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
\(3 × 2\sqrt{3} = 6\sqrt{3}\)
