The vertices of a triangle are (6,0), (0,6) a | vedclass

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(B) Let the vertices of the triangle be $A(6,0), B(0,6)$ and $C(6,6)$.First,we calculate the lengths of the sides:$AB = \sqrt{(0-6)^2 + (6-0)^2} = \sq

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$\sqrt{2}$

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(B) Let the vertices of the triangle be $A(6,0), B(0,6)$ and $C(6,6)$.First,we calculate the lengths of the sides:$AB = \sqrt{(0-6)^2 + (6-0)^2} = \sqrt{36+36} = 6\sqrt{2}$$BC = \sqrt{(6-0)^2 + (6-6)^2} = \sqrt{36+0} = 6$$CA = \sqrt{(6-6)^2 + (0-6)^2} = \sqrt{0+36} = 6$Since $AB^2 = BC^2 + CA^2$ $(72 = 36 + 36)$,the triangle is a right-angled triangle at vertex $C(6,6)$.For a right-angled triangle,the circumcentre is the midpoint of the hypotenuse $AB$.Circumcentre $O = \left(\frac{6+0}{2}, \frac{0+6}{2}\right) = (3,3)$.The centroid $G$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) = \left(\frac{6+0+6}{3}, \frac{0+6+6}{3}\right) = (4,4)$.The distance between the circumcentre $(3,3)$ and the centroid $(4,4)$ is $\sqrt{(4-3)^2 + (4-3)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.

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

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