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

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The NCERT Solutions for Class 9 Mathematics Chapter 13 Surface Areas And Volumes have been provided in this article. Surface Areas and Volumes deals with the calculation of surface area, total area and volume of various 3-D figures such as cube, cuboid, cone, cylinder, prism, etc. Surface area is the area that is occupied by the surface of the object while volume is the total space that is occupied by the whole object.

Class 9 Maths Chapter 13 Surface Areas and Volumes belong to Unit 5 Mensuration which has a weightage of 13 marks in the Class 9 Maths Examination. Class 9 Mathematics Chapter 13 has the following important concepts: 

  1. Area of Prism
  2. Surface Area of a Cylinder
  3. Surface Area of a Cone Formula
  4. Cuboid

Download: NCERT Solutions for Class 9 Mathematics Chapter 13 pdf


NCERT Solutions for Class 9 Mathematics Chapter 13

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Important Topics in Class 9 Maths Chapter 13 Surface Areas and Volumes

Important Topics in Class 9 Maths Chapter 13 Surface Areas and Volumes are given below:

Area of Prism

Prism refers to a three-dimensional solid figure which has two ends that are identical. A Prism is a combination of plane faces, identical bases and equivalent cross-sections. 

Example: Express the Area of a Prism.

Solution: Considering the area of the lateral surface = S1xh + S2xh +S3xh + .........+ Snxh
= h (S1 + S2 + S3 …+ Sn)
Thus,
= height x perimeter of the base
Area of the total bases = 2 x Area of the base
Hence, the total area of the prism is = (height x perimeter of the base) + (2 x Area of the base)

Surface Area of a Cylinder

The total amount of area a cylinder is seen to cover is its surface are. It is further classified into two segments: The Curved Surface Area (CSA) and the Total Surface Area (TSA).

Important Formulae to Remember:

  • Surface Area of Cylinder = 2πr (h + r) sq. unit
  • Curved surface area (CSA) of cylinder = 2πrh sq. units.
  • Total surface area (TSA) = 2π × r × h + 2πr2 = 2πr (h + r) sq. units
  • Area of the circular bases of cylinder = 2 (πr2)

Surface Area of a Cone Formula

A cone, in geometry, is a three-dimensional object which narrows gradually from a circular base to an upward point, otherwise known as the apex or vertex.

Total Surface Area of a Cone = πr2 + πrl
Here,

  • I = Slant height
  • r = The base circle's radius
  • SA = Total surface area
  • h = Height of cone

Cuboid

A cuboid is a three-dimensional figure that comprises six rectangular faces placed at right angles. The volume of a cuboid is calculated considering its length, breadth and height.

Example: What is the total surface area of a plywood box which is 40 cm long, 35 cm wide, and 20 cm high?

Solution: We are already aware of the formula required to determine the total surface area of the plywood box, 
Total surface area of cuboid = 2 (lb + bh + lh)
= 2 (40 x 35 + 35 x 20 + 40 x 20)
= 2 (1400 + 700 + 800)
= 2 (2900)
= 5800 sq. cm.


NCERT Solutions for Class 9 Maths Chapter 13 Exercises:

The detailed solutions for all the NCERT Solutions for Surface Areas and Volumes under different exercises are:

Also check:

Also Read:

CBSE X Related Questions

  • 1.
    There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?

      • 144
      • 2
      • 420
      • 272

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


        • 3.
          The HCF of 960 and 432 is :

            • 48
            • 54
            • 72
            • 36

          • 4.
            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.


              • 5.
                In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.


                  • 6.
                    PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

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