question_answer

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)  \[\frac{\sqrt{5}}{2}\]

Correct Answer: E

Solution :

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 \[=\sqrt{{{R}^{2}}-2Rr}\] \[=\sqrt{{{\left( \frac{5}{2} \right)}^{2}}-2.\frac{5}{2}.1}\] \[=\sqrt{\frac{5}{2}}\sqrt{\frac{5}{2}-2}=\frac{\sqrt{5}}{2}\]