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`        Prove that the area of equilateral triangle described on side of a square is half the area of equilateral triangle described on its diagonals`
one year ago

```							Dear Vaibhav I am not attaching any figure.but you can assume related to my answer. 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.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 ~ △AEBAr(ΔDBF) / Ar(ΔAEB) = DB2 /AB2 -----------1We 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} -----2Substitute equation 2 in equation 1, 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)
```
one year ago
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