square inscribed in a right triangle

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A square of maximum possible area is circumscribed by a right angle triangle ABC in such a way that one of its side just lies on the hypotenuse of the triangle.
What is the area of the square? actually the answer is given as $(abc/(a^2+b^2+ab))^2$ Please provide the approach to solve the problem.

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4 Answers

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Hint:

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The red solid line is the height dropped onto the hypotenuse, i.e. $h = \frac{ab}{c}$ and the red dotted lines are of the same length. The green parallel lines are unnecessary, but might get you some intuitions.

Good luck! ;-)

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Consider - a,b as right legs and c as the hypotenuse.

Let side of square = s AC = b, BC = a, AB = c.

FB = as/b and AE = bs/a as the colored triangles are similar to the bigger triangle.

Steps to calculate area (S^2) :

1)Calculate GB and AD using right angle triangle rule for triangles GBF and ADE.

2)Calculate GD using right angle triangle rule for triangle GCD.

3)GD^2 = s^2. You get a quadratic equation in s which can be factorized. You get s = (abc)/(a^2 + b^2 +a.b)

If still not clarified will post the answer then.

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Hint: The square formed some similar right-angled triangles, make use of the ratio of the sides.

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The question can be easily solved by subtracting the two sides of a right triangle from the bigger triangle i.e. subtract 'as/b' and 'bs/a' from the hypotenuse of larger triangle 'c'.and it will be equal to the side of a square. S=c-(as/b+bs/a). And thus s=abc/a^2+b^2+ab

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