prove that the area f an equilateral triangle described on one side of square is equal to half the area of the equilateral triangle described on one of its diagonal
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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.
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.
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.
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.
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