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(D) Let $ABCD$ be the rectangle inscribed in a circle of radius $r$.Since the rectangle is inscribed in the circle,its diagonal is the diameter of the

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$\sqrt{2} r$ units,$\sqrt{2} r$ units

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(D) Let $ABCD$ be the rectangle inscribed in a circle of radius $r$.Since the rectangle is inscribed in the circle,its diagonal is the diameter of the circle.$\Rightarrow AC = BD = 2r = 	ext{diameter}$.Let $x$ and $y$ be the length and breadth of the rectangle.By Pythagoras theorem,$x^{2} + y^{2} = (2r)^{2} = 4r^{2}$.$\Rightarrow y = \sqrt{4r^{2} - x^{2}}$.Now,the area of the rectangle $A = xy = x\sqrt{4r^{2} - x^{2}}$.To find the maximum area,differentiate $A$ with respect to $x$:$\frac{dA}{dx} = \sqrt{4r^{2} - x^{2}} + x \cdot \frac{1}{2\sqrt{4r^{2} - x^{2}}} \cdot (-2x) = \sqrt{4r^{2} - x^{2}} - \frac{x^{2}}{\sqrt{4r^{2} - x^{2}}} = \frac{4r^{2} - 2x^{2}}{\sqrt{4r^{2} - x^{2}}}$.For maximum area,set $\frac{dA}{dx} = 0$:$4r^{2} - 2x^{2} = 0 \Rightarrow x^{2} = 2r^{2} \Rightarrow x = \sqrt{2}r$.Substituting $x = \sqrt{2}r$ in the expression for $y$:$y = \sqrt{4r^{2} - (\sqrt{2}r)^{2}} = \sqrt{4r^{2} - 2r^{2}} = \sqrt{2r^{2}} = \sqrt{2}r$.Thus,the dimensions are $\sqrt{2}r$ units and $\sqrt{2}r$ units.

If rectangles are inscribed in a circle of radius $r$ units,then the dimensions of the rectangle which has maximum area are:

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