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:

  1. Surface Area of Cuboid
  2. Sphere Formula
  3. Volume of a Pyramid Formula
  4. Surface Area of a Cylinder
  5. Area of Hollow Cylinder
  6. Surface Area of a Cone Formula
  7. Surface Area of a Cylinder Formula
  8. Surface Areas and Volumes Revision Notes

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.

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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.
    A trader has three different types of oils of volume \(870 \text{ l}\), \(812 \text{ l}\) and \(638 \text{ l}\). Find the least number of containers of equal size required to store all the oil without getting mixed.


      • 2.
        The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :

          • 0
          • 1
          • 3
          • 2

        • 3.
          Assertion (A) : \((\sqrt{3} + \sqrt{5})\) is an irrational number.
          Reason (R) : Sum of the any two irrational numbers is always irrational.

            • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
            • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
            • Assertion (A) is true, but Reason (R) is false.
            • Assertion (A) is false, but Reason (R) is true.

          • 4.
            PQ is tangent to a circle with centre O. If \(OQ = a\), \(OP = a + 2\) and \(PQ = 2b\), then relation between \(a\) and \(b\) is

              • \(a^2 + (a + 2)^2 = (2b)^2\)
              • \(b^2 = a + 4\)
              • \(2a^2 + 1 = b^2\)
              • \(b^2 = a + 1\)

            • 5.
              A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use \(\pi = 3.14\))

                • \(314 \sqrt{2}\) \(\text{cm}^{2}\)
                • \(314\) \(\text{cm}^{2}\)
                • \(\frac{3140}{3}\) \(\text{cm}^{2}\)
                • \(3140 \sqrt{2}\) \(\text{cm}^{2}\)

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

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