2013 AMC 10B Problem 7

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

Answer: B

Concepts:regular polygon special right triangle triangle area

Difficulty rating: 1280

Small hint:

The only allowed non-isosceles triangle from six equally spaced points is a

30

-

60

-

90

triangle

Big hint:

Use the diameter as the hypotenuse

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 .