Question

In a triangle ABC, a straight line parallel to BC intersects AB and AC at point D and E respectively. If the area of ADE is one-fifth of the area of ABC and $BC=10$ cm, then DE equals

• A. $2$ cm
• B. $2\sqrt{5}$
• C. $4$ cm
• D. $4\sqrt{5}$

Answer 1

Answer: (B)

$2\sqrt{5}$

It is given that area of $△ADE=\frac{1}{5}\times$ area of $△ABC$

Let the height of the $△ABC$ be $h$ and height of the $△ADC$ be $x$ as shown in the diagram. Therefore, we have:

$\frac{1}{2}\times x\times DE=\frac{1}{5}\times \frac{1}{2}\times h\times 10\Rightarrow \frac{1}{2}xDE=\frac{10}{10}h\Rightarrow \frac{1}{2}xDE=h\Rightarrow \frac{x}{h}=\frac{2}{DE}$

Area of trapezium $DECB=$Area of $△ABC-$Area of$△ADE$. Thus,

$\frac{h-x}{2}(10+DE)=\frac{4}{5}\times \frac{1}{2}\times h\times 10$

$\Rightarrow (1-\frac{x}{h})(10+DE)=8$

$\Rightarrow (1-\frac{2}{DE})(10+DE)=8$

$\Rightarrow 10+DE-\frac{20}{DE}-2=8$

$\Rightarrow DE-\frac{20}{DE}+8=8$

$\Rightarrow DE-\frac{20}{DE}=8-8$

$\Rightarrow DE-\frac{20}{DE}=0\Rightarrow \frac{{DE}^2-20}{DE}=0$

$\Rightarrow {DE}^2-20=0\Rightarrow {DE}^2=20\Rightarrow DE=\sqrt{20}\Rightarrow DE=\sqrt{2\times 2\times 5}\Rightarrow DE=2\sqrt{5}$

Hence, $DE=2\sqrt{5}$.