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If the sum of the lengths of the hypotenuse and a side of a right triangle is given, show that the area of the triangle is maximum, when the angle between them is $60^{\circ}$.


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Let the hypotenuse of the right triangle be x, and the height be y
Hence its base is $ \sqrt{x^2-y^2}$ by applying phythagorous theorem.
Hence its are = $ \large\frac{1}{2} \times base \times height$
Area = $ \large\frac{1}{2}$$ \times \sqrt{x^2-y^2} \times y$
But it is given $x+y = p(say)$
Substituting this in the area we get
Area = $ \large\frac{1}{2}$$ \times \sqrt{(p-y)^2} - y^2 \times y$
$ \large\frac{1}{2}$$ y \sqrt{p^2+y^2-2py-y^2}$
$ = \large\frac{1}{2}$$y \sqrt{p^2-2py}$
Squaring on both the sides we get
$ ( Area)^2 = \large\frac{1}{4}$$y^2(p^2-2py)$
i.e., $A = \large\frac{1}{4}$$ y^2(p^2-2py)$
$ = \large\frac{1}{4}$$ p^2y^2 = \large\frac{1}{2}py^3$
For maximum or miniumu area
$ \large\frac{dy}{dA} $$= 0$
Here the area of the triangle is maximum when $ x =\large\frac{2p}{3}\: and \: y = \large\frac{p}{3}$
$ \cos \theta = \large\frac{y}{x}$
$ = \large\frac{\Large\frac{p}{3}}{2\Large\frac{p}{3}}$
$ \therefore \cos \theta = \large\frac{1}{2}$
$ \Rightarrow \theta = \Large\frac{\pi}{3}$ or $60^{\circ}$
Hence the area is maximum if the angle between the hypotenuse and the side is $60^{\circ}$
answered Mar 31, 2014 by thanvigandhi_1
edited Apr 14, 2014 by thanvigandhi_1

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