NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.4 Solutions

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.4 Solutions are based on the concept of Surface area of a sphere.

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Check out NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.4 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

Also Check:

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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Math Study Guides
Maths Important Formulas	Maths MCQs	Comparison Articles in Maths
Difference Between Topics in Maths	Math Study Material	Algebra
Trigonometry	Number Systems	Mensuration
Math Formula	NCERT Solutions for Class 9 Maths	Geometry

CBSE X Related Questions

1.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.
View Solution
2.
Find the H.C.F. and L.C.M. of 1530 and 2040.
View Solution
3.
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
4.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.
View Solution
5.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).
View Solution
6.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)
View Solution

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.4 Solutions

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

CBSE X Related Questions

1.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

2.
Find the H.C.F. and L.C.M. of 1530 and 2040.

3.
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.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.

5.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

6.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

Similar Mathematics Concepts

Comments

dOGVmW

NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.4 Solutions

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

CBSE X Related Questions

1.D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

2.Find the H.C.F. and L.C.M. of 1530 and 2040.

3.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.A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.

5.In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

6.Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

Similar Mathematics Concepts

Comments

dOGVmW

1.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

2.
Find the H.C.F. and L.C.M. of 1530 and 2040.

3.
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.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.

5.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

6.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)