The vertices of a triangle are (6. 6), (0, 6) and (6, 0) are the vertices of a right triangle as shown in the figure. Find the distance between its centroid and circumcentre.

$\sqrt{3}$

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$\sqrt{6}$

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$\sqrt{2}$

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$\sqrt{5}$

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Solution

The correct option is C
$\sqrt{2}$

Circumcentre = midpoint of AB
$= (\frac{6 + 0}{2}, \frac{6 + 0}{2})$
D = (3,3)

Centroid (G) $= (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}) = (\frac{6 + 0 + 6}{3}, \frac{0 + 6 + 6}{3}) = (4, 4)$

$∴ distance DG = \sqrt{(4 - 3)^{2} + (4 - 3)^{2}} = \sqrt{1 + 1} = \sqrt{2}$

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The vertices of a triangle are (6. 6), (0, 6) and (6, 0) are the vertices of a right triangle as shown in the figure. Find the distance between its centroid and circumcentre.

The correct option is C √2 Circumcentre = midpoint of AB =(6+02, 6+02) D = (3,3) Centroid (G) =(x1+x2+x33, y1+y2+y33)=(6+0+63, 0+6+63)=(4,4) ∴distance DG=√(4−3)2+(4−3)2=√1+1=√2