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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.

Download PDF: NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.4 Solutions

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

Read More: NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas and Volumes

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

Also Check:

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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.
The following data shows the number of family members living in different bungalows of a locality:

Number of Members 0−2 2−4 4−6 6−8 8−10 Total
Number of Bungalows 10 p 60 q 5 120

If the median number of members is found to be 5, find the values of p and q.
View Solution
2.
Two identical cones are joined as shown in the figure. If radius of base is 4 cm and slant height of the cone is 6 cm, then height of the solid is
- 8 cm
- \(4\sqrt{5}\) cm
- \(2\sqrt{5}\) cm
- 12 cm
View Solution
3.
Using prime factorisation, find the HCF of 144, 180 and 192.
View Solution
4.
In the adjoining figure, TP and TQ are tangents drawn to a circle with centre O. If $\angle OPQ = 15^\circ$ and $\angle PTQ = \theta$, then find the value of $\sin 2\theta$.
View Solution
5.
Solve the equation \(4x^2 - 9x + 3 = 0\), using quadratic formula.
View Solution
6.
In a right triangle ABC, right-angled at A, if $\sin B = \dfrac{1}{4}$, then the value of $\sec B$ is
- 4
- $\dfrac{\sqrt{15}}{4}$
- $\sqrt{15}$
- $\dfrac{4}{\sqrt{15}}$
View Solution

Similar Mathematics Concepts

Surface Areas and Volumes Revision Notes

Area of Hollow Cylinder: Total Surface Area & Lateral Surface Area

Surface Area of a Cylinder Formula: Definition and Solved Examples

Area of Square: Formula & How to Find Area of Square?

Volume of a Sphere: Formula, Derivation & Solved Examples

NCERT Solutions for class 10 Mathematics Chapter 13: Surface Areas and Volumes

NCERT Solutions for Class 9 Maths Chapter 13 : Surface Areas And Volumes

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MCQ On Surface Area And Volume

Surface Areas and Volumes MCQs

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Mathematics NCERT Solutions

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