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- 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 ahemisphere. The base of the block is a cube with edge 5 cm, and the hemispherefixed 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 hemisphereis 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 entirerocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has adiameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portionis to be painted orange and the cylindrical portion yellow, find the area of the rocket painted witheach of these colours. (Take π = 3.14)Denote radius of cone by r, slant height of cone by lheight 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 6.5cmThe 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 cm2The 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 thecone is 4cm and diameter of the base is 6cm, Calculate:1. The volume of the toy2. 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 5Volume of the toy = volume of cone + Surface area of the toy = Curve surface area ofvolume 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

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