In $$\triangle ABC$$ and  $$\triangle PQR$$, $$\angle B=\angle Q, \angle C=\angle R$$. $$M$$ is  the midpoint of side $$QR$$. If $$AB : PQ = 7 : 4$$, then $$\frac{area(\triangle ABC)}{area(\angle PMR)}$$ is:

=> $$\angle B=\angle Q$$

and $$ \angle C=\angle R$$

Thus, $$\triangle ABC$$ $$\sim$$ $$\triangle PQR$$  (By AA criterion)

In $$\triangle$$ PQR, PM is the median, => It divides the triangle in two parts of equal areas.

=> $$ar(\triangle PMR)=\frac{1}{2}\times ar(\triangle PQR)$$ -------------(i)

Let $$AB=7$$ cm and $$PQ=4$$ cm

Now, ratio of areas of two similar triangles is equal to the square of ratio of their corresponding sides.

$$\therefore$$ $$\frac{ar(\triangle ABC)}{ar(\angle PMR)}=$$ $$\frac{2\times ar(\triangle ABC)}{ar(\angle PQR)}$$     [Using equation (i)]

= $$2\times(\frac{7}{4})^2=2\times\frac{49}{16}=\frac{49}{8}$$

=> Ans - (C)