In the figure above (not drawn to scale), triangle ABC is inscribed in

From the figure it can be seen that the triangle ACB is inscribed within a semicircle. So angle C is 90 degree.
Lets take OB = x , so OA =x (both radius) 2x, Also it is given that AC = OB = x.
For triangle OAC (draw a line ) all sides are equal (OA = AC= OC) so it is an equilateral triangle. So <A = 60degree

The right triangle ACB is thus a 30 60 90 triangle so sides are x, x√3and 2x.
Area of triangle ACB = 0.5 * x * x√3 = 8√3 So x = 4 i.e. radius of the circle is 4

Area of circle = π x * x = 16π

Answer C

Here, \(AC = OB = OA = r\), we can say that \(AB = 2r\).

Given the inscribed triangle property, we can get that triangle ABC is a right triangle and so \(AB^2 = AC^2 + BC^2\).

So, \((2r)^2 = r^2 + BC^2\), which means that \(BC = \sqrt{3}r\).

Area of the triangle ABC = \(\frac{1}{2}(AC)(BC) = 8\sqrt{3}\), means \(\frac{1}{2}(r)(\sqrt{3}r) = 8\sqrt{3}\), which simplifies to \(r^2 = 16\), so Area of the circle = 16\(\pi\)

Answer is C.

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800score Official Solution:

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.

This is a 30-60-90 triangle and area is given we use x as the common multiplier in the ratio X:root 3X:2X, And find the value of the side.

Given, AB is the diameter of the circle= 2r and AC = OB =r

Let BC = h

Thus is right triangle ABC, right angled at C,

\(r^2+h^2 = (2r)^2\) and from area of the triangle = \(8\sqrt{3}\) = \(0.5*h*r\) ----> \(h = \frac{16\sqrt{3}}{r}\)

Thus, \(r^2+ (\frac{16\sqrt{3}}{r})^2 = 2r^2\) ----> \(r^4 = 256\) ---> \(r = 4\)

Thus the area of the circle = \(\pi*4^2\) = \(16\pi\). C is the correct answer.

OB is the radius, thus, AC has the same length as the radius.

since we have a triangle inscribed in a circle, and one side is the diagonal, the triangle MUST be a right triangle.
since the hypotenuse is 2R,and since one leg is R, it must be true that the triangle is 30-60-90. It must also be true, from the property of 30-60-90 triangle, that the other leg must be R*sqrt(3).
we then have the area of the triangle:
R*R(sqrt3)/2 = 8*sqrt(3)
r^2(sqrt3) = 16*sqrt(3)
r=4.
Area of the circle:
pi*r^2 = 16pi.