Question

Prove that the area of an equilateral teiangle descirbed on one side of the sqaure is equal to half the area of the equilateral triangle descirbed on one of its diagonal

Answer 1

Here ABCD is a square, AEB is an equilateral triangle described on the side of the square and DBF is an equilateral triangle described on diagonal BD of the square.

To Prove: Ar(ΔDBF) / Ar(ΔAEB) = 2 / 1

Proof: If two equilateral triangles are similar then all angles are = 60 degrees.

Therefore, by AAA similarity criterion , △DBF ~ △AEB

Ar(ΔDBF) / Ar(ΔAEB) = DB2 / AB2 --------------------(i)

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

But, we have DB = √2AB {But diagonal of square is √2 times of its side} -----(ii).

Substitute equation (ii) in equation (i), we get

Ar(ΔDBF) / Ar(ΔAEB) = (√2AB )2 / AB2 = 2 AB2 / AB2 = 2

∴ Area of equilateral triangle described on one side os square is equal to half the area of the equilateral triangle described on one of its diagonals.