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.)

$\frac{31}{60}$

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

= \(\frac{M}{N}\) = \(\frac{31sinB }{60sinB }\) = \(\frac{31}{60}\)