Question

Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle?

$\dfrac{\sqrt{3}}{3}$

$\dfrac{\sqrt{3}}{2}$

$1$

$\sqrt{2}$

$2$

Solution:

Six equally spaced points on the circle form a regular hexagon. A triangle using three of them is not equilateral or isosceles only when its angles are $30^\circ,60^\circ,90^\circ$ .

The hypotenuse is a diameter of the unit circle, so it has length $2$ . The legs of the $30$ - $60$ - $90$ triangle are $1$ and $\sqrt3$ .

The area is $\frac12\cdot1\cdot\sqrt3=\frac{\sqrt3}{2}$ .

Thus, the correct answer is B .

2013 AMC 10B Problem 7

Solution:

Problem 7 in Other Years