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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.
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.
CBSE | Class 10 | Coordinate Geometry | Exercise 16.1 | Maths | RS Aggarwal
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.

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

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

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

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

Therefore, $O A=A B=0 B=2 \sqrt{3}$ units

Thus, the points $0(0,0) \mathrm{A}(3, \sqrt{3})$ and $\mathrm{B}(3,-\sqrt{3})$ are the vertices of an equilateral triangle

the area of the triangle $0 \mathrm{AB}=\frac{\sqrt{3}}{4} \times(\text { side })^{2}$

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

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

$=3 \sqrt{3}$ square units.

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