NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.6 Solutions

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.6 Solutions are based on the Volume of a cylinder.

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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.
In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD.
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2.
In the adjoining figure, $ AP = 1 \, \text{cm}, \ BP = 2 \, \text{cm}, \ AQ = 1.5 \, \text{cm}, \ AC = 4.5 \, \text{cm} $ Prove that $ \triangle APQ \sim \triangle ABC $.
Hence, find the length of $ PQ $, if $ BC = 3.6 \, \text{cm} $.
View Solution
3.
PA and PB are tangents drawn to a circle with centre O. If $\angle AOB = 120^\circ$ and OA = 10 cm, then

(i) Find $\angle OPA$.
(ii) Find the perimeter of $\triangle OAP$.
(iii) Find the length of chord AB.
View Solution
4.
$\alpha, \beta$ are zeroes of the polynomial $3x^2 - 8x + k$. Find the value of $k$, if $\alpha^2 + \beta^2 = \dfrac{40}{9}$
View Solution
5.
Find the zeroes of the polynomial: \[ q(x) = 8x^2 - 2x - 3 \] Hence, find a polynomial whose zeroes are 2 less than the zeroes of $q(x)$
View Solution
6.
Prove that: \[ \frac{\cos \theta - 2 \cos^3 \theta}{\sin \theta - 2 \sin^3 \theta} + \cot \theta = 0 \]
View Solution

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.6 Solutions

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

CBSE X Related Questions

1.
In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD.

2.
In the adjoining figure, \( AP = 1 \, \text{cm}, \ BP = 2 \, \text{cm}, \ AQ = 1.5 \, \text{cm}, \ AC = 4.5 \, \text{cm} \) Prove that \( \triangle APQ \sim \triangle ABC \).
Hence, find the length of \( PQ \), if \( BC = 3.6 \, \text{cm} \).

3.
PA and PB are tangents drawn to a circle with centre O. If \(\angle AOB = 120^\circ\) and OA = 10 cm, then

(i) Find \(\angle OPA\).
(ii) Find the perimeter of \(\triangle OAP\).
(iii) Find the length of chord AB.

4.
\(\alpha, \beta\) are zeroes of the polynomial \(3x^2 - 8x + k\). Find the value of \(k\), if \(\alpha^2 + \beta^2 = \dfrac{40}{9}\)

5.
Find the zeroes of the polynomial: \[ q(x) = 8x^2 - 2x - 3 \] Hence, find a polynomial whose zeroes are 2 less than the zeroes of \(q(x)\)

6.
Prove that: \[ \frac{\cos \theta - 2 \cos^3 \theta}{\sin \theta - 2 \sin^3 \theta} + \cot \theta = 0 \]

Similar Mathematics Concepts

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.6 Solutions

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

CBSE X Related Questions

1.In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD.

2.In the adjoining figure, \( AP = 1 \, \text{cm}, \ BP = 2 \, \text{cm}, \ AQ = 1.5 \, \text{cm}, \ AC = 4.5 \, \text{cm} \) Prove that \( \triangle APQ \sim \triangle ABC \). Hence, find the length of \( PQ \), if \( BC = 3.6 \, \text{cm} \).

3.PA and PB are tangents drawn to a circle with centre O. If \(\angle AOB = 120^\circ\) and OA = 10 cm, then (i) Find \(\angle OPA\). (ii) Find the perimeter of \(\triangle OAP\). (iii) Find the length of chord AB.

4.\(\alpha, \beta\) are zeroes of the polynomial \(3x^2 - 8x + k\). Find the value of \(k\), if \(\alpha^2 + \beta^2 = \dfrac{40}{9}\)

5.Find the zeroes of the polynomial: \[ q(x) = 8x^2 - 2x - 3 \] Hence, find a polynomial whose zeroes are 2 less than the zeroes of \(q(x)\)

6.Prove that: \[ \frac{\cos \theta - 2 \cos^3 \theta}{\sin \theta - 2 \sin^3 \theta} + \cot \theta = 0 \]

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD.

2.
In the adjoining figure, \( AP = 1 \, \text{cm}, \ BP = 2 \, \text{cm}, \ AQ = 1.5 \, \text{cm}, \ AC = 4.5 \, \text{cm} \) Prove that \( \triangle APQ \sim \triangle ABC \).
Hence, find the length of \( PQ \), if \( BC = 3.6 \, \text{cm} \).

3.
PA and PB are tangents drawn to a circle with centre O. If \(\angle AOB = 120^\circ\) and OA = 10 cm, then

(i) Find \(\angle OPA\).
(ii) Find the perimeter of \(\triangle OAP\).
(iii) Find the length of chord AB.

4.
\(\alpha, \beta\) are zeroes of the polynomial \(3x^2 - 8x + k\). Find the value of \(k\), if \(\alpha^2 + \beta^2 = \dfrac{40}{9}\)

5.
Find the zeroes of the polynomial: \[ q(x) = 8x^2 - 2x - 3 \] Hence, find a polynomial whose zeroes are 2 less than the zeroes of \(q(x)\)

6.
Prove that: \[ \frac{\cos \theta - 2 \cos^3 \theta}{\sin \theta - 2 \sin^3 \theta} + \cot \theta = 0 \]