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?
(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, 2x. 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 × radius².
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 = 2x 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.
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 (2x) and one leg. Could you explain why you do not use x(x√3) instead?