Question

An equilateral triangle is drawn with its vertices 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

$\frac{4\pi -3\sqrt{3}}{4\pi }$

If the radius of the circle is r, then the side of the triangle will be $\sqrt{3}r$

Area of the circle = $\pi r^2$

Area of the equilateral triangle =$\frac{\sqrt{3}}{4}\times (\sqrt{3}r{)}^2$ = $\frac{\sqrt{3}}{4}\times 3r^2$

The probability of the dot being inside the triangle = $\frac{\frac{\sqrt{3}}{4}\times 3r^2}{\pi r^2}$ = $\frac{\sqrt{3}\times 3r^2}{4\pi r^2}$ = $\frac{3\sqrt{3}}{4\pi }$

The probability of the dot being outside the triangle = $1-\frac{3\sqrt{3}}{4\pi }$ = $\frac{4\pi -3\sqrt{3}}{4\pi }$