Question

prove that the area of equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on side of its diagonal​

Answer 1

Step-by-step explanation:

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