Thank you for another great geometry question, emmak. Now let's solve it properly
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The first thing you'll notice in the figure above is that the hypothenuse (the biggest side in a triangle - in this case AC) is the diameter of the circle.
Which automatically means that the ABC triangle is a right triangle. Therefore, the Pythagorean theorem become applicable.
We know the lengths of AB and BC, we can therefore deduce the length of AC by using the Pythagorean theorem.
First of all : \(AC^2 = AB^2 + BC^2\)
By plugging in the values of AB and BC, we get :
\(AC^2 = 12^2 + (4\sqrt{3})^2\)
\(AC^2 = 144 + 16*3 = 144 + 48 = 192\)
\(AC = \sqrt{192}\)
By breaking down 192 into prime factors we get : \(192 = 2^6 * 3\)
So \(AC = 8\sqrt{3}\)
Since O is the center of the circle, then \(AO = OC = OB = \frac{AC}{2} = 4\sqrt{3}\)
Notice that the triangle AOB is an isoceles triangle (since AO = OB) and the formula for calculating the area of a triangle is : \(Area = \frac{base * height}{2}\)
The base here is AB. We need to find the height. Since AOB is an isoceles triangle, then the height will cut the base in half. So, noting H as the height of AOB, we can apply the Pythagorean theorem to deduce the value of the height :
\(H^2 + \frac{AB^2}{4} = OB^2\)
\(H = \sqrt{OB^2 - \frac{AB^2}{4}}\)
By plugging in the values of AB and OB we get :
\(H = \sqrt{16*3 - 6^2} = \sqrt{48 - 36} = \sqrt{12} = 2*\sqrt{3}\)
We can finally apply the formula to calculate the area of AOB :
\(Area of AOB = \frac{H*AB}{2} = 2*12*\sqrt{3}/2\) = \(12*\sqrt{3}\)
Which corresponds to answer choice A.
Hope that helped