Question : D and E are points on sides AB and AC of $\Delta ABC$. DE is parallel to BC. If AD : DB = 2 : 3. What is the ratio of the area of $\Delta ADE$ and the area of quadrilateral BDEC?
Option 1: 4 : 21
Option 2: 4 : 25
Option 3: 4 : 29
Option 4: 4 : 9
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Correct Answer: 4 : 21
Solution :
AD : DB = 2 : 3
Let AD =$2x$
DB = $3x$
Now, the ratio of the area of the similar triangle is equal to the square of the ratio of their corresponding sides.
Hence, in similar $\triangle ABC$ and $\triangle ADE$
$\frac{\text{Area of}\triangle ABC}{\text{Area of}\triangle ADE}=\frac{AB^2}{AD^2}$
⇒ $\frac{\text{Area of}\triangle ABC}{\text{Area of}\triangle ADE}=\frac{(5x)^2}{(2x)^2} = \frac{25}{4}$
Let the area of$\triangle ABC$ be 25k and the area of $\triangle ADE$ be 4k
$\therefore$ Area of quadrilateral BDEC = Area of $\triangle ABC-$Area of $\triangle ADE$
= 25k – 4k = 21k
The ratio of the area of $\Delta ADE$ and the area of quadrilateral BDEC = 4 : 21.
Hence, the correct answer is 4 : 21.
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