Question

AB is the diameter of a circle and C is any point on the circle. Show that the area of $△ABC$ is maximum, When it is an isosceles triangle.

Answer 1

$\angle ACB={90}^{\circ }$
By Pythagoras theorem
$x^2+y^2=4a^2$
$y=\sqrt{4a^2-x^2}$
Area will be maximum
Area $=\frac{1}{2}\times xy$
$A=\frac{1}{2}\times x\times \sqrt{4a^2-x^2}$
$\frac{dA}{dx}=\frac{1}{2}\left[x\frac{1}{2\sqrt{4a^2-x^2}}\right(-2x)+\sqrt{4a^2-x^2}.1]$
$=\frac{1}{2}[\frac{-x^2}{\sqrt{4a^2-x^2}}+\sqrt{4a^2-x^2}]$
For max. $\frac{dA}{dx}=0$
$\frac{1}{2}[\frac{-x^2}{\sqrt{4a^2-x^2}}+\sqrt{4a^2-x^2}]=0$
$\frac{x^2}{\sqrt{4a^2-x^2}}=\sqrt{4a^2-x^2}$
$x^2=4a^2-x^2$
$2x^2=4a^2$
$x^2=2a^2$
$x=\sqrt{2}a$
Local maxima
$y=\sqrt{4a^2-x^2}=\sqrt{4a^2-2a^2}=\sqrt{2}a$
$x=y=\sqrt{2}a$
Isosceles triangle