Question

ABC is an equilateral triangle and D is the mid-point of side AC. ∆ BDP is an other equilateral triangle, show that ar(BDP) = ¼ar(ABC)

Answer 1

Answer:

Given :

ΔABC and ΔBDE are equilateral triangles. D is the point of BC.

ΔABC∼ΔBDE (By AAA criteria of similarity) [all angles of the equilateral triangles are equal]

We know that the ratio of the areas of the two similar triangles is equal to the ratio of squares of their corresponding sides.

arΔABC/ arΔBDE = (BC/BD)²

BD = DC as D is the mid point of BC.

Hence

=>arΔABC/ arΔBDE = ((BD+DC)/ BD)²

=>arΔABC/ arΔBDE = ((BD+ BD )/ BD)²

=>arΔABC/ arΔBDE = (2BD/BD)²

=>arΔABC/ arΔBDE = (2/1)²

=>arΔABC/ arΔBDE = 4/1

=>arΔABC : arΔBDE = 4 : 1

Hence, the ratio of areas of ΔABC  and ΔBDE is 4 : 1 .