NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 Solutions

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 Solutions are based on the volume of a sphere.

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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.
Prove that :
$\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$.
View Solution
2.
$ABCD$ 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
3.
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
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.
The first term of an AP is $p$ and the common difference is $q$, then its 10th term is :
- $q - 9p$
- $p - 9q$
- $p + 9q$
- $2p + 9q$
View Solution
6.
The HCF of 960 and 432 is :
- 48
- 54
- 72
- 36
View Solution

View All

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 Solutions

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

CBSE X Related Questions

1.
Prove that :
\(\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta\).

2.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

3.
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.

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.
The first term of an AP is $p$ and the common difference is $q$, then its 10th term is :

6.
The HCF of 960 and 432 is :

Similar Mathematics Concepts

Comments

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 Solutions

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

CBSE X Related Questions

1.Prove that : \(\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta\).

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

3.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.

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.The first term of an AP is $p$ and the common difference is $q$, then its 10th term is :

6.The HCF of 960 and 432 is :

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Prove that :
\(\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta\).

2.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

3.
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.

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.
The first term of an AP is $p$ and the common difference is $q$, then its 10th term is :

6.
The HCF of 960 and 432 is :