In the given figure, AD = 4, DE = 6, AB = 15, BF = 2, FG = 8, BC = 13, then find the ratio between Area of quadrilateral FGDE to Area of ΔABC.
Answer & explanation
Correct answer: option 1
Area = \(\frac{1}{2}\) × adjacent side × Sin (angle between sides)
Area of ΔEBF = \(\frac{1}{2}\) × EB × BF × Sin B
Area of ΔEBF = \(\frac{1}{2}\) × 5 × 2 × Sin B = 5 Sin B
Area of ΔBDG = \(\frac{1}{2}\) × BD × BG × Sin B
Area of ΔBDG = \(\frac{1}{2}\) × 11 × 10 × Sin B = 55 Sin B
Area of ΔABC (P) = \(\frac{1}{2}\) × BA × BC × Sin B
Area of ΔABC (P) = \(\frac{1}{2}\) × 15 × 13 × Sin B = \(\frac{195}{2}\) Sin B
Area of quard. FGDE (Q) = Area of ΔBDG - Area of ΔBEF
= 55 Sib B - 5 Sin B = 50 Sin B
Now,
⇒ \(\frac{Q}{P}\) = \(\frac{50 Sin B}{195/2 Sin B}\) = \(\frac{100}{195}\) = \(\frac{20}{39}\)