Question

An equilateral triangle is drawn inside a circle as shown. If we put a dot in it without looking into the picture, find the probability of the dot being outside the triangle?

• A. $\frac{4\pi -3\sqrt{3}}{4}$
• B. $\frac{4\pi -3\sqrt{3}}{4\pi }$
• C. $\frac{\pi -3\sqrt{3}}{4\pi }$
• D. $\frac{4\pi -3}{4\pi }$

Answer 1

The correct option is B $\frac{4\pi -3\sqrt{3}}{4\pi }$

OB and OC are perpendiculars . In triangle OAC and OAB;
OA(common side)
AB=AC(perpendicualar from the centre of circle bisect the chord.)
$\angle OBA=\angle OCA={90}^{\circ }$
Triangles OAB and OAC are congruent
So OA is the angle bisectors of $\angle BAC$
$\angle OAB=\frac{{60}^{\circ }}{2}={30}^{\circ }$
$cos{30}^{\circ }=\frac{AB}{OA},\frac{\sqrt{3}}{2}=\frac{AB}{OA},AB=\frac{\sqrt{3}}{2}r$
Side of the triangle$=2\times AB=\sqrt{3}r$
Area of Circle $=\pi r^2$
Area of equilateral triangle $\frac{\sqrt{3}}{4}\times (\sqrt{3}r{)}^2=\frac{3\sqrt{3}{r}^2}{4}$
The probability of the dot being inside the triangle $=\frac{\sqrt{3}\times 3r^2}{4\pi r^2}=\frac{3\sqrt{3}}{4\pi }$
The probability of the dot being inside the triangle = $1-\frac{3\sqrt{3}}{4\pi }=\frac{4\pi -3\sqrt{3}}{4\pi }$