Question

$ABC$ is an isosceles triangle inscribed in a circle of radius $r$. If $AB=AC$ and $h$ is the altitude from $A$ to $BC$ then the triangle has perimeter $P=2(\sqrt{2hr-h^2}+\sqrt{2hr})$ and area $A$ is equal to

• A. $h\sqrt{3rh-h^2}$
• B. $h\sqrt{2rh-h^2}$
• C. $2h\sqrt{2rh-h^2}$
• D. None of these

Answer 1

Answer: (B)

$h\sqrt{2rh-h^2}$

In $△ABC$ $AB=AC$

$\Rightarrow AD\perp BC$

$\Rightarrow r=$radius of circumcenter

$\Rightarrow OA=OB=OC=r$

$\Rightarrow BD=\sqrt{B{O}^2+O{D}^2}$

$\Rightarrow BD=\sqrt{r^2+(h-r{)}^2}$

$\Rightarrow BD=\sqrt{2rh-h^2}$

$\Rightarrow BC=2\sqrt{2rh-h^2}$

$\Rightarrow Area=\frac{1}{2}\left(BC\right)\left(AD\right)$

$\Rightarrow Area=\frac{1}{2}\times 2\sqrt{2rh-h^2}\times h$

$\Rightarrow Area=h\sqrt{2rh-h^2}$