Question

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

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

Answer 1

The correct option is D

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

Solve for the maximum area of triangle:

The height of the triangle is $h$

$$\begin{gathered} \therefore h=AP \\ =AO+OP \\ =r+r\sin \left(\theta \right) \end{gathered}$$

The base of the triangle is $b$

$$\begin{gathered} \therefore b=BC \\ =BP+PC \\ =2BP \\ =2r\cos \left(\theta \right) \end{gathered}$$

Area of the $∆\mathrm{ABC}=\frac{1}{2}\times \left(\mathrm{b}\right)\times \left(\mathrm{h}\right)$

$$\begin{gathered} \Rightarrow △=\frac{1}{2}\left(2r\cos \left(\theta \right)\right)\left(r+r\sin \left(\theta \right)\right) \\ =r^2\cos \left(\theta \right)\left(1+\sin \left(\theta \right)\right) \end{gathered}$$

Now to find the maximum differentiate with respect to $\theta$

$$\begin{gathered} \Rightarrow \frac{d△}{d\theta }=r^2\left({\cos }^2\left(\theta \right)-\sin \left(\theta \right)-{\sin }^2\left(\theta \right)\right) \\ =r^2\left(1-\sin \left(\theta \right)-2{\sin }^2\left(\theta \right)\right) \\ =r^2\left(1+\sin \left(\theta \right)\right)\left(1-2\sin \left(\theta \right)\right) \end{gathered}$$

For extremum $\frac{d△}{d\theta }=0$

$$\begin{gathered} \Rightarrow r^2\left(1+\sin \left(\theta \right)\right)\left(1-2\sin \left(\theta \right)\right)=0 \\ \Rightarrow \sin \left(\theta \right)=-1,\frac{1}{2} \\ \Rightarrow \theta =\frac{\pi }{6} \end{gathered}$$

We neglect $\sin \left(\theta \right)=-1$ , since in this case $\theta =\frac{3\pi }{2}$ ,we know sum of the angles of triangle is $\pi$.

Therefore,

$$\begin{gathered} {△}_{max}=r^2\left(\cos \left(\frac{\pi }{6}\right)\right)\left(1+\sin \left(\frac{\pi }{6}\right)\right) \\ =r^2\left(\frac{\sqrt{3}}{2}\right)\left(1+\frac{1}{2}\right) \\ =r^2\left(\frac{3\sqrt{3}}{4}\right) \end{gathered}$$

Therefore ${△}_{max}$ is a area of equilateral $△$ with base $\sqrt{3}r$

Hence, the correct option is (D).