In the figure above (not drawn to scale), triangle

*ABC* is inscribed in the circle with the center

*O* and

*AB* is a diameter of the circle. Segments AC and OB are equal. If the area of triangle

*ABC* is 8√3, then what is the area of the circle?

A. π

B. 8π

C. 16π

D. 48π

E. 64π

(C) There are four 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 the sides of a 30⁰-60⁰-90⁰ triangle are

*x*,

*x*√3, 2

*x*. The third concept is that the area of a right triangle is equal to one half the product of its legs: (1/2)

*bh*. Finally, the fourth concept is that the area of a circle is π ×

*radius*².

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. AB is the diameter so AB = 2 × OB. We know that OB = AC. Therefore AB = 2 × AC. Therefore it must be a 30°-60°-90° triangle where AC =

*x*, AB = 2

*x* and CB = √3 ×

*x* (This can also be calculated using the Pythagorean Theorem).

We are now ready to solve this problem. Using the third concept from above, the area must be equal to (1/2) × √3 ×

*x* ×

*x* = 8√3.

The solution of the equation is

*x* = 4. So the radius of the circle is 4. Therefore the area of the circle is π × 4² = 16π. The correct answer is C.

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Hi, I am not sure why for calculation of area of a triangle 30-60-90 we need to use the formula that multiplies lengths of two legs (base and height). In the answer you are using the length of hypotenuse (2*x*) and one leg. Could you explain why you do not use *x*(*x*√3) instead?

thanks