D and E are points on the sides. AB and AC respectively of ΔABC such that DE is parallel to BC and AD : DB = 2 : 3. CD and BE intersect each other at F. Then the ratio of the area of ΔDEF and ΔCBF is :
Answer & explanation
Correct answer: option 1
ΔADE ∼ ΔABC
\(\frac{AD}{DE}\) = \(\frac{AB}{BC}\) ⇒ \(\frac{2}{DE}\) = \(\frac{2+3}{BC}\)
⇒ \(\frac{DE}{BC}\) = \(\frac{2}{5}\)
ΔDEF ∼ ΔCBF
\(\frac{Area\;of\;ΔDEF}{Area\;of\;ΔCBF}\) = \(\frac{DE^2}{BC^2}\) = \(\frac{(2)^2}{(5)^2}\) = \(\frac{4}{25}\)
Req. Ratio = 4 : 25