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Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of an equilatral triangle described on one of its diagonals
Dear NickyHere 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 ~ ?AEBAr(ΔDBF) / Ar(ΔAEB) = DB2 / AB2 --------------------(i)We know that the ratio of the areas of two similar triangles is equal tothe 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 getAr(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.RegardsArun (askIITians forum expert)
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