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## 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

Answer: $\frac{30}{11}$

AMC-10A (2001) Problem 21

Pre College Mathematics

## Try with Hints

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 ……….

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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## 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

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=$\frac {\frac {9h}{25}}{h} \times 100=36$%