In figure, DE || BC. If DE = 3 cm, BC = 6 cm and area of ΔADE = 15 sq cm, then the area of ΔABC is

It is given thatDE = 3 cm, BC = 6 cm

Let area of $$\triangle$$ ABC = $$x$$ sq cm and area of $$\triangle$$ ADE = 15 sq cm

In $$\triangle$$ ADE and $$\triangle$$ ABC

$$\angle$$ DAE = $$\angle$$ BAC  (common)

$$\angle$$ ADE = $$\angle$$ ABC   (Alternate interior angles)

$$\angle$$ AED = $$\angle$$ ACB    (Alternate interior angles)

=> $$\triangle$$ ADE $$\sim$$ $$\triangle$$ ABC

=> Ratio of Area of $$\triangle$$ ADE : Area of $$\triangle$$ ABC = Ratio of square of corresponding sides = $$(DE)^2$$ : $$(BC)^2$$

=> $$\frac{15}{x}=\frac{(3)^2}{(6)^2}$$

=> $$\frac{15}{x} = \frac{9}{36}$$

=> $$\frac{15}{x}=\frac{1}{4}$$

=> $$x=15\times4=60$$ $$cm^2$$

=> Ans - (D)