This is a great math puzzle and critical thinking challenge that is sure to get you thinking!

## Challenge:

What is the perimeter of the inscribed equilateral triangle, if the diameter of the circle above is 4?

**A-** \(4\sqrt{2}\)

**B-** \(4\sqrt{3}\)

**C-** \(6\sqrt{2}\)

**D-** \(6\sqrt{3}\)

**E-** 12

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The correct answer is D.

Draw the bisector of the angle A perpendicular to line BC.

D is the center of the circle and CD is equal to the radius. The diameter of the circle above is 4. So, CD is 2.

Triangle CDE is a 30-60-90 degree triangle and angle DCE is 30.

Since, CD is 2 (the hypotenuse of the triangle CDE), DE is 1 and CE is \(\sqrt{3}\). Why?

Therefore, BC is \(2\sqrt{3} \) and the perimeter of the triangle ABC is

\(3 × 2\sqrt{3} = 6\sqrt{3}\)