D and E are points on side AB and AC of ∆ABC. DE is parallel to BC. If AD:DB = 1:2 and area of ∆ABC is 45 sq cm, what is the area (in sq cm) of quadrilateral BDEC?

It is given that AD : DB = 1 : 2

Let AD = 1 cm and DB = 2 cm

Let area of $$\triangle$$ ADE = $$x$$ 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 = $$(AD)^2$$ : $$(AB)^2$$

= $$\frac{(1)^2}{(1 + 2)^2} = \frac{x}{(45)}$$

=> $$\frac{x}{45} = \frac{1}{9}$$

=> $$x=\frac{45}{9}=5$$

$$\therefore$$ Area of quadrilateral BDEC = ar($$\triangle$$ ABC) - ar($$\triangle$$ ADE)

= $$45-5=40$$ $$cm^2$$

=> Ans - (B)