Question

A right triangle ABC circumscribes a circle of radius r. If AB and BC are of length 8 cm and 6 cm respectively, find the value of r?

Answer 1

AB, BC and CA are tangents to the circle at P, N and M.
$\therefore OP=ON=OM=r$ (radius of the circle)
Area of $\Delta ABC=\frac{1}{2}\times 6\times 8=24c{m}^2$
By pythagoras theorem, we have
$C{A}^2=A{B}^2+B{C}^2$

$\Rightarrow C{A}^2=8^2+6^2$

$\Rightarrow C{A}^2=100$
$\Rightarrow CA=10cm$
Area of $\Delta ABC$ $=$ Area of $\Delta OAB+$ Area $\Delta OBC+$ Area $\Delta OCA$
$\Rightarrow 24=\frac{1}{2}\times r\times AB+\frac{1}{2}\times r\times BC+\frac{1}{2}\times r\times CA$
$\Rightarrow 24=\frac{1}{2}r(AB+BC+CA)$
$\Rightarrow r=\frac{2\times 24}{AB+BC+CA}$
$\Rightarrow r=\frac{48}{8+6+10}$
$\Rightarrow r=\frac{48}{24}$
$\Rightarrow r=2cm$