Question

The vertices of a triangle are $(6,0),(0.6)$ and $(6,6)$. The distance between its circumcentre and centroid is

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

Answer 1

Answer: (C)

Let the vertices of a triangle be $A(6,0),B(0,6)$ and $C(6,6)$
Now,$AB=\sqrt{6^2+6^2}=6\sqrt{2}$
$BC=\sqrt{6^2+0}=6$
and $CA=\sqrt{0+6^2}=6$
Also,$A{B}^2=B{C}^2+C{A}^2$
Therefore, $\Delta ABC$ is right angled at $C$. So, mid point of $AB$ is the circumcentre of $△ABC$.
$\therefore$ Coordinate of circumcentre are $(3,3)$.
Coordinate of centroid are,
$G\left(\frac{6+0+6}{3},\frac{0+6+6}{3}\right),ie,(4,4)$
$\therefore$Required distance $=\sqrt{(4-3{)}^2+(4-3{)}^2}=\sqrt{2}$