In the figure above, showing circle with center O and points

Area of ABC = 1/2*AB*BC = 1/2*12*4\sqrt{3} = 24\sqrt{3}
Area of AOB = 1/2 Area of ABC = 12\sqrt{3}

Hence, A is correct.

Thank you for another great geometry question, emmak. Now let's solve it properly


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

Hey Bunuel,
Each median divides the triangle into two smaller triangles which have the same area

is it that according to this property as mentioned above by you mean that the two smaller triangles are similiar?

Where did I say that the two smaller triangles are similar? They are not.

AOC is the diameter so the angle subtended in semi circle ABC will be 90 degrees. So triangle ABC is a right triangle at B.

If \(AB = 12\) and \(BC = 4*\sqrt{3}\), this is a ratio of \(\sqrt{3}:1\) which means that the hypotenuse will be 2 on the ratio scale. So \(AC = 2*4*\sqrt{3} = 8\sqrt{3}\)

Area of triangle ABC = (1/2) * AB * BC = (1/2) * Altitude from B to AC * AC
Altitude from B to AC \(= 12 * (4\sqrt{3}) / (8\sqrt{3}) = 6\)

Area of triangle AOB = (1/2) * AO * Altitude from B to AC = \((1/2) * (8\sqrt{3})/2 * 6 = 12\sqrt{3}\)

Answer (A)

I only calculated the area of triangle ABC which came out to be 24sqrt3 . Now triangle AOB will have an area smaller than triangle ABC .Went through the options and 12sqrt3 is the only value that is less than 24sqrt3 . Took less than 30 secs.

AB=12 and BC=4\sqrt{3}

AC is the diameter of the circle and also the side of triangle ABC. So the angle at B is 90.

AC^2=AB^2 + BC^2
from this we get AC=8\sqrt{3}

Area of ABC= 1/2 * AB * BC

= 24\sqrt{3}

now

BC=OC=OB=4\sqrt{3}

Area of triangle OBC is \sqrt{3}/4 * (4\sqrt{3})^2..i.e., 12\sqrt{3}

Area of AOB = Area of ABC- Area of OBC

= 24\sqrt{3} - 12\sqrt{3}

= 12\sqrt{3}

IMO: option A

Question is good but tge option make it 600 difficulty one. Because for the whole triangle ABC is are is 24√3 so any how area of AOB will be less than 24√3 and only one smaller than that option A

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could you please elaborate on how BO is the median?
"Now, notice that since OA=OC=radius, then BO turns out to be the median of triangle ABC"

The median of a triangle is a line from a vertex to the midpoint of the opposite side. O is the midpoint of AC, so BO is the median.

Shouldn't A0=B0=OC. In that case since the length of AC=8 root 3 then A0=OC=4 root 3 and also B0=4 root 3 ?

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