In the given figure, if AD = 3, DE = 4, AB = 12, BF = 2, FG = 6, BC = 10, then the value of $$\frac{{\text{M}}}{{\text{N}}}$$ is: (Assume: M is the area of the quadrilateral FGDE and N is the area of the triangle ABC.)

Solution (By Examveda Team)

$$\eqalign{ & {\text{Area of }}\Delta ABC = \frac{1}{2}\sin \theta \times 12 \times 10 \cr & {\text{Let }}\frac{1}{2}\sin \theta = x \cr & {\text{Area of }}\Delta ABC = 120x \cr & {\text{Same as,}} \cr & {\text{Area of }}\Delta BEF = 10x \cr & {\text{Area of }}\Delta BDG = 9 \times 8x = 72x \cr & M = 72x - 10x = 62x \cr & N = 120x \cr & \frac{M}{N} = \frac{{62x}}{{120x}} = \frac{{31}}{{60}} \cr} $$