Question

Vertices of a triangle $ABC$ are the points $(0,0),(a,0),(\frac{a}{2},\frac{a\sqrt{3}}{2})$. Then its centroid is the point $(\frac{a}{2},\frac{a\sqrt{3}}{6})$

• A. True
• B. False

Answer 1

The correct option is A True

Given vertices of triangle are $A\left(0.0\right)$, $B(a,0)$ and $C(\frac{a}{2},\frac{a\sqrt{3}}{2})$

Centriod of a triangle whose vertices are given is $G(x,y)$ is:

$x=\frac{0+a+a/2}{3}=\frac{a}{2}$ and

$y=\frac{0+0+a\sqrt{3}/2}{3}=\frac{a\sqrt{3}}{6}$.

Hence given satement is true.