NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.2 Solutions

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NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.2 Solutions are based on parallelograms with the same base and parallel lines.

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Check out NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.2 Solutions

Exercise Solutions of Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles

Also check other Exercise Solutions of Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles

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Important Chapter Related Topics
Area of Triangle	Perimeter of a Parallelogram
Area of a Trapezoid Formula	Trapezoids

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CBSE X Related Questions

1.
Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is
- 34
- 43
- 38.5
- 41.5
View Solution
2.
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
3.
In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is
- $6$
- $4$
- $\pm 4$
- $3$
View Solution
4.
Assertion (A) : H.C.F. $(36 m^{2}, 18 m) = 18 m$, where $m$ is a prime number.
Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
View Solution
5.
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
6.
Arc $PQ$ subtends an angle $\theta$ at the centre of the circle with radius $6.3 \text{ cm}$. If $\text{Arc } PQ = 11 \text{ cm}$, then the value of $\theta$ is
- $10^{\circ}$
- $60^{\circ}$
- $45^{\circ}$
- $100^{\circ}$
View Solution

View All

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NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.2 Solutions

Exercise Solutions of Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles

CBSE X Related Questions

1.
Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is

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

3.
In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

4.
Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number.
Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

5.
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

6.
Arc \(PQ\) subtends an angle \(\theta\) at the centre of the circle with radius \(6.3 \text{ cm}\). If \(\text{Arc } PQ = 11 \text{ cm}\), then the value of \(\theta\) is

Similar Mathematics Concepts

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NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.2 Solutions

Exercise Solutions of Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles

CBSE X Related Questions

1.Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is

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

3.In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

4.Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number. Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

5.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

6.Arc \(PQ\) subtends an angle \(\theta\) at the centre of the circle with radius \(6.3 \text{ cm}\). If \(\text{Arc } PQ = 11 \text{ cm}\), then the value of \(\theta\) is

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is

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

3.
In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

4.
Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number.
Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

5.
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

6.
Arc \(PQ\) subtends an angle \(\theta\) at the centre of the circle with radius \(6.3 \text{ cm}\). If \(\text{Arc } PQ = 11 \text{ cm}\), then the value of \(\theta\) is