The triangle of maximum area that can be inscribed in a given circle of radius 'r' is:
Option: 1 An isosceles triangle with base equal to 2r.
Option: 2 A right angle triangle having two of its sides of length 2r and r
Option: 3 n equilateral triangle having each of its side of length \sqrt{3} r
Option: 4 An equilateral triangle of height \frac{2r}{3}

Answers (1)

AP is the height (h) of the triangle ABC

\\\mathrm{h}=\operatorname{rsin} \theta+\mathrm{r} \\ \text { base }=\mathrm{BC}=2 \mathrm{r} \cos \theta \\ \theta \in\left[0, \frac{\pi}{2}\right)

\\\text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}(\mathrm{BC}) \cdot \mathrm{h} \\ \Delta=\frac{1}{2}(2 \mathrm{r} \cos \theta) \cdot(\mathrm{r} \sin \theta+\mathrm{r}) \\ =\mathrm{r}^{2}(\cos \theta) \cdot(1+\sin \theta) \\ \frac{\mathrm{d} \Delta}{\mathrm{d} \theta}=\mathrm{r}^{2}\left[\cos ^{2} \theta-\sin \theta-\sin ^{2} \theta\right] \\ =\mathrm{r}^{2}\left[1-\sin \theta-2 \sin ^{2} \theta\right] \\=r^{2}[1+\sin \theta][1-2 \sin \theta]=0

\sin \theta\;\text{ is always positive in }\left [ 0,\frac{\pi}{2}\right ]

\\\text{so, }1-2\sin\theta=0\\\Rightarrow\theta=\frac{\pi}{6}

\\\Rightarrow \Delta \text { is maximum where } \theta=\frac{\pi}{6}\\ \Delta_{\max }=\frac{3 \sqrt{3}}{4} \mathrm{r}^{2}=\text { area of equilateral } \Delta \text { with }\mathrm{BC}=\sqrt{3} \mathrm{r}

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