proof for heron's formula for equilateral triangle.

we know that area of a triangle is = 1/2 x base x height now lets us take an equilateral triangle ABC with side = m units and altitude AD to base BC.let AD = n units. now in right angled triangle ABD , AB = m , BD = m/2. now by Pythagoras therom AD-square + BD-square = AB-square i.e. n/2-square + m/2 square = m- square i.e (n-square +m-square )/4 = m-square i.e n-square + m-square = 4(m-square) i.e n-square = 3(m-square) i.e n = m x root 3 now area of triangle ABC = 1/2 x base x height = 1/2 x m/2 x m x root 3 = root 3/4 x m-square.

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Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
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