Problem Solving

Vertex A of equilateral triangle ABC is the center of the circle above. If the side of the triangle is 2, what is the radius of the cirlce?

A. 1/(2√3)
B. 1/2
C. (√3)/2
D. 1
E. √3

The OA is E.

Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.

Vertex A of equilateral triangle ABC is the center of the circle above. If the side of the triangle is 2, what is the radius of the cirlce?

A. 1/(2√3)
B. 1/2
C. (√3)/2
D. 1
E. √3

The OA is E.

Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.

Hi AAPL,
Let's take a look at your question.

We can see that if an altitude is drawn from the vertex A of the triangle to BC, it will represent the radius of the circle.
It means to find the radius of the circle, we need to find the length of Altitude of the triangle ABC.
Let the altitude from the vertex A meet BC at point D then ADB will be a right triangle.
$$AB=2$$
The altitude AD will bisect BC therefore,
$$BD=1$$
Now we can find the length of AD using Pythagorean theorem,
$$\left(AB\right)^2=\left(AD\right)^2+\left(DB\right)^2$$
$$\left(2\right)^2=\left(AD\right)^2+\left(1\right)^2$$
$$4=\left(AD\right)^2+1$$
$$\left(AD\right)^2=4-1$$
$$\left(AD\right)^2=3$$
$$AD=\sqrt{3}$$

Which is the radius of the circle.
Therefore, Option E is correct.

Hope it helps.
I am available if you'd like any follow up.