MCQ
$\triangle D E F \sim \triangle A B C$; If $D E: A B=2: 3$ and $\operatorname{ar}(\triangle D E F)$ is equal to $44 $ square units, then area $(\triangle A B C)$ in square units is
- ✓$99$
- B$120$
- C$\frac{176}{9}$
- D$66$
✓
Answer
Correct option: A.
$99$
We know that,
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\frac{\operatorname{ar}(\triangle D E F)}{\operatorname{ar}(\triangle A B C)}=\frac{D E^2}{A B^2} \ldots \ldots(1)$
We have,
$ar(\triangle D E F)=44$ square units, and $ \frac{D E}{A B}=\frac{2}{3}$
Substituting above values in equation $(1),$
$\frac{44}{\triangle A B C}=\frac{2^2}{3^2}$
$\Rightarrow \frac{44}{\triangle A B C}=\frac{4}{9}$
$\Rightarrow \triangle A B C=\frac{44 \times 9}{4}$
$\Rightarrow \triangle A B C=11 \times 9=99$
The area $(\triangle A B C)$ in square units is $99$ .
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\frac{\operatorname{ar}(\triangle D E F)}{\operatorname{ar}(\triangle A B C)}=\frac{D E^2}{A B^2} \ldots \ldots(1)$
We have,
$ar(\triangle D E F)=44$ square units, and $ \frac{D E}{A B}=\frac{2}{3}$
Substituting above values in equation $(1),$
$\frac{44}{\triangle A B C}=\frac{2^2}{3^2}$
$\Rightarrow \frac{44}{\triangle A B C}=\frac{4}{9}$
$\Rightarrow \triangle A B C=\frac{44 \times 9}{4}$
$\Rightarrow \triangle A B C=11 \times 9=99$
The area $(\triangle A B C)$ in square units is $99$ .
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