If each edge of a cube is increased by 50 %,t | vedclass

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(A) Let the original edge of the cube be $a$.Original surface area $S_1 = 6a^2$.New edge length $a' = a + 0.50a = 1.5a$.New surface area $S_2 = 6(1.5a

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(A) Let the original edge of the cube be $a$.Original surface area $S_1 = 6a^2$.New edge length $a' = a + 0.50a = 1.5a$.New surface area $S_2 = 6(1.5a)^2 = 6(2.25a^2) = 13.5a^2$.Increase in surface area $= S_2 - S_1 = 13.5a^2 - 6a^2 = 7.5a^2$.Percentage increase $= \left(\frac{7.5a^2}{6a^2}ight) 	imes 100 = 1.25 	imes 100 = 125 \%$.Alternatively,using the successive percentage formula for area change: $	ext{Percentage change} = (x + y + \frac{xy}{100}) \%$,where $x = y = 50$.Percentage increase $= (50 + 50 + \frac{50 	imes 50}{100}) \% = (100 + 25) \% = 125 \%$.

If each edge of a cube is increased by $50 \%$,the percentage increase in surface area is (in $\%$)

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