In triangle PQR, points E and F are on sides PQ and PR respectively such that EF is parallel to QR. If PE = 2 cm and EQ = 3 cm, then area($$\triangle$$ PQR) : area($$\triangle$$ PEF) is equal to:

$$\triangle$$PQR and $$\triangle$$PEF are similar triangles.

$$\Rightarrow$$  $$\frac{\text{Area of triangle PQR}}{\text{Area of triangle PEF}}$$ = $$\left(\frac{PQ}{PE}\right)^2$$

$$\Rightarrow$$  $$\frac{\text{Area of triangle PQR}}{\text{Area of triangle PEF}}$$ = $$\left(\frac{PE+EQ}{PE}\right)^2$$

$$\Rightarrow$$ $$\frac{\text{Area of triangle PQR}}{\text{Area of triangle PEF}}$$ = $$\left(\frac{2+3}{2}\right)^2$$

$$\Rightarrow$$ $$\frac{\text{Area of triangle PQR}}{\text{Area of triangle PEF}}$$ = $$\frac{25}{4}$$

$$\Rightarrow$$  Area of triangle PQR : Area of triangle PEF = 25:4

Hence, the correct answer is Option B