Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal.
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Sol:
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.
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.
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