Question

Show that the points O(0, 0) A(3, $\sqrt{3}$) and B(3, −$\sqrt{3}$) are the vertices of an equilateral triangle. Find the area of this triangle.

Answer 1

The given points are O(0, 0) A(3, $\sqrt{3}$) and B(3, − $\sqrt{3}$).
$$\begin{gathered} OA=\sqrt{{\left(3-0\right)}^2+{\left\{\left(\sqrt{3}\right)-0\right\}}^2}=\sqrt{{\left(3\right)}^2+{\left(\sqrt{3}\right)}^2}=\sqrt{9+3}=\sqrt{12}=2\sqrt{3}\mathrm{units} \\ AB=\sqrt{{\left(3-3\right)}^2+{\left(-\sqrt{3}-\sqrt{3}\right)}^2}=\sqrt{0+{\left(2\sqrt{3}\right)}^2}=\sqrt{4\left(3\right)}=\sqrt{12}=2\sqrt{3}\mathrm{units} \\ OB=\sqrt{{\left(3-0\right)}^2+{\left(-\sqrt{3}-0\right)}^2}=\sqrt{{\left(3\right)}^2+{\left(\sqrt{3}\right)}^2}=\sqrt{9+3}=\sqrt{12}=2\sqrt{3}\mathrm{units} \\ \mathrm{Therefore},OA=AB=OB=2\sqrt{3}\mathrm{units} \end{gathered}$$
Thus, the points O(0, 0) A(3, $\sqrt{3}$)and B(3, − $\sqrt{3}$) are the vertices of an equilateral triangle.
Also, the area of the triangle OAB = $\frac{\sqrt{3}}{4}\times {\left(\mathrm{side}\right)}^2$
$$\begin{gathered} \frac{\sqrt{3}}{4}\times {\left(2\sqrt{3}\right)}^2 \\ =\frac{\sqrt{3}}{4}\times 12 \\ =3\sqrt{3}\mathrm{square}\mathrm{units} \end{gathered}$$