(A)OR(B)

In the given figure, $\displaystyle \Delta \mathrm{ABC}$ is a right triangle in which $\displaystyle \angle \mathrm{B}=90^{\circ}, \mathrm{AB}=4 \mathrm{~cm}$ and $\displaystyle \mathrm{BC}=3 \mathrm{~cm}$. Find the radius of the circle inscribed in the triangle ABC.

\[\begin{array}{l} \mathrm{AC}=\sqrt{3^{2}+4^{2}}=5 \mathrm{~cm} \text { Let } \mathrm{BE}=\mathrm{BD}=x \mathrm{~cm} \mathrm{AD}=4-x=\mathrm{AF}, \mathrm{CE}=3-x=\mathrm{CF} \mathrm{AF}+\mathrm{CF}=\mathrm{AC} \Rightarrow 4-x+3-x=5 \therefore x=1 \mathrm{BD}=\mathrm{BE}=1 \text { and } \angle \mathrm{B}=90^{\circ} \end{array} \] Hence radius of circle $\displaystyle =x=1 \mathrm{~cm}$ ALTERNATE SOLUTION: \[\mathrm{AC}=\sqrt{3^{2}+4^{2}}=5 \mathrm{~cm} \] Let $\displaystyle r$ be the radius of the circle \[\operatorname{ar}(\triangle A B C)=\frac{1}{2} \times 4 \times 3=6 \mathrm{~cm}^{2} \] Also, $\displaystyle \operatorname{ar}(\triangle A B C)=\left(\frac{1}{2} \times r \times 4\right)+\left(\frac{1}{2} \times r \times 3\right)+\left(\frac{1}{2} \times r \times 5\right)$ \[\Rightarrow 6 r=6 \Rightarrow r=1 \] Hence the radius of the circle is $\displaystyle 1$ cm.\[\mathrm{PM}=\frac{1}{2}(\mathrm{PQ}+\mathrm{QR}+\mathrm{PR}) \] \[\begin{array}{l} \mathrm{PM}=\mathrm{PN} \mathrm{QS}=\mathrm{QM} \mathrm{RS}=\mathrm{RN} \mathrm{PM}+\mathrm{PN}=\mathrm{PQ}+\mathrm{QM}+\mathrm{PR}+\mathrm{RN} \qquad \begin{aligned} 2 \mathrm{PM} & =\mathrm{PQ}+\mathrm{QS}+\mathrm{PR}+\mathrm{RS} & =\mathrm{PQ}+\mathrm{QS}+\mathrm{RS}+\mathrm{PR} \end{aligned} \end{array} \] $\displaystyle \begin{array}{c}=\mathrm{PQ}+\mathrm{QR}+\mathrm{PR} \therefore \mathrm{PM}=\frac{1}{2}(\mathrm{PQ}+\mathrm{QR}+\mathrm{PR})\end{array}$