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(A) Let the diameter of the semicircle be $AB = 2R = 1$. Let the right triangle be $	riangle ABC$ with the right angle at $C$,where $C$ lies on the d

Vedclass · Accepted Answer

$\frac{1}{4}$

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(A) Let the diameter of the semicircle be $AB = 2R = 1$. Let the right triangle be $	riangle ABC$ with the right angle at $C$,where $C$ lies on the diameter $AD$. Let $AC = x$. Then $CD = 1 - x$. From the property of the circle,the altitude $BC$ to the diameter satisfies $BC^2 = AC 	imes CD = x(1 - x)$. Thus,$BC = \sqrt{x(1 - x)}$. The area $A$ of the triangle $	riangle ABD$ is given by $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AD 	imes BC = \frac{1}{2} 	imes 1 	imes \sqrt{x(1 - x)} = \frac{1}{2} \sqrt{x - x^2}$. To maximize $A$,we maximize $f(x) = x - x^2$. Taking the derivative,$f'(x) = 1 - 2x$. Setting $f'(x) = 0$,we get $x = \frac{1}{2}$. The maximum area is $A = \frac{1}{2} \sqrt{\frac{1}{2}(1 - \frac{1}{2})} = \frac{1}{2} \sqrt{\frac{1}{4}} = \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{4}$.

$A$ right triangle is drawn in a semicircle of radius $R = \frac{1}{2}$ with one of its legs along the diameter. The maximum area of the triangle is

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