Question : If $D$ and $E$ are points on sides $AB$ and $AC$ of $\Delta ABC$. $DE$ is parallel to $BC$. If $AD: DB = 2:3$ and the area of $\Delta ADE$ is 4 sq. cm, what is the area (in sq. cm) of quadrilateral $BDEC$?
Option 1: 25
Option 2: 21
Option 3: 5
Option 4: 9
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Correct Answer: 21
Solution :
We have,
$AD:DB = 2:3$, area($\triangle ADE$) = 4 sq. cm
Let $AD = 2x$ and $DB = 3x$
So, $AB = AD + DB = 2x + 3x = 5x$
Two triangles $ABC$ and $ADE$ such that $\triangle ABC$ ∼$\triangle ADE$
So, $\frac{\text{area of} (\triangle ABC)}{\text{area of } (\triangle ADE)}=(\frac{AB}{AD})^2$
⇒ $\frac{\text{area of} (\triangle ABC)}{4}=(\frac{5x}{2x})^2$
⇒ $\frac{\text{area of} (\triangle ABC)}{4}=\frac{25}{4}$
⇒ area of ($\triangle ABC$) = 25 sq. cm
Area of quadrilateral $BDEC$ = area of ($\triangle ABC$)–area of ($\triangle ADE$) = 25 – 4 = 21 sq.cm
Hence, the correct answer is 21.
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