Question : In the given $\triangle{ABC}, DE \parallel BC$. If $BC=8 \text{ cm}$, $DE=6 \text{ cm}$, and the area of $\triangle ADE=90 \text{ cm}^2$, what is the area of $\triangle ABC$?
Option 1: 140
Option 2: 120
Option 3: 160
Option 4: 180
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Correct Answer: 160
Solution :
Given: $DE\parallel BC$ and $DE = 6\text{ cm}$, $BC = 8\text{ cm}$, and area of ($\triangle ADE) = 90 \text{ cm}^2$
Since $DE\parallel BC$, $\triangle ADE$ and $\triangle ABC$ are similar.
So, $\frac{(Area \ of \ \triangle ADE)}{(Area \ of \ \triangle ABC)} = \frac{DE^2}{BC^2}$
$⇒\frac{90}{(Area \ of \ \triangle ABC)} = \frac{6^2}{8^2}$
$⇒\frac{90}{Area\ of \ (\triangle ABC)}=\frac{36}{64}$
⇒ Area of $(\triangle ABC)=160 \text{ cm}^2$
Hence, the correct answer is 160.
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