Question #75eb0

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Question #75eb0

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Answer 1 · 2018-01-03T04:56:03

$96 %$ $96%$

Explanation:

Suppose the side of square is $x$ $x$
Then its area = $x^{2}$ $x^2$

When there is 40% increment in the length of the side, the side length would become = $1.4 x$ $1.4x$

( $40 %$ $40%$ = $\frac{40}{100}$ $40/100$ = $0.4$ $0.4$ and when $0.4 x$ $0.4x$ is added to initial length $x$ $x$ , the increased or new length becomes $1.4 x$ $1.4x$ )

New Area of square with increased length = ${(1.4 x)}^{2}$ $(1.4x)^2$ = $1.96 x^{2}$ $1.96x^2$

Now, percentage change in area of resultant square =

$(\frac{1.96 x^{2} - x^{2}}{x^{2}}) \cdot 100$ $((1.96x^2-x^2)/x^2)*100$ = $96 %$ $96%$

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Question #75eb0

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