Question

$△ABC$ is a right-angled triangle right angled at$\mathrm{B}$ such that$\mathrm{BC}=6\mathrm{cm}$ and $\mathrm{AB}=8\mathrm{cm}$. A circle with center $O$ is inscribed in triangle $△ABC$. The radius of the circle is.

• A. $1cm$
• B. $2cm$
• C. $3cm$
• D. $4cm$

Answer 1

The correct option is B

$2cm$

Finding the radius of the circlre:

Given:

$\mathrm{BC}=6\mathrm{c}m$ and $\mathrm{AB}=8\mathrm{cm}$

By using Pythagoras Theorem, we can find$\mathrm{AC}.$

$\begin{array}{rcl}{\mathrm{AC}}^2& =& {\mathrm{BC}}^2+{\mathrm{AB}}^2\\ {\mathrm{AC}}^2& =& 6^2+8^2\\ \mathrm{AC}& =& \sqrt{(36+64)}\\ & =& \sqrt{100}\\ & =& 10\mathrm{cm}\end{array}$

Let $\mathrm{OP}=\mathrm{OR}=\mathrm{OQ}=\mathrm{r}$

We know that two tangents drawn from an external point to a circle are equal in measure.

therefore

$$\begin{gathered} \mathrm{BP}=\mathrm{BQ} \\ \mathrm{CQ}=\mathrm{CR} \\ \mathrm{AP}=\mathrm{AR} \end{gathered}$$

$\mathrm{OP}$ and $\mathrm{OQ}$ are radii of the circle.

$\mathrm{OP}\perp \mathrm{AB}$ and $\mathrm{OQ}\perp \mathrm{BC}$ and $\angle \mathrm{B}=90°$ (given)

Hence, $\square \mathrm{BPOQ}$ is a square.

Thus,$\mathrm{BP}=\mathrm{BQ}=\mathrm{r}$ (sides of a square are equal)

So, $\mathrm{AR}=\mathrm{AP}=\mathrm{AB}–\mathrm{BP}$

$=8–\mathrm{r}$_______________$\left[1\right]$

$\mathrm{CR}=\mathrm{CQ}=\mathrm{BC}-\mathrm{BQ}$

$=6–\mathrm{r}$________________$\left[2\right]$

From equation $1$ and $2$

$\begin{array}{rcl}& \Rightarrow & \mathrm{AR}+\mathrm{CR}=\mathrm{AC}\\ 8–\mathrm{r}+6–\mathrm{r}& =& 10\\ 14–2\mathrm{r}& =& 10\\ 14-10& =& 2\mathrm{r}\\ 4& =& 2\mathrm{r}\\ \mathrm{r}& =& 2\mathrm{cm}\end{array}$

Therefore the radius of the circle is $\begin{array}{rcl}\mathrm{r}& =& 2\mathrm{cm}\end{array}$