Question

What is area of a circle that is inscribed in a right triangle with sides of length $8m$, $15m$ and $17m$?

• A. $9\pi m^2$
• B. $11\frac{1}{9}\pi m^2$
• C. $14\pi m^2$
• D. $10\pi m^2$

Answer 1

The correct option is A $9\pi m^2$

Area of $△ABC=\frac{1}{2}\times b\times h$
$=\frac{1}{2}\times 8\times 15=60m^2$
Let r be radius of circle $\therefore OE=OF=OD=r$
$\therefore$ Area of $△AOB=\frac{1}{2}\times AB\times OD=\frac{1}{2}\times 15\times r⟶1$
Area of $△AOC=\frac{1}{2}\times AC\times OF=\frac{1}{2}\times 17\times r⟶2$
Area of $△BOC=\frac{1}{2}\times BC\times OE=\frac{1}{2}\times 8\times r⟶3$
Area of $△AOB+$ Area of $△AOC+$ Area of $△BOC=$ Area of $△ABC$
$\frac{15}{2}r+\frac{17}{2}r+\frac{8}{2}r=60$
$40r=120\Rightarrow r=3m$
$\therefore$ Area of circle$=\pi r^2=9\pi m^2$