prove that area of equilateral triangle to solve on side of square equal to half of the area of equilateral triangle described one of its diagonal
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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.
We have 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 -----------1
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} -----2
Substitute equation 2 in equation 1, 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.
We have 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 -----------1
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} -----2
Substitute equation 2 in equation 1, 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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