CBSE Notes Class 10 Maths Chapter 12 Surface Areas And Volumes

1.0Introduction to Surface Area and Volume

This chapter relates to the measurement of the surface area and volume of various geometrical solids, such as cubes, cuboids, cylinders, cones, spheres, and hemispheres. It gives an understanding of how to calculate the surface areas and volumes directly and applied problems for those shapes.

2.0Download CBSE Notes for Class 10 Maths Chapter 12: Surface Area and Volume - Free PDF!!

Hey there! Ready to ace Surface Area and Volume? Grab these free CBSE Class 10 Maths - Chapter 12 notes in PDF format and make understanding those formulas and concepts a breeze!

3.0CBSE Class 10 Maths Chapter 12 Surface Areas and Volume - Revision Notes

Surface Area of Solids

Surface area measures the total area of all the outer surfaces of any 3D object. It is used to describe the surface external to the object, and a measurement is done in square units, that is cm² or m². Surface areas of solids are of two types: 

1. CSA (Curved Surface Area): It is the area of a 3D figure's lateral or curved surface, excluding the base or top. For example, in a cylinder, CSA means the curved surface area excluding the circular bases.
2. TSA (Total Surface Area): It is the sum of all the outer surface areas of the 3D figure, both the curved/lateral surface and the bases. For example, in a cylinder, TSA comprises both the curved side and the area of the two circular bases.

Volume of Solids 

Volume is a measure of the amount of space occupied by a 3D object. Volume is measured in cubic units-for example, cm³ or m³, which therefore represent how much capacity or content an object has.

Formulas Related to the Surface Area and Volume of Solids

Shape	Formulas related to the solids
	Cube: A Cube has six identical square faces.  Total Surface Area of cube = 6a2 Curved Surface area of cube = 4a2 Volume of Cube = a3
	Rectangle: A Rectangle has six rectangular faces.  Total Surface Area of Rectangle  = 2(lb+bh+hl) Curved surface area of Rectangle  = 2(l+b)h Volume of Rectangle = lbh
	Cylinder: A Cylinder consists of two circular base joints with a curved surface.  The Total Surface Area of the Cylinder $=\pi r(r+h)$ Curved Surface area of the cylinder $=\pi rh$ The volume of the cylinder = $\pi r^{2}h$
	Cone: In maths, a cone has a circular base and slanted surface.  The Total Surface Area of Cone = $\pi r(l+r)$ Curved Surface area of cone = $\pi rl$ The volume of cone $=1/3\pi r^{2}h$ Slant height $(l)=\sqrt{h^2+r^2}$
	Sphere: Sphere has only one curved surface in the shape of a ball.  The Total Surface Area of the Sphere = $4\pi r^2$ The volume of the Sphere $=\frac{4}{3}\pi r^3$
	Hemisphere: A hemisphere is half of a sphere and consists of one curved surface and one base.  The total Surface Area of the Hemisphere $=3\pi r^2$ Curved Surface area of the Hemisphere  $=2\pi r^2$ The volume of the Hemisphere $=\frac{2}{3}\pi r^3$

4.0Solved Problems

Question 1: A canal is 300 cm wide and 120 cm deep. The water in the canal is flowing at a speed of 20km/h. How much area will it irrigate in 20 minutes if 8 cm of standing water is desired? 

Answer: Volume of water flows in the canal in one hour = width of the canal $\times$ depth of the canal $\times$speed of the canal water =31.21000=72000m3

In 20 minutes the volume of $\text{ water }=\frac{72000\times 20}{60}{\mathrm{m}}^3=24000{\mathrm{m}}^3$

Area irrigates in 20 minutes

$=\frac{24000}{0.08}=300000{\mathrm{m}}^2=30\text{ hectares }$ .

Question 2: A cone of radius 4 cm is divided into two parts by drawing a plane through the midpoint of its axis and parallel to its base. Compare the volumes of the smaller cone and the bigger cone. 

Answer: Let the height of the cone = h Therefore,

Answer: Let the height of the cone = h 

As the plane cut through the midpoint of the cone’s axis then,

The height of the smaller cone will be = h/2 

Therefore, 

Volume of the cone $=\frac{1}{3}\pi r^{2}h$ ………….(1)

Volume of the smaller cone $=\frac{1}{3}\pi r^2\frac{h}{2}$ ……………..(2)

Comparing equations (1) and (2), 

$\frac{Volumeofthesmallercone}{Volumeofthecone}=\frac{1}{2}$

Question 3: A cone of maximum size is carved out from a cube of edge 14 cm. Find the surface area of the cone and the remaining solid after the cone is carved out.

Solution: The maximum-sized cone that can be carved out of the cube has a base radius of 7cm and a height of 14cm. So, the slant height of the cone $=\sqrt{14^2+7^2}=7\sqrt{5}$

The surface area of the remaining solid = Surface area of the cube - the surface area of a cone $6a^2-\pi r(l+r)$

$=6\times 14\times 14-22/7\times 7(7\sqrt{5}+7)=[1176-154(\sqrt{5}+7)]c{m}^2$

Question 4: Three cubes of a metal whose edges are in the ratio 3:4:5 are melted and converted into a single cube whose diagonal is $12\sqrt{3}cm$. Find the edges of the three cubes.

Answer: Let the edges of 3 Cubes 3x, 4x, and 5x. 

Diagonal of single resultant cube $=a\sqrt{3}=12\sqrt{3}$

The side of single resulted in a cube after melting = a = 12cm 

The sum of the volume of 3 cubes = Volume of a single cube. 

(a₁)³+(a₂)³+(a₃)³ = a³

(3x)³+(4x)³+(5x)³ = 12³

27x³ + 64x³ + 125x³ = 1728

216x³ = 1728

x³ = 8

x = 2 cm 

Hence, the edges of cubes are 6 cm, 8 cm, and 10 cm. 

5.0Key Features of CBSE Maths Notes for Class 10 Chapter 12

• The notes are aligned with the latest pattern of the CBSE Class 10 curriculum. 
• Visual aids are provided with every concept to get a better understanding of surface area and volumes of solid. 
• The notes are easy to understand, making it ideal for self-learning.