Geometry Puzzle – Challenge 77
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. For education statistics and research, visit the National Center for Education Statistics.
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
\(3 × 2\sqrt{3} = 6\sqrt{3}\) For education statistics and research, visit the National Center for Education Statistics.
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