Page 292 - Start Up Mathematics_8 (Non CCE)

Lateral surface area or curved surface area (CSA) = Area of the rectangle (assuming the cylinder is cut
vertically and rolled out)
= Area of the rectangle with length ‘2pr’ and breadth ‘h’
= 2prh sq. units
2
Base surface area = pr sq. units
Total surface area (TSA) = Curved surface area + Area of the two bases
2
= (2prh + 2pr ) sq. units
= 2pr (h + r) sq. units

Surface area of a hollow cylinder
Objects like piece of rubber tube, piece of iron pipe, etc. are hollow cylinders (Fig. 18.4). A solid bounded by
two co-axial cylinders of same height and different radii is called a hollow cylinder.
In Fig. 18.4, R and r are the external and internal radii respectively and h is the height of the hollow cylinder.
2
2
(i) Surface area of each base = p(R – r ) sq. units
(ii) Lateral surface area or curved surface area (CSA)
= External surface area + Internal surface area r
= (2pRh + 2prh) sq. units h
= 2ph(R + r) sq. units
(iii) Total surface area (TSA) = CSA + Surface area of two bases R
2
2
= {2ph(R + r) + 2p(R – r )} sq. units
= {2ph(R + r) + 2p(R + r)(R – r)} sq. units Fig. 18.4
= {2p(R + r)(h + R – r)} sq. units
2
Example 25: The diameter of a cylinder is 8 cm. Its curved surface area is 1,320 cm . Find the height of
the cylinder.
8
Solution: Diameter of cylinder = 8 cm, Radius of cylinder (r) = 2 = 4 cm

Curved Surface Area (CSA) = 1,320 cm 2
22
fi 2prh = 1,320 fi 2 × × 4 × h = 1,320
7
1 320 7, ¥ 9 240,
fi h = = = 52.5 cm
2224¥ ¥ 176

Hence, the height of the cylinder is 52.5 cm.
Example 26: Find the area levelled by a cylindrical roller of diameter 60 cm and length 2.1 m in 150
revolutions.
Solution: Diameter = 60 cm
60 30 1
Radius (r) = = 30 cm = = 0.3 m ( 1 cm = m)
2 100 100
Height (h) = 2.1 m

22
Area levelled in 1 revolution = 2prh = 2 × × 0.3 × 2.1 m 2
7
22 3 21
2
= 2 × × × m = 3.96 m 2
7 10 10
\ Area levelled in 150 revolutions = 150 × 3.96 = 594 m 2

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