The least area of a circle circumscribing any | vedclass

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(A) Let the legs of the right triangle be $x$ and $y$. The area of the triangle is $S = \frac{1}{2}xy$,so $xy = 2S$.The hypotenuse of the right triang

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$\pi S$

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(A) Let the legs of the right triangle be $x$ and $y$. The area of the triangle is $S = \frac{1}{2}xy$,so $xy = 2S$.The hypotenuse of the right triangle is $h = \sqrt{x^2 + y^2}$.$A$ circle circumscribing a right triangle has the hypotenuse as its diameter. Thus,the diameter $d = \sqrt{x^2 + y^2}$,and the radius $r = \frac{\sqrt{x^2 + y^2}}{2}$.The area of the circle is $A = \pi r^2 = \pi \left( \frac{\sqrt{x^2 + y^2}}{2} \right)^2 = \frac{\pi}{4}(x^2 + y^2)$.Since $y = \frac{2S}{x}$,we substitute this into the area formula:$A(x) = \frac{\pi}{4} \left( x^2 + \left( \frac{2S}{x} \right)^2 \right) = \frac{\pi}{4} \left( x^2 + \frac{4S^2}{x^2} \right)$.To find the minimum area,we differentiate $A(x)$ with respect to $x$ and set it to zero:$A'(x) = \frac{\pi}{4} \left( 2x - \frac{8S^2}{x^3} \right) = 0$.$2x = \frac{8S^2}{x^3} \implies x^4 = 4S^2 \implies x^2 = 2S$.Substituting $x^2 = 2S$ back into the area equation:$A = \frac{\pi}{4} \left( 2S + \frac{4S^2}{2S} \right) = \frac{\pi}{4} (2S + 2S) = \frac{\pi}{4} (4S) = \pi S$.Thus,the least area of the circle is $\pi S$.

The least area of a circle circumscribing any right triangle of area $S$ is

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