In the given figure, in triangle ABC, BC = 8 cm, AC = 9 cm and AB = 2 cm. AE = 24 cm, BF = 32 cm and DC = 27 cm. What is the ratio of the area of triangle DEA and area of triangle DLF ?
Answer & explanation
Correct answer: option 2
Let, ∠DAE = θ
then, ∠BAC = 180° - θ
and let ∠FCD = ∝
then, ∠BCA = 180° - ∝
∴ Ratio of area of
\(\frac{ΔDAE}{ΔABC}\) = \(\frac{\frac{1}{2}× 18 × 24 × sinθ}{\frac{1}{2} × 12 × 9 × sin (180 - θ)}\)
\(\frac{ΔDAE}{ΔABC}\) = \(\frac{4sinθ}{sinθ}\) = \(\frac{4}{1}\)
\(\frac{ΔDCE}{ΔABC}\) = = \(\frac{9sin∝}{sin∝}\) = \(\frac{9}{1}\)
Ratio of area,
ΔDAE : ΔABC : ΔDCF
4 : 1
1 : 9
4 : 1 : 9
∴ Ratio of area \(\frac{ΔDAE}{ΔDCF}\) = \(\frac{4}{9}\)
= 4 : 9