NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.4 Solutions

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NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volume Exercise 13.4 Solutions are based on the following concepts:

Height of the cylinder based on given condition
Radius of the resulting sphere
Number of cones for a given situation.
Number of coins formed by melting a cuboid structured object.

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Exercise Solutions of Class 10 Maths Chapter 13 Surface Areas and Volume

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Chapter Related Topics
Surface Areas And Volumes Revision Notes	Area Of Prism	Volume Of A Rectangular Prism Formula
Volume Of A Square Pyramid Formula	Volume Of A Triangular Prism Formula	Surface Area Of A Prism Formula
Surface Area Of A Pyramid Formula	Surface Area Of A Rectangular Prism Formula	Surface Area Of A Square Pyramid Formula
Surface Area of a Cylinder	Area of Hollow Cylinder	Surface Area of a Prism Formula

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Math Formula	NCERT Solutions for Class 10 Maths	Algebra

CBSE X Related Questions

1.
A peacock sitting on the top of a tree of height 10 m observes a snake moving on the ground. If the snake is $10\sqrt{3}$ m away from the base of the tree, then angle of depression of the snake from the eye of the peacock is
- $60^\circ$
- $45^\circ$
- $30^\circ$
- $90^\circ$
View Solution
2.
From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$
View Solution
3.
Find the smallest value of $p$ for which the quadratic equation $x^2 - 2(p + 1)x + p^2 = 0$ has real roots. Hence, find the roots of the equation so obtained.
View Solution
4.
In $\triangle ABC, \angle B = 90^\circ$. If $\frac{AB}{AC} = \frac{1}{2}$, then $\cos C$ is equal to
- $\frac{3}{2}$
- $\frac{1}{2}$
- $\frac{\sqrt{3}}{2}$
- $\frac{1}{\sqrt{3}}$
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.
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

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NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.4 Solutions

Exercise Solutions of Class 10 Maths Chapter 13 Surface Areas and Volume

CBSE X Related Questions

1.
A peacock sitting on the top of a tree of height 10 m observes a snake moving on the ground. If the snake is $10\sqrt{3}$ m away from the base of the tree, then angle of depression of the snake from the eye of the peacock is

2.
From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$

3.
Find the smallest value of $p$ for which the quadratic equation $x^2 - 2(p + 1)x + p^2 = 0$ has real roots. Hence, find the roots of the equation so obtained.

4.
In \(\triangle ABC, \angle B = 90^\circ\). If \(\frac{AB}{AC} = \frac{1}{2}\), then \(\cos C\) is equal to

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.
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 \( BC = 3.6 \, \text{cm} \).

Similar Mathematics Concepts

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NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.4 Solutions

Exercise Solutions of Class 10 Maths Chapter 13 Surface Areas and Volume

CBSE X Related Questions

1.A peacock sitting on the top of a tree of height 10 m observes a snake moving on the ground. If the snake is $10\sqrt{3}$ m away from the base of the tree, then angle of depression of the snake from the eye of the peacock is

2.From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$

3.Find the smallest value of $p$ for which the quadratic equation $x^2 - 2(p + 1)x + p^2 = 0$ has real roots. Hence, find the roots of the equation so obtained.

4.In \(\triangle ABC, \angle B = 90^\circ\). If \(\frac{AB}{AC} = \frac{1}{2}\), then \(\cos C\) is equal to

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.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 \( BC = 3.6 \, \text{cm} \).

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
A peacock sitting on the top of a tree of height 10 m observes a snake moving on the ground. If the snake is $10\sqrt{3}$ m away from the base of the tree, then angle of depression of the snake from the eye of the peacock is

2.
From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$

3.
Find the smallest value of $p$ for which the quadratic equation $x^2 - 2(p + 1)x + p^2 = 0$ has real roots. Hence, find the roots of the equation so obtained.

4.
In \(\triangle ABC, \angle B = 90^\circ\). If \(\frac{AB}{AC} = \frac{1}{2}\), then \(\cos C\) is equal to

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.
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 \( BC = 3.6 \, \text{cm} \).