Question

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

Answer 1

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

$OA=\sqrt{(3-0{)}^2+(\sqrt{3}-0{)}^2}$

$=\sqrt{(3{)}^2+(\sqrt{3}{)}^2}$

$=\sqrt{9+3}=\sqrt{12}=2\sqrt{3}units$

$AB=\sqrt{(3-3{)}^2-(-\sqrt{3}-\sqrt{3}{)}^2}$

$=\sqrt{0+(2\sqrt{3}{)}^2}$

$=\sqrt{4\times 3}$

$=\sqrt{12}$

$=2\sqrt{3}units$

$OB=\sqrt{(3-0{)}^2+(-\sqrt{3}-0{)}^2}$

$=\sqrt{(3{)}^2+(\sqrt{3}{)}^2}$

$=\sqrt{9+3}=\sqrt{12}=2\sqrt{3}$

Therefore, $OA=AB=OB=2\sqrt{3}units$

The given points are $O(0,0),A(3,\sqrt{3})andB(3,-\sqrt{3})$ are the vertices of an equilateral triangle

Also, the area of the equilateral triangle OAB

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

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

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

$=3\sqrt{3}$ square units