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Q. Maximum area of a right angled triangle of given hypotenuse $= a$ is....

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Solution:

A right angled triangle of given hypotenuse has maximum area when it is isosceles.
$\therefore x=y$
$\therefore x^{2}+y^{2}=a^{2}$
$\Rightarrow 2 x^{2}=a^{2}$
$\therefore x=a / \sqrt{2}=y$
$\therefore \Delta=\frac{1}{2} x y=\frac{(a / \sqrt{2})^{2}}{2}$
$=\frac{a^{2}}{4}$

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