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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 equilateral triangle described on one of its diagonals. |
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Answer» Let the side length of square ABCD is a. The length of diagonal of square is \(\sqrt{a^2+a^2}\) = \(\sqrt 2 a\). Let A1 and A2 are areas of equilateral triangles described on one side of square and one diagonal of that square, respectively. And we also know that area of equilateral triangle whose side length is a is \(\frac{\sqrt 3}{4}\)a2 ∴ A1 = \(\frac{\sqrt 3}{4}\) a2. And A2 = \(\frac{\sqrt 3}{4}\) (\(\sqrt 2 a\))2 = \(\frac{\sqrt 3}{4}\) × 2a2 = \(\frac{2\sqrt 3}{4}\) a2 ∴\(\frac{A_1}{A_2}\) = \(\frac{1}{2}\) ⇒ A1 = \(\frac{1}{2}\)A2. Hence Proved. |
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