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

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NCERT Solutions for Class 10 Mathematics Chapter 13 Surface Areas and Volumes are provided in this article. Some of the important topic in Surface areas and volumes chapter are:

Expected no of questions: 3 to 4 questions of total 6 to 7 marks

Download PDF: NCERT Solutions for Class 10 Mathematics Chapter 13 pdf

NCERT Solutions for Class 10 Mathematics Chapter 13

NCERT Solutions for Class 10 Mathematics Chapter 13 Surface area and volumes is given below.

Class 10 Mathematics Chapter 13 Surface Areas and Volumes – Important Topics

Surface area and volume are calculated for any three-dimensional geometrical shape. The surface area of any given object is the area or region occupied by the surface of the object. Whereas Volume is the amount of space available in an object.

In geometry, there are different shapes and sizes such as sphere, cube, cuboid, cone, cylinder, etc. Each shape has its surface area as well as volume. But in the case of two dimensional figures like square, circle, rectangle, triangle, etc., we can measure only the area covered by these figures and there is no volume available.

Some of the important formulas in this chapter include:

Cube:

The volume of cube = a³

Lateral Surface Area of cube = 4a²

Total Surface Area of cube = 6a²

Cuboid:

Volume of cuboid = l x b x h

Lateral Surface Area of cuboid = 2h(l+b)

Total Surface Area of cuboid = 2(lb + lh + bh)

Sphere:

Volume of sphere = (4/3) πr³

Curved Surface Area of sphere = 4πr²

Total Surface Area of sphere = 4πr²

Hemisphere:

Volume of hemisphere = (2/3) πr³

Curved Surface Area of hemisphere = 2πr²

Total Surface Area of hemisphere = 3πr²

Cone:

Volume of cone = (1/3) πr²h

Curved Surface Area of cone = πrl

Total Surface Area of cone = πr(r+l)

Cylinder:

Volume of cylinder = πr²h

Curved Surface Area of cone = 2πrh

Total Surface Area of cone = 2πrh + 2πr²

NCERT Solutions for Class 10 Maths Chapter 13 Exercises

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercises is given below.

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CBSE X Related Questions

1.
Through the mid-point Q of side CD of a parallelogram ABCD, the line AR is drawn which intersects BD at P and produced BC at R. Prove that \(AQ = QR\).
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In the given figure, \(TP\) and \(TQ\) are tangents to a circle with centre \(M\), touching another circle with centre \(N\) at \(A\) and \(B\) respectively. It is given that \(MQ = 13 \text{ cm}\), \(NB = 8 \text{ cm}\), \(BQ = 35 \text{ cm}\) and \(TP = 80 \text{ cm}\).
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