Question

In a $∆ABC,\text{if}a=3,b=4,c=5$, then the distance between its incenter and circumcenter is

• A. $\frac{1}{2}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{3}{2}$
• D. $\frac{\sqrt{5}}{2}$

Answer 1

The correct option is D

$\frac{\sqrt{5}}{2}$

Determine the distance between its incenter and circumcenter.

Given, $a=3,b=4,c=5$.

Step 1: Determine the area and semi-perimeter of the triangle.

We can observe that $a^2+b^2=c^2$, therefore by the converse of Pythagoras theorem, We can say the triangle is a right-angle triangle.

We know that for right-angle triangle : $\text{Area}=\frac{1}{2}\times \text{Base×Height}$

$$\begin{gathered} \therefore ar(△ABC)=\frac{1}{2}\times 3\times 4 \\ \Rightarrow ar(△ABC)=6\text{unit sq.} \end{gathered}$$

We know that Semi-perimeter is given by : $s=\frac{(a+b+c)}{2}$.

$$\begin{gathered} \therefore s=\frac{3+4+5}{2} \\ \Rightarrow =\frac{12}{2} \\ \Rightarrow s=6\text{units} \end{gathered}$$

Step 2: Determine value of $R$ and $r$.

We know that the distance between incenter and circumcenter is given by $\sqrt{(R^2–2rR)}...\left(1\right)$,

where, $R$ is Circumradius, and $r$ is inner circle radius.

we know that, $R=\frac{a\times b\times c}{4\times \text{ar}(△\text{ABC)}}$

Substituting the values we get,

$$\begin{gathered} \Rightarrow R=\frac{3\times 4\times 5}{4\times 6} \\ \Rightarrow R=\frac{5}{2}...\left(2\right) \end{gathered}$$

For value of $r$, we know $r=\frac{ar(△ABC)}{s}$

Substituting the values we have,

$$\begin{gathered} \Rightarrow r=\frac{6}{6} \\ \Rightarrow r=1...\left(3\right) \end{gathered}$$

Step 3: Determine the distance between its incenter and circumcenter

We know that the distance between incenter and circumcenter is given by $\sqrt{(R^2–2rR)}$,

Therefore, substituting the values of $\left(2\right)\text{and}\left(3\right)\text{in}\text{equation}\left(1\right)$,

$$\begin{gathered} \Rightarrow \sqrt{{\left(\frac{5}{2}\right)}^2–2\times 1\times \left(\frac{5}{2}\right)} \\ \Rightarrow \sqrt{\left(\frac{25}{4}\right)-5} \\ \Rightarrow \sqrt{\frac{25-20}{4}} \\ \Rightarrow \frac{\sqrt{5}}{2} \end{gathered}$$

Therefore, the distance between its incenter and circumcenter is $\frac{\sqrt{5}}{2}\text{units}$

Hence, option (D) is correct.