# Show that the points O(0, 0) A

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.

Solution:

The given points are $O(0,0) A(3, \sqrt{3})$ and $B(3,-\sqrt{3})$.

$O A=\sqrt{(3-0)^{2}+\{(\sqrt{3})-0\}^{2}}=\sqrt{(3)^{2}+(\sqrt{3})^{2}}=\sqrt{9+3}=\sqrt{12}=2 \sqrt{3}$ units

$A B=\sqrt{(3-3)^{2}+(-\sqrt{3}-\sqrt{3})^{2}}=\sqrt{0+(2 \sqrt{3})^{2}}=\sqrt{4(3)}=\sqrt{12}=2 \sqrt{3}$ units

$O B=\sqrt{(3-0)^{2}+(-\sqrt{3}-0)^{2}}=\sqrt{(3)^{2}+(\sqrt{3})^{2}}=\sqrt{9+3}=\sqrt{12}=2 \sqrt{3}$ units

Therefore, $O A=A B=O B=2 \sqrt{3}$ units

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 $O A B=\frac{\sqrt{3}}{4} \times(\text { side })^{2}$

$\frac{\sqrt{3}}{4} \times(2 \sqrt{3})^{2}$

$=\frac{\sqrt{3}}{4} \times 12$

$=3 \sqrt{3}$ square units