Question : In $\triangle$ABC, D and E are two points on the sides AB and AC, respectively, so that DE $\parallel$ BC and $\frac{AD}{BD}=\frac{2}{3}$. Then $\frac{\text{Area of trapezium DECB}}{\text{Area of $\triangle$ABC}}$ is equal to:
Option 1: $\frac{5}{9}$
Option 2: $\frac{21}{25}$
Option 3: $1\frac{4}{5}$
Option 4: $5\frac{1}{4}$
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Correct Answer: $\frac{21}{25}$
Solution :
Given: $\frac{AD}{BD}=\frac{2}{3}$ and DE $\parallel$ BC
So, $\triangle$ABC and $\triangle$ADE are similar.
⇒ $\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{(2+3)^2}{(2)^2}=\frac{25}{4}$
Let the area of $\triangle$ABC and $\triangle$ADE be 25k and 4k, respectively.
So, the area of trapezium DECB = (Area of $\triangle$ABC) – (Area of $\triangle$ADE) = 25k – 4k = 21k
Therefore, $\frac{\text{Area of trapezium DECB}}{\text{Area of $\triangle$ABC}}$ = $\frac{21k}{25k}$ = $\frac{21}{25}$
Hence, the correct answer is $\frac{21}{25}$.
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