Question

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

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

Answer 1

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

Let $BC=17cm,AB=8cm$ and $AC=15cm$

Area of $△ABC=\frac{1}{2}\times 15\times 8$

$=60c{m}^2$

Let r be the radius of a circle

$\Rightarrow OD=OE=OF=r$

$\therefore$ Area of $△AOB=\frac{1}{2}\times AB\times OD$

$=\frac{1}{2}\times 8\times r$

$=4rc{m}^2$

Similarly,

Area of $△AOC=\frac{1}{2}\times AC\times OE$

$=\frac{1}{2}\times 15\times r$

$=\frac{15r}{2}c{m}^2$

And

Area of $△BOC=\frac{1}{2}\times BC\times OF$

$=\frac{1}{2}\times 17\times r$

$=\frac{17r}{2}c{m}^2$

Now,

Area of $△ABC=$ Area of $△AOB+$ Area of $△AOC+$ Area of $△BOC$

$\Rightarrow 60=4r+\frac{15r}{2}+\frac{17r}{2}$

$\Rightarrow 120=8r+15r+17r$

$\Rightarrow 40r=140$

$\Rightarrow r=3cm$

$\therefore$ Area of circle $=\pi r^2$

$=\pi (3{)}^2$

$=9\pi c{m}^2$