The triangle of maximum area that can be insc | vedclass

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(C) Let the triangle be $\Delta ABC$ inscribed in a circle of radius $r$ with center $O$.Let $	heta$ be the angle $\angle OBP$,where $P$ is the midpo

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An equilateral triangle having each of its sides of length $\sqrt{3}r$.

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(C) Let the triangle be $\Delta ABC$ inscribed in a circle of radius $r$ with center $O$.Let $	heta$ be the angle $\angle OBP$,where $P$ is the midpoint of $BC$.The height of the triangle is $h = r \sin 	heta + r$.The base of the triangle is $BC = 2r \cos 	heta$.The area of the triangle $\Delta$ is given by $\Delta = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} (2r \cos 	heta) (r \sin 	heta + r) = r^2 \cos 	heta (1 + \sin 	heta)$.To find the maximum area,differentiate $\Delta$ with respect to $	heta$:$\frac{d\Delta}{d	heta} = r^2 [-\sin 	heta (1 + \sin 	heta) + \cos 	heta (\cos 	heta)] = r^2 [-\sin 	heta - \sin^2 	heta + \cos^2 	heta] = r^2 [-\sin 	heta - \sin^2 	heta + (1 - \sin^2 	heta)] = r^2 [1 - \sin 	heta - 2 \sin^2 	heta]$.Setting $\frac{d\Delta}{d	heta} = 0$,we get $2 \sin^2 	heta + \sin 	heta - 1 = 0$,which factors as $(2 \sin 	heta - 1)(\sin 	heta + 1) = 0$.Since $	heta \in [0, \pi/2)$,$\sin 	heta = 1/2$,so $	heta = \pi/6$.At $	heta = \pi/6$,the side length of the triangle is $s = 2r \cos(\pi/6) = 2r (\sqrt{3}/2) = \sqrt{3}r$.Thus,the triangle of maximum area is an equilateral triangle with side length $\sqrt{3}r$.

The triangle of maximum area that can be inscribed in a given circle of radius $r$ is ...... .

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