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Problem on Circumscribed Circle | AMC-10A, 2003 | Problem 17

Try this beautiful problem from Geometry:Radius of a circle.AMC-10A, 2003. You may use sequential hints to solve the problem

Try this beautiful problem from Geometry based on Circumscribed Circle

Problem on Circumscribed Circle – AMC-10A, 2003- Problem 17


The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?

  • \(\frac{5\sqrt3}{\pi}\)
  • \(\frac{3\sqrt3}{\pi}\)
  • \(\frac{3\sqrt3}{2\pi}\)

Key Concepts


Geometry

Triangle

Circle

Check the Answer


Answer: \(\frac{3\sqrt3}{\pi}\)

AMC-10A (2003) Problem 17

Pre College Mathematics

Try with Hints


Circumscribed circle figure

Let ABC is a equilateral triangle which is inscribed in a circle. with center \(O\). and also given that perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle.so for find out the peremeter of Triangle we assume that the side length of the triangle be \(x\) and the radius of the circle be \(r\). then the side of an inscribed equilateral triangle is \(r\sqrt{3}\)=\(x\)

Can you now finish the problem ……….

circumscribed circle

The perimeter of the triangle is=\(3x\)=\(3r\sqrt{3}\) and Area of the circle=\(\pi r^2\)

Now The perimeter of the triangle=The Area of the circle

Therefore , \(3x\)=\(3r\sqrt{3}\)=\(\pi r^2\)

can you finish the problem……..

Now \(3x\)=\(3r\sqrt{3}\)=\(\pi r^2\) \(\Rightarrow {\pi r}=3\sqrt 3\) \(\Rightarrow r=\frac{3\sqrt3}{\pi}\)

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