Question

A point P is given on the circumference of a circle of radius r the chord QR is parallel to the tangent line at P and the maximum area of $\Delta PQR$ is $\frac{3\sqrt{3}{r}^2}{b}$ where r is radius of circle, evaluate b.

Answer 1

$ON\perp QR$
$QR=2QN=2rsin\theta$
$ON=rcos\theta$
area of $\Delta PQR=\frac{1}{2}\left(2rsin\theta \right)\times (r+rcos\theta )$
$A=r^2\{sin\theta +sin\theta cos\theta \}$
$A=r^2[sin\theta +\frac{sin2\theta }{2}]$
$\frac{dA}{d\theta }=r^2(cos\theta +cos2\theta )$
$\frac{dA}{d\theta }=r^2(cos\theta +cos2\theta )$
$\frac{d^{2}A}{d{\theta }^2}=r^2(-sin\theta -2sin2\theta )$
$Nowformaximumarea\frac{dA}{d\theta }=0,\frac{d^{2}A}{d{\theta }^2}<0$
$cos2\theta +cos\theta =0$
$cos2\theta =-cos\theta$
$2\theta =\pi -\theta \Rightarrow \theta =\frac{\pi }{3}$
clearly $\frac{d^{2}A}{d{\theta }^2}<0$
$maximumarea=\frac{3\sqrt{3}{r}^2}{4}$
$b=4$