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AMC 8 Math Olympiad USA Math Olympiad

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

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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AMC 8 Math Olympiad USA Math Olympiad

Problem on Cylinder | AMC-10A, 2004 | Problem 11

Try this beautiful problem from AMC 10A, 2004 based on Mensuration: Cylinder

Problem on Cylinder – AMC-10A, 2004- Problem 11


A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by \(25\%\) without altering the volume, by what percent must the height be decreased?

  • \(16\)
  • \(18\)
  • \(20\)
  • \(36\)
  • \(25\)

Key Concepts


Mensuration

Cylinder

Percentage

Check the Answer


Answer: \(36\)

AMC-10A (2004) Problem 11

Pre College Mathematics

Try with Hints


Let the radius of the jar be \(x\) and height be \(h\).then the volume (V) of the jar be\(V\)= \(\pi (x)^2 h\). Diameter of the jar increase \(25 \)% Therefore new radius will be \(x +\frac{x}{4}=\frac{5x}{4}\) .Now the given condition is “after increase the volume remain unchange”.Let new height will be \(h_1\).Can you find out the new height….?

can you finish the problem……..

Let new height will be \(H\).Therefore the volume will be \(\pi (\frac{5x}{4})^2 H\).Since Volume remain unchange……

\(\pi (x)^2 h\)=\(\pi (\frac{5x}{4})^2 H\) \(\Rightarrow H=\frac{16h}{25}\).

height decrease =\(h-\frac{16h}{25}=\frac{9h}{25}\).can you find out the decrease percentage?

can you finish the problem……..

Decrease Percentage=\( \frac {\frac {9h}{25}}{h} \times 100=36\)%

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