Question

In a triangle $ABC$, if $a=3,b=4,c=5,$ then the distance between its incentre and circumcentre is

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

Answer 1

Answer: (E)

Given, $a=3,b=4,c=5$
$\Rightarrow$ $c^2=a^2+b^2$
Therefore, it is a right angled triangle at C.
$\therefore$ $R=\frac{1}{2}c=\frac{5}{2}$ and $r=\frac{\Delta }{s}=\frac{\frac{1}{2}\times 3\times 4}{\frac{12}{2}}=1$
$\therefore$ Distance between incentre and circumcentre
$=\sqrt{R^2-2Rr}$
$=\sqrt{{\left(\frac{5}{2}\right)}^2-2.\frac{5}{2}.1}$
$=\sqrt{\frac{5}{2}}\sqrt{\frac{5}{2}-2}=\frac{\sqrt{5}}{2}$