NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.2 Solutions

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.2 Solutions are based on the concept of Surface area of a right circular cylinder.

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Check out NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.2 Solutions

Exercise Solutions of Class 9 Maths Chapter 13 Surface Areas and Volumes

Also check other Exercise Solutions of Class 9 Maths Chapter 13 Surface Areas and Volumes

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Important Chapter Related Topics
Surface Areas And Volumes	Cuboid	Area of Prism
Surface Area of a Cylinder	Area of Hollow Cylinder	Surface Area of a Cone Formula
Surface Area of a Cylinder Formula

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CBSE X Related Questions

1.
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
2.
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
3.
If roots of the quadratic equation $x^2 - k\sqrt{3}x + 2 = 0$ are real and equal, then value of k is
- $-2$
- $\sqrt{\frac{8}{3}}$
- $1$
- $2$
View Solution
4.
The value of $\left(\frac{1}{2} \tan^2 45^\circ - \cos^2 60^\circ\right)$ is :
- $0$
- $-\frac{1}{2}$
- $\frac{1}{4}$
- $-\frac{1}{4}$
View Solution
5.
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
6.
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

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You can also check

NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.2 Solutions

Exercise Solutions of Class 9 Maths Chapter 13 Surface Areas and Volumes

CBSE X Related Questions

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

2.
If for a data, median is 5 and mode is 4, then mean is equal to :

3.
If roots of the quadratic equation \(x^2 - k\sqrt{3}x + 2 = 0\) are real and equal, then value of k is

4.
The value of $\left(\frac{1}{2} \tan^2 45^\circ - \cos^2 60^\circ\right)$ is :

5.
Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is

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

Similar Mathematics Concepts

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.2 Solutions

Exercise Solutions of Class 9 Maths Chapter 13 Surface Areas and Volumes

CBSE X Related Questions

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

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

3.If roots of the quadratic equation \(x^2 - k\sqrt{3}x + 2 = 0\) are real and equal, then value of k is

4.The value of $\left(\frac{1}{2} \tan^2 45^\circ - \cos^2 60^\circ\right)$ is :

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

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

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

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

2.
If for a data, median is 5 and mode is 4, then mean is equal to :

3.
If roots of the quadratic equation \(x^2 - k\sqrt{3}x + 2 = 0\) are real and equal, then value of k is

4.
The value of $\left(\frac{1}{2} \tan^2 45^\circ - \cos^2 60^\circ\right)$ is :

5.
Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is

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