Question

In a $△ABC$, 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. $2\sqrt{5}$ cm
• C. $4$ cm
• D. $4\sqrt{5}$ cm

Answer 1

Answer: (B)

$2\sqrt{5}$ cm

$\frac{\text{Area of}\Delta ADE}{\text{Area of}\Delta ABC}=\frac{D{E}^2}{B{C}^2}$

$\frac{\frac{1}{5}\times \text{Area of}\Delta ABC}{\text{Area of}\Delta ABC}=\frac{D{E}^2}{{10}^2}$

$\frac{1}{5}=\frac{D{E}^2}{100}$

$D{E}^2=\frac{100}{5}=20$

$\therefore DE=\sqrt{20}=2\sqrt{5}cm$