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,$A B=\sqrt{6^{2}+6^{2}}=6 \sqrt{2}$
$B C=\sqrt{6^{2}+0}=6 $
and $C A=\sqrt{0+6^{2}}=6$
Also,$A B^{2}=B C^{2}+C A^{2}$
Therefore, $\Delta A B C$ is right angled at $C$. So, mid point of $A B$ is the circumcentre of $\triangle A B C$.
$\therefore $ Coordinate of circumcentre are $(3,3)$.
Coordinate of centroid are,
$G\left(\frac{6+0+6}{3}, \frac{0+6+6}{3}\right), i e,(4,4)$
$\therefore $Required distance $=\sqrt{(4-3)^{2}+(4-3)^{2}}=\sqrt{2}$