In the figure above, the triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. If side CB equals 2√3 and the segment OB equals 2, what is the area of triangle ABC?

A. √3

B. 2√3

C. π√3

D. 2π√3

E. 8√3

(B) There are three main concepts that must be understood in order to solve this problem.

The first concept is that any triangle inscribed within a circle such that any one side of the triangle is a diameter of the circle, is a right triangle.

The second concept is that the proportions of a 30-60-90 triangle are

*x*,

*x*√3, 2

*x*.

Finally, the third concept is that the area of a triangle is equal to one half the product of its base and height: (1/2)

*bh*.

Since the triangle is inscribed in the circle and one of its sides constitutes a diameter of the circle, the triangle must be a right triangle. Since we are told that OB, the radius of the circle, is equal to 2, the longest side of the triangle must be equal to 4. And, since its a right triangle with a hypotenuse equal to 4 and the second longest leg is equal to 2√3, it must be a 30-60-90 triangle where the shortest side is equal to 2.

We are now ready to solve this problem. Using the third concept from above, the area must be equal to (1/2)

*bh*. Taking the shortest side to be the base,

*b*, and the longest of the two legs to be the height,

*h*, we have:

(1/2)

*bh* = (1/2)(2)(2√3) = 2√3.

So, the correct answer choice is (B).

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**How do you get B?**