Equilateral triangle ABC is inscribed in a circle with center O, as sh

A helpful property to know about triangles is that the intersection of the 3 medians (lines from a vertex to the center of the opposite side) all intersect at the centroid, and the centroid divides each median in the ratio of 2:1. More easily explained with a diagram:


Here c is the centroid of triangle ABC, and c divides the median AD into two parts with ratio 2:1. That is, Ac:cD = 2:1. This is true of the other medians as well.

Ok, back to our problem:


In the problem, because we have an equilateral triangle, CO is the radius and CD is both the median and the altitude (height) of the triangle. So CO=2 (given), and OD=1 (from the ratio CO:OD = 2:1), therefore the height, CD = 2+1 = 3.

From here, we can use the property of a 30-60-90 triangle to find the length of the sides. In our 30-60-90 triangle CAD, we know that the ratio of the sides AD:CD:CA is 1:\(\sqrt{3}\):2. And we know that side CD is 3, so our side lengths are \(\sqrt{3}:3:2\sqrt{3}\)

Alternatively, if we know the height of an equilateral triangle, then the length of a side is \(\frac{2}{\sqrt{3}}*h\). So in our case, the length of a side is \(\frac{2}{\sqrt{3}}*3=2\sqrt{3}\)

CA = \(2\sqrt{3}\) which is the base of the triangle.

The area of the triangle is therefore \(\frac{bh}{2}=\frac{2\sqrt{3}*3}{2}=3\sqrt{3}\)

Answer: D


A shortcut method that requires memorizing a formula: For an equilateral triangle, if we know the height,

\(A=\frac{h^2}{\sqrt{3}}\)

\(A=\frac{3^2}{\sqrt{3}}=\frac{9}{\sqrt{3}}=3\sqrt{3}\)

Another straightforward answer:
O is the centre of the circle. Therefore Angle AOC = 2*Angle ABC.
Hence Angle AOC = 120.

Length of the line segment AC = 2r Sin(theta/2)
=> 2*2*sin(60)
=>AC=2*root3

Area=root3/4*(AC^2)
=> Area=3*root3

Radius of circle = 2
So , OC =2
Now since triangle is equilateral and O is the centre of circle. It divides equilateral triangle in ratio 2: 1

So, Height of equilateral triangle = 3/2 * 2 = 3

Let side of equilateral triangle be a
H = sqrt {3}/2 *a
-> a= (2/sqrt{3} )* H
a = (2/sqrt{3} )* 3 = 2 sqrt {3}

so area =( sqrt{3}/4 )* a^2 = \(3 \sqrt{3}\)

Wouldn't the base be 4 cm and height be 2* sqrt(3) cm?
Then the area of ABC would be 3*2*0.5*4*2*sqrt(3)

Bunuel, can you share the official solution?