Presentation by Pratima Nayak to popularise Maths if you like it message pnpratima@gmail.com

1. SUFACE AREA AND VOLUMES OF SOLID SHAPES

2. SUFACE AREA AND VOLUME OF CUBE
• Total Surface Area : 6a2
• Volume: a3
• Lateral Surface Area : 4a2

3. SUFACE AREA AND VOLUME OF CUBOID
• Total Surface Area: 2(lb+ bh +hl)
• Lateral Surface Area:2h(l + b)
• Volume: l x b x h

4. SUFACE AREA AND VOLUME OF
CYLINDER
• TOTAL SURFACE AREA: 2 r ( h+ r)
• CURVED SURFACE AREA: 2 rh
• VOLUME: r2h

5. SUFACE AREA AND VOLUME OF CONE
• Total Surface Area : r(l +r )
• Lateral Surface Area: rl
• Volume: 1/3 r 2h

6. SUFACE AREA AND VOLUME OF SPHERE
• Total Surface Area: 4 π r2
• Volume : 4/3 π r 3

7. SUFACE AREA AND VOLUME OF
HEMISPHERE
• Total Surface Area: 3 π r 2
• Volume : 2/3 π r3

8. SUFACE AREA AND VOLUME OF FRUSTUM
• Volume: 1/3πh(r1 2+ r2 2+r1r2)
• Curved Surface Area: πl(r1+r2)
• Total Surface Area : πl(r1+r2)+π r1 2+π r2 2

9. Combination of Solids

10. Find the volume of the above combination of solid.
Here the above solid is a combination of cone and a cylinder
We know the formula for volume of a cylinder V1 = π r2 h
And the Volume of cone V2 = 1/3 π r2 h
Volume of a combination of the above solid = V1 + V2
Volume V1 = 3.14 x 22 x 5
V1 = 3.14 x 2 x 2 x 5 = 62.8m3
Volume V2 =1/3π r2 h
(Here height of the cone is 8 - 5 = 3)
V2 = 1/3 x 3.14 x 2 x2 x 3 = 12.56 m3
So volume of a combination of solids V = = (62.8 + 12.56) m3
Volume V= 75.36 m3

11. Find the volume of the above combination of solid.
Here the above solid is a combination of cone and a hemisphere
We know the formula for volume of a hemisphere V1 =2/3π r3
And the Volume of cone V2 = 1/3 π r2 h
Volume of a combination of the above solid = V1 + V2
Volume V1 =2/3π r3
V1 = 2/3 x 3.14 x 3 x 3 x 3
V1 = 56.52 cm3
Volume V2 =1/3π r2 h
=1/3 x 3.14 x 3 x 3 x4
V2 = 37.68 cm3
So volume of a combination of solids V = (56.52 + 37.68) cm3
Volume V= 94.20 cm3

12. The decorative block shown. is made of two solids — a cube and a
hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere
fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block.
(Take π =22 /7)
The total surface area of the cube
= 6 (edge) 2 = 6 5 5 cm2 = 150 cm2.
Note that the part of the cube where the hemisphere
is attached is not included in the surface area.
So, the surface area of the block
= TSA of cube – base area of hemisphere + CSA of hemisphere
= 150 –πr2+ 2πr2 = (150 +πr2) cm2
= 150+(3.14 2.1 2.1) = (150 + 13.86) cm2 = 163.86 cm2

13. A wooden toy rocket is in the shape of a cone mounted on a cylinder. The height of the entire
rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a
diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion
is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with
each of these colours. (Take π = 3.14)
Denote radius of cone by r, slant height of cone by l
height of cone by h, radius of cylinder by r′
height of cylinder by h′. The r = 2.5 cm ,h = 6 cm, r′ = 1.5 cm,
h′= 26 – 6 = 20 cm ,
l 62 (2.5) 2 36 6.25 42.25 6.5cm
The area to be painted orange
= CSA of the cone + base area of the cone – base area of the cylinder
=πrl +πr2– π(r ′)2 =π[(2.5 × 6.5) + (2.5) 2 – (1.5) 2] cm2
=π[20.25] cm2 = 3.14 20.25 cm2
= 63.585 cm2
The area to be painted yellow
= CSA of the cylinder + area of one base of the cylinder
= 2πr′h′ +π(r′) = πr′ (2h′ +r′) = (3.14 × 1.5) (2 × 20 + 1.5) cm
2 2
= 4.71 41.5 cm = 195.465 cm
2 2

14. A solid toy is in the form of a hemisphere surmounted by a right circular cone. If height of the
cone is 4cm and diameter of the base is 6cm, Calculate:
1. The volume of the toy
2. The surface area of the toy.
Solution:- Radius, r of cone = 6/2 = 3cm
Height, h of cone = 4cm
Radius, r of hemisphere = 3cm
Slant height = 32 42 5
Volume of the toy = volume of cone + Surface area of the toy = Curve surface area of
volume of hemisphere cone + curved surface area of hemisphere
= 3.14 X 3 X (5 + 2 X 3)
= 103.62cm2

15. Acknowledgement
• http//www.tutorvista.com
• http//www.wikipedia.com
• http//www.youtube.com
• http//www.google.com
• http//www.thinkquest.org