Page 289 - Start Up Mathematics_8 (Non CCE)

Area of roof = l × b = 12 m × 6 m = 72 m 2
Area of 1 door = 2.5 m × 1.5 m = 3.75 m 2

2
\ Area of 2 doors = (2 × 3.75) m = 7.5 m 2
Area of 1 window = 1.5 m × 1.5 m = 2.25 m 2
2
\ Area of 4 windows = (4 × 2.25) m = 9 m 2
Area to be whitewashed = Area of 4 walls + Area of roof – Area of 2 doors – Area of 4 windows
2
= (108 + 72 – 7.5 – 9) m = 163.5 m 2
2
Cost of whitewashing 1 m area = ` 6
2
\ Cost of whitewashing 163.5 m area = ` (6 × 163.5) = ` 981
Example 19: If the edge of a cube is doubled, what will be the change in its surface area?

Solution: Let the edge of the cube be x unit.
2
2
Surface area = 6(edge) = 6x sq. units
When the edge of the cube is doubled, new edge = 2x units
2
2
2
New surface area = 6(2x) = 6(4x ) = 4(6x ) sq. units
Thus, the surface area of a cube becomes 4 times when its edge is doubled.
Example 20: A suitcase of measurement 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth.
How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
(NCERT)

Solution: Length (l) = 80 cm = 0.8 m, breadth (b) = 48 cm = 0.48 m, height (h) = 24 cm = 0.24 m
\ Total surface area of suitcase = 2(lb + bh + hl)

= 2 (0.8 × 0.48 + 0.48 × 0.24 + 0.24 × 0.8) m 2
2
2
= 2 (0.384 + 0.1152 + 0.192) m = 2 (0.6912) m = 1.3824 m 2
Width of cloth = 96 cm = 0.96 m
1 3824.
\ Length of cloth required to cover 1 suitcase = = 1.44 m
096.
Hence, the length of cloth required to cover 100 suit cases = 1.44 m × 100 = 144 m
Example 21: The ratio of surface areas of two cubes is 1 : 9. Find the ratio of their volumes.

Solution: Let the surface area of first cube with edge x be A and the surface area of second cube with
1
1
edge x be A . Let V and V be their respective volumes.
2
2
1
2
A 1 6x 2 1
2
Then, 1 = fi 1 = {  Surface area of cube = 6 × (edge) }
A 2 9 6x 2 2 9
x 2 1 x 1
fi 1 = fi 1 =
x 2 2 9 x 2 3
V x 3 Ê x ˆ 3 Ê 1ˆ 3 1
Now, 1 = 1 = Á 1 ˜ = Á ˜ =
V 2 x 2 3 Ë x 2 ¯ Ë 3¯ 27

\ V : V = 1 : 27
2
1
Hence, their volumes are in the ratio 1 : 27.

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