An equilateral triangle ABC with side length 6\sqrt{3} is inscribed in a circle with cente… — GRE General Test

An equilateral triangle $ABC$ with side length $6\sqrt{3}$ is inscribed in a circle with center $O$ . What is the area of sector $AOB$ , divided by $\pi$ ?

Answer: 12

Answer

The area of sector

AOB

divided by

\pi

is 12.

Since triangle

ABC

is equilateral, its three vertices divide the circle into three congruent arcs of

120^\circ

each. Thus, central angle

\angle AOB = 120^\circ

. The relationship between the side length

s

of an inscribed equilateral triangle and the radius

R

of its circumscribed circle is

s = R\sqrt{3}

. Given

s = 6\sqrt{3}

, we solve for

R

to find

R = 6

. The area of sector

AOB

\frac{120^\circ}{360^\circ} \pi R^2 = \frac{1}{3} \pi (6^2) = 12\pi

. Dividing this area by

\pi

yields

12

Step-by-Step Solution

Find the central angle

\angle AOB

corresponding to side

AB

of the inscribed equilateral triangle.

\angle AOB = 120^\circ

An inscribed equilateral triangle divides the

360^\circ

circle into three equal central angles.

Calculate the radius

R

of the circumscribed circle from the given side length

s = 6\sqrt{3}

R = 6

In an inscribed equilateral triangle,

s = R\sqrt{3}

. Substituting

6\sqrt{3} = R\sqrt{3}

gives

R = 6

Compute the area of sector

AOB

and divide by

\pi

\text{Sector Area} = \frac{120^\circ}{360^\circ} \times \pi \times 6^2 = 12\pi

. Dividing by

\pi

leaves

12

Key Concept

Relationship between inscribed shapes, circle radii, and sector area

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