$(a)\;minimum\qquad(b)\;maximum\qquad(c)\;equal\qquad(d)\;None\;of\;these$

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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 $60^{\circ}$

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