question_answer

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)         2/5 cm  

C)  4 cm                     

D)         4/5 cm

Correct Answer: B

Solution :

 \[\frac{Area\,of\,\Delta \,ADE}{Area\,of\,\Delta \,ABC}=\frac{D{{E}^{2}}}{B{{C}^{2}}}\] or            \[\frac{\frac{1}{5}\times Area\,of\,\Delta \,ABC}{Area\,of\,\Delta \,ABC}=\frac{D{{E}^{2}}}{{{10}^{2}}}\]                                                 or            \[\frac{1}{5}=\frac{D{{E}^{2}}}{100}\]                 or            \[D{{E}^{2}}=\frac{100}{5}=20\]                 \[\therefore \]  \[DE=\sqrt{20}\]                                 \[=2\sqrt{5}\,CM\]