Question

In fig. 4 triangle ABC is a right triangle with AB$=6$ cm, AC$=8$ cm and $\angle A={90}^{\circ }$. A circle with center O is inscribed inside the triangle. Find the radius 'r'

Answer 1

Since $AB,BC$ and $CA$ are tangents to the circle

$\Rightarrow$ $OP,OM$ and $ON$ are radius of the incircle

To find : radius 'r' of the incircle.

$ABC$ is a right angled triangle

${AB}^2+{AC}^2={BC}^2$ (Using pythagoras theorem)

$\Rightarrow$ ${\left(6\right)}^2+{\left(8\right)}^2={BC}^2$

$\Rightarrow$ ${BC}^2=36+64$

$=100$

$\Rightarrow$ $BC=\sqrt{100}=10cm$

Since area of right anged triangle $=\frac{1}{2}\times base\times height$

$\Rightarrow$ Area of $△ABC=\frac{1}{2}\times 6\times 8$

$=3\times 8=24{cm}^2$

$\Rightarrow$ Area of $△ABC=24{cm}^2$....(1)

Also, Area of $△ABC=$ area of $△OAC+$ area of $△OBC+$ area of $△OBA$

$=\frac{1}{2}\times r\times 8+\frac{1}{2}\times r\times 10+\frac{1}{2}\times 6\times r=\frac{1}{2}r(8+10+6)=\frac{1}{2}r\left(24\right)=12r....\left(2\right)$

Equating (1) and (2) $12r=24$

$\Rightarrow$ $r=\frac{24}{12}=2cm$