In ABC , DE is drawn parallel to BC cutting AB in the ratio 2: 3.… - Vidyadip

Question

In $\triangle ABC , DE$ is drawn parallel to $BC$ cutting $AB$ in the ratio $2: 3$. Calculate:$ \text { (i) } \frac{\operatorname{area}(\triangle ADE )}{\operatorname{area}(\triangle ABC )};\text { (i) } \frac{\operatorname{area}(\text {trapezium EDBC})}{\operatorname{area}(\triangle ABC)}$

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Answer

$ AD : DB =2: 3$
$= AB + DB$
$=2+3$
$=5$
$\text { (i) } \frac{\operatorname{area}(\Delta ADE )}{\operatorname{area}(\Delta ABC )}=\frac{ AD ^2}{ AB ^2}$
$\Rightarrow \frac{\operatorname{area}(\Delta ADE )}{\operatorname{area}(\Delta ABC )}=\frac{2^2}{5^2}$
$\Rightarrow \frac{\operatorname{area}(\Delta ADE )}{\operatorname{area}(\Delta ABC )}=\frac{4}{25} .$
$\text { (ii) } \frac{\operatorname{area}(\text { trapezium EDBC) }}{\operatorname{area}(\Delta ABC )}=\frac{\operatorname{area}(\Delta ABC )-\operatorname{area}(\Delta ADE )}{\operatorname{area}(\text { trapezium EDBC })}$
$\Rightarrow \frac{\operatorname{area}(\Delta ABC )}{\operatorname{area}(\Delta ABC )}=\frac{25-4}{25}$
$\Rightarrow \frac{\operatorname{arapezium EDBC})}{25} .$

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