Class 10 - Surface Areas and Volumes Formula Sheet

1. Cube

Surface Area

� Total Surface Area = 6a�
� Lateral Surface Area = 4a�

Volume

Volume = a�

Diagonal

Face diagonal = av2
Space diagonal = av3

2. Cuboid

Surface Area

� TSA = 2(lb + bh + hl)
� LSA = 2h(l + b)

Volume

Volume = l � b � h

Diagonal

Diagonal = v(l� + b� + h�)

3. Right Circular Cylinder

Surface Area

� TSA = 2pr(h + r)
� CSA = 2prh

Volume

Volume = pr�h

Hollow Cylinder

Volume = ph(R� - r�)
TSA = 2ph(R + r) + 2p(R� - r�)

4. Right Circular Cone

Surface Area

� TSA = pr(l + r)
� CSA = prl
� l = v(h� + r�)

Volume

Volume = (1/3)pr�h

Slant Height

l = v(h� + r�)
h = v(l� - r�)

5. Sphere and Hemisphere

Shape	Surface Area	Volume
Sphere	4pr�	(4/3)pr�
Hemisphere (Solid)	3pr�	(2/3)pr�
Hemisphere (Hollow)	2pr�	(2/3)pr�

6. Frustum of a Cone

� Slant height: l = v[h� + (R - r)�]
� CSA = pl(R + r)
� TSA = pl(R + r) + p(R� + r�)
� Volume = (1/3)ph(R� + r� + Rr)

7. Combination of Solids

General Approach:

Identify individual solids in combination
Calculate surface area/volume of each
Add for total volume
For surface area: Subtract common areas where joined

8. Common Combinations

Cone on Cylinder

Total height = h1 + h2
Volume = pr�h1 + (1/3)pr�h2

Hemisphere on Cylinder

Total height = r + h
Volume = pr�h + (2/3)pr�

Cone on Hemisphere

Same radius r
Volume = (2/3)pr� + (1/3)pr�h

9. Conversion of Solids

Important Principle: Volume remains constant when solid is melted and recast into different shape.

If solid A is melted to form solid B:
Volume of A = Volume of B
Number of smaller solids = Volume of bigger solid � Volume of smaller solid

10. Quick Reference Table

Solid	Volume	TSA
Cube (side a)	a�	6a�
Cuboid (l,b,h)	lbh	2(lb+bh+hl)
Cylinder (r,h)	pr�h	2pr(h+r)
Cone (r,h)	(1/3)pr�h	pr(l+r)
Sphere (r)	(4/3)pr�	4pr�
Hemisphere (r)	(2/3)pr�	3pr�

11. Example Problems (NCERT Style)

Problem 1: A cone of height 24 cm and radius 6 cm is made. Find its volume and curved surface area.

Solution:
Volume = (1/3)pr�h = (1/3) � (22/7) � 6� � 24
= (1/3) � (22/7) � 36 � 24 = (22 � 36 � 8)/7 = 6336/7 � 905.14 cm�

Slant height l = v(h� + r�) = v(576 + 36) = v612 = 6v17 cm
CSA = prl = (22/7) � 6 � 6v17 � 132v17/7 cm�

Problem 2: A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is to be filled into cylindrical bottles of radius 3 cm and height 6 cm. How many bottles are needed?

Solution:
Radius of hemisphere = 18 cm
Volume of hemisphere = (2/3)pr� = (2/3) � p � 18� = 3888p cm�
Volume of one bottle = pr�h = p � 3� � 6 = 54p cm�
Number of bottles = 3888p � 54p = 72 bottles

Problem 3: A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many liters of water can it hold?

Solution:
Volume = l � b � h = 6 � 5 � 4.5 = 135 m�
1 m� = 1000 liters
Capacity = 135 � 1000 = 135,000 liters

12. Important Points to Remember

TSA = Total Surface Area, CSA/LSA = Curved/Lateral Surface Area
For hollow objects: Use outer and inner radii
When solids are combined: Subtract common areas
Volume remains constant during melting/recasting
1 m� = 1000 liters, 1 liter = 1000 cm�
p � 22/7 or 3.14 as per question requirement
Always include units in final answer
For frustum: R = larger radius, r = smaller radius

13. Conversion Units

� 1 cm� = 1 mL
� 1 liter = 1000 cm� = 1000 mL
� 1 m� = 1000 liters = 106 cm�
� 1 km = 1000 m, 1 m = 100 cm
� 1 hectare = 10000 m�