The Sulvasutras give the following formula for constructing a square whose area

Question

The Sulvasutras give the following formula for constructing a square whose area equals n times the area of a given square:
If the given square has side (alpha), construct an isosceles triangle with base (n-1)(alpha) and two sides equal to (n+1)(alpha)/2 in length. The altitude (i.e., the perpendicular bisector) of the triangle is the side of a square with area n(alpha)^2.
Here n is an integer at least 2. Prove this algorithm is correct.

Wouldn't the formula be something like L^2=n*(alpha)^2. I do not understand how to apply the formulas for isosceles triangle into the proof.

Explanation / Answer

n *a ^2 = ((n+1)a/2)^2 - ((n-1)a/2)^2

which will give

n^2 a^2 = n^2 a^2

is the phythagoras theorom that thealtitude^2= twinsideside^2-half of linegth of base^2

should be satisfied

which is satifiesd in the above formula

so the result is true

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