Question

Consider a right angled triangle $ABC$


If the sides of the triangle are $a=6,b=10,c=8$ units, then the distance between incentre and circumcentre will be

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

Answer 1

The correct option is C $\sqrt{5}$

Consider the triangle at vertex $A(0,8),B(0,0),C(6,0)$


Incentre is $(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})$
$=(2,2)$
Circumcentre will be mid point of hypotenuse
$=(3,4)$

Using distance formula
Distance $\sqrt{(3-2{)}^2+(4-2{)}^2}=\sqrt{5}$