Δ ABC is an equilateral triangle and D, E are midpoints of AB and AC respectively. Then the area of Δ ABC : the area of the trapezium BDEC is

Given : ABC is an equilateral triangle and D and E are mid points of AB and AC respectively.

To find : Area of Δ ABC : the area of the trapezium BDEC

Solution : Let the side of triangle AB = 2 cm

=> AD = DB = 1 cm

Clearly, $$\triangle$$ ADE $$\sim \triangle$$ ABC

Ratio of areas of two similar triangles is equal to the ratio of squares of corresponding sides.

=> $$\frac{ar(ADE)}{ar(ABC)}=(\frac{AD}{AB})^2$$

=> $$\frac{ar(ADE)}{ar(ABC)}=(\frac{1}{2})^2 = \frac{1}{4}$$

Let ar$$(\triangle ADE) = x$$ and ar$$(\triangle ABC)=4x$$

=> ar(BDEC) = ar$$(\triangle ABC) - $$ar$$(\triangle ADE)$$

= $$4x-x=3x$$

$$\therefore$$ $$\frac{ar(ABC)}{ar(BDEC)}=\frac{4x}{3x}$$

= 4 : 3

=> Ans - (D)