This document defines and provides formulas for calculating the surface area and volume of various 3D shapes. It discusses total surface area, curved surface area, and volume. Formulas are provided for calculating these values for cubes, cuboids, cylinders, cones, spheres, hemispheres, and frustums. An example problem is also worked out calculating the number of spherical balls that can fit in an inverted conical vessel based on the volume of water displaced. The value shown in this example is judicious use of water resources.

1. SURFACE AREA & VOLUME
MADE BY :- TARUN KUMAR CLASS – X –C ROLL NO. – 37

2. The area occupied by a three-dimensional object by its outer surface is called surface
area. It is measured in square units.
Area can be of two types:
1. Total surface area
Total surface area means adding up the areas of all the surfaces.
2. Curved surface area (lateral surface area)
Curved surface area refers to the area of only the curved part of the shape excluding
its base(s).
What is Surface area ?

3. WHAT IS VOLUME ?
The amount of space that an object or substance occupies is called volume.
It is measured in cubic units.

4. TYPES OF SOLIDS
cube cylinder cone
sphere hemisphere frustum

5. CUBE
Total Surface Area = 6 (Side)2
Curve Surface Area = 4 (Side)2
Volume =(Side)3
Diagonal of a cube = a√3

6. CUBOID
TSA= 2 ((l x b + b x h + l x h)
CSA=2(l + b)h
Volume = l x b x h
Diagonal = √(l2 + b2 +h2)
height

7. CYLINDER
TSA = 2πr (h + r)
CSA = 2πrh
Volume = πr²h
radius

8. CONE
TSA = πr (l + r)
CSA= πrl
Where l stand for slant height
l = √h2+ r2
Volume1/3 πr2h
radius

9. SPHERE
radius
SPHERE
TSA = 4 π r2
CSA = 4 π r2
Volume = 4/3 π r3

10. HEMISPHERE
HEMISPHERE
TSA= 3πr2
CSA= 2π2
Volume = 2/3πr3
radius

11. FRUSTUM
L
R1
h
FRUSTUM
TSA= πl(r1 + r2) + πr1
2 + πr2
2
CSA= πl( r1 + r2)
where r1 = radius of the larger circular
face of the frustum, r2 = radius of the
smaller circular face of the frustum, and
l = slant height of the frustum.
l =√ h2+(r1-r2)2
Volume= (1/3) πh (r1
2 + r2
2 + r1 r2)

12. 11cm
2.5 cm
Sushant has a vessel, in the form of an inverted cone, open at the top, of height 11 cm and
radius of top as 2.5 cm and is full of water. Metallic spherical balls each of diameter 0.5 cm are put in the vessel
due to which 2
5
th of the water in the vessel flows out. Find how many balls were put in the vessel. Sushant made
the arrangement so that the water that flows out irrigates the flower beds. What value has been shown by
Sushant?
According to the question,
h = 11 cm, Radius of a conical vessel = r1 = 2.5 cm
radius of a metallic spherical ball = r2= 0.5/2 = 0.25 cm
Let n be the number of spherical balls that were dropped in the vessel.
Volume of water spilled = Volume of spherical balls dropped
2/5×volume of cone=n×Volume of each spherical ball
2/5×1/3πr1
2h=n×4/3πr2
3
r1
2 =n×10r2
3
2.52×11=n×10(0.253)
n=440
Hence, the number of spherical balls that were dropped in the vessel is 440. Sushant has made such an arrangement
so that the water that flows out, irrigates the flower beds. It is judicious usage of water.
Question1

13. A Rocket is in the Form of a Circular Cylinder Closed at the Lower End with a Cone of the
Same Radius Attached to the Top. the Cylinder is of Radius 2.5m and Height 21m and the
Cone Has a Slant Height 8m. Calculate Total Surface Area and Volume of the Rocket?
Solution
Total surface area = Area of base of cylinder+ Area of CSA of cylinder +Area of CSA of cone
A=π×r2 +2πrh+πrl
h=height of cylindrical part
l=slant height of conical part
r=radius of cylinder=radius of cone
A=π×2.5 2+2π×2.5×21+π×2.5×8
A=412.334m2
Volume of tent=π r2 h+31π r2 h
V=π×2.52 ×21+31π×2.52 ×7.6
V=462.076m3
Total Surface Area =412.334m2 and Volume of the Rocket= V=462.076m3
Question 2

14. THANK
YOU