INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More

Contents

[hide]

Try this beautiful problem from Geometry based on Circular Cylinder.

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.

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

Geometry

Cylinder

cone

But try the problem first...

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

Source

Suggested Reading

AMC-10A (2001) Problem 21

Pre College Mathematics

First hint

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 ..........

Second Hint

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........

Final Step

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

- https://www.cheenta.com/area-of-hexagon-problem-amc-10a-2014-problem-13/
- https://www.youtube.com/watch?v=fRj9NuPGrLU&t=269s

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIAL
Google