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In the above figure, what is the ratio of area
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08 Aug 2025, 09:19
08 Aug 2025, 09:19
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In the above figure, what is the ratio of area of triangle $A D E$ to the area of square $A B C D$ ?
(A) $\(\frac{5-x}{10}\)$
(B) $\(\frac{x}{10}\)$
(C) $\(\frac{5+x}{10}\)$
(D) $\(\frac{5-x}{2}\)$
(E) $\(5-x\)$
Re: In the above figure, what is the ratio of area
[#permalink]
20 Aug 2025, 10:48
20 Aug 2025, 10:48
Expert Reply
In the square ABCD above, $\(\mathrm{DE}=\mathrm{DC}-\mathrm{EC}=5-\mathrm{x}\)$, so the area of right triangle ADE is $\(\frac{1}{2} \times\)$ Base $\(\times\)$ Height $\(=\frac{1}{2} \times \mathrm{AD} \times \mathrm{DE}=\frac{1}{2} \times(5-\mathrm{x}) \times 5=\frac{5(5-\mathrm{x})}{2}\)$
Next the area of square $\(\mathrm{ABCD}=(\text { Side })^2=5^2=25\)$
So, the ratio of the area of triangle ADE and square ABCD is $\(\frac{5(5-\mathrm{x}) / 2}{25}=\frac{5(5-\mathrm{x})}{50}=\frac{(5-\mathrm{x})}{10}\)$ Hence the answer is (A).
Next the area of square $\(\mathrm{ABCD}=(\text { Side })^2=5^2=25\)$
So, the ratio of the area of triangle ADE and square ABCD is $\(\frac{5(5-\mathrm{x}) / 2}{25}=\frac{5(5-\mathrm{x})}{50}=\frac{(5-\mathrm{x})}{10}\)$ Hence the answer is (A).
gmatclubot
Re: In the above figure, what is the ratio of area [#permalink]
20 Aug 2025, 10:48
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