In the given figure, if AD = 3, DE = 4, AB = 12, BF = 2, FG= 6, BC = 10, then the value of $\frac{M}{N}$ is:
(Assume: M is the area of the quadrilateral FGDE and N is the area of the triangle ABC.)
Answer & explanation
Correct answer: option 1
We know that,
Area of triangle = \(\frac{1}{2}\) × Multiple of adjacent sides × sin(Angle between those sides)
We have,
AD = 3, DE = 4, AB = 12, BF = 2, FG = 6, BC = 10
M is the area of the quadrilateral FGDE and N is the area of the triangle ABC.
Area of ΔEBF = \(\frac{1}{2}\) × BE × BF × sinB
= \(\frac{1}{2}\) × 5 × 2 × sinB = 5 × sinB
Area of ΔDBG = \(\frac{1}{2}\) × BD × BG × sinB
= \(\frac{1}{2}\) × 9 × 8 × sinB = 36 × sinB
Area of ΔABC = \(\frac{1}{2}\) × BA × BC × sinB
= \(\frac{1}{2}\) × 12 × 10 × sinB = 60 × sinB
Area of quadrilateral FGDE(M) = Area of ΔDBG - Area of ΔEBF
= 36sinB - 5sinB = 31sinB
Area of ΔABC(N) = 60sinB
= \(\frac{M}{N}\) = \(\frac{31sinB }{60sinB }\) = \(\frac{31}{60}\)