Question

The triangle of maximum area that can be inscribed in a given circle of radius ${}^{\prime }{r}^{\prime }$ is :

• A. An isosceles triangle with base equal to $2r$.
• B. An equilateral triangle of height $\frac{2r}{3}$.
• 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 $2r$ and $r$.

Answer 1

Answer: (C)

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

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

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