Question

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

• A. An isosceles triangle with base equal to $2 r$.
• B. An equilateral triangle of height $\frac{2 r}{3}$.
• C. An equilateral triangle having each of its side of length $\sqrt{3} r$.
• D. A right angle triangle having two of its sides of length $2 r$ and $r$.

Answer 1

Answer: (C)

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

Area of $\Delta ABC =\frac{1}{2}( BC ) \cdot h$
$\Delta=\frac{1}{2}(2 r \cos \theta) \cdot( r \sin \theta+ r )$
$= r ^{2}(\cos \theta) \cdot(1+\sin \theta)$
$\frac{ d \Delta}{ d \theta}= r ^{2}\left[\cos ^{2} \theta-\sin \theta-\sin ^{2} \theta\right]$
$= r ^{2}\left[1-\sin \theta-2 \sin ^{2} \theta\right]$
$=\underbrace{r^{2}[1+\sin \theta]}_{\text {positive }}[1-2 \sin \theta]=0$
$\Rightarrow \theta=\frac{\pi}{6}$

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