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

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(D) Let the original edge of the cube be $a$. The original surface area is $S_1 = 6a^2$.If the edge is increased by $50 \%$,the new edge $a' = a + 0.5

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(D) Let the original edge of the cube be $a$. The original surface area is $S_1 = 6a^2$.If the edge is increased by $50 \%$,the new edge $a' = a + 0.5a = 1.5a$.The new surface area is $S_2 = 6(1.5a)^2 = 6(2.25a^2) = 13.5a^2$.The increase in surface area is $S_2 - S_1 = 13.5a^2 - 6a^2 = 7.5a^2$.The percentage increase is $\frac{7.5a^2}{6a^2} 	imes 100 = 1.25 	imes 100 = 125 \%$.Alternatively,using the successive percentage change formula for area (two dimensions): $x + y + \frac{xy}{100} = 50 + 50 + \frac{50 	imes 50}{100} = 100 + 25 = 125 \%$.

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

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