{7} \text{ cm}\)
- \(5.5 \text{ cm}\)
View Solution
2.
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
3.
If the quadratic equation \(9x^2 + 8kx + 16 = 0\) has real and equal roots, then the value of k is
- 3
- –3
- –4
- \(\pm 3\)
View Solution
4.
If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).
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.
If \(\alpha, \beta\) are the zeroes of the polynomial \(p(x) = x^2 - 3x - 1\), then find the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\).
View Solution

View All

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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.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

2.
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.

3.
If the quadratic equation \(9x^2 + 8kx + 16 = 0\) has real and equal roots, then the value of k is

4.
If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).

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

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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.\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

2.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.

3.If the quadratic equation \(9x^2 + 8kx + 16 = 0\) has real and equal roots, then the value of k is

4.If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).

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

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

2.
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.

3.
If the quadratic equation \(9x^2 + 8kx + 16 = 0\) has real and equal roots, then the value of k is

4.
If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).

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