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Circular Cylinder Problem | AMC-10A, 2001 | Problem 21

Try this beautiful problem from Geometry: circular cylinder from AMC-10A, 2001. You may use sequential hints to solve the problem.

Try this beautiful problem from Geometry based on Circular Cylinder.

Circular Cylinder Problem – AMC-10A, 2001- Problem 21


A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter $10$ and altitude $12$, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.

  • \(\frac{30}{23}\)
  • \(\frac{30}{11}\)
  • \(\frac{15}{11}\)
  • \(\frac{17}{11}\)
  • \(\frac{3}{2}\)

Key Concepts


Geometry

Cylinder

cone

Check the Answer


Answer: \(\frac{30}{11}\)

AMC-10A (2001) Problem 21

Pre College Mathematics

Try with Hints


Circular Cylinder Problem

Given that the diameter equal to its height is inscribed in a right circular cone.Let the diameter and the height of the right circular cone be \(2r\).And also The cone has diameter $10$ and altitude $12$, and the axes of the cylinder and cone coincide.we have to find out the radius of the cylinder.Now if we can show that \(\triangle AFE \sim \triangle AGC\), then we can find out the value of \(r\)

Can you now finish the problem ……….

Circular cylinder problem

Given that \(Bc=10\),\(AG=12\),\(HL=FG=2r\). Therefore \(AF=12-2r\),\(FE=r\),\(GC=5\)

Now the \(\triangle AFE \sim \triangle AGC\), Can you find out the radius from from this similarity property…….?

can you finish the problem……..

Since \(\triangle AFE \sim \triangle AGC\), we can write \(\frac{AF}{FE}=\frac{AG}{GC}\)

\(\Rightarrow \frac{12-2r}{r}=\frac{12}{5}\)

\(\Rightarrow r=\frac{30}{11}\)

Therefore the radius of the cylinder is \(\frac{30}{11}\)

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