Question

$△ABC$ is a right-angled triangle with $AB=12cm$ and $AC=13cm$. A circle with centre $O$ has been inscribed inside the triangle. The value of $x$, the radius of the inscribed circle is

• A. 3cm
• B. 10cm
• C. 2cm
• D. 6cm

Answer 1

The correct option is C 2cm

$BMOP$ is a square since all angles are ${90}^0$ and adjacent sides are given to be of equal length and opposite sides are obssiously of same length since all angels are ${90}^0$.

$\Rightarrow BM=BP=x$

$\because ABC$ is right angled.

$\Rightarrow BC=\sqrt{A{C}^2-A{B}^2}$

$=\sqrt{169-144}$

$=5cm$

$PC+B{C}_{B}P$

$=5-x$

$CP=CN$ ($\because$ Tangents from same external points.)

$\Rightarrow CN=5-x$

$\Rightarrow AN=AC-CN=13-(5-x)$

$=8+x$

$AM=AB-BM=12-x$

$A{O}^2=A{M}^2+M{O}^2=A{N}^2+N{O}^2$

$\Rightarrow (12-x{)}^2+x^2=(8+x{)}^2+x^2$

$\Rightarrow 12-x=8+x$

$\Rightarrow 2x=4$

$\Rightarrow x=2cm$

$\therefore C)$ Answer.