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AMC 10 Geometry Math Olympiad USA Math Olympiad

Problem based on Cylinder | AMC 10A, 2015 | Question 9

Try this beautiful problem from Mensuration: Problem based on Cylinder from AMC 10A, 2015.

Cylinder – AMC-10A, 2015- Problem 9


Two right circular cylinders have the same volume. The radius of the second cylinder is $10 \%$ more than the radius of the first. What is the relationship between the heights of the two cylinders?

  • (A) The second height is $10 \%$ less than the first.
  • (B) The first height is $10 \%$ more than the second.
  • (C) The second height is $21 \%$ less than the first.
  • (D) The first height is $21 \%$ more than the second.
  • (E) The second height is $80 \%$ of the first.

Key Concepts


Mensuration

Cylinder

Check the Answer


Answer: (D) The first height is $21 \%$ more than the second.

AMC-10A (2015) Problem 9

Pre College Mathematics

Try with Hints


Let the radius of the first cylinder be $r_{1}$ and the radius of the second cylinder be $r_{2}$. Also, let the height of the first cylinder be $h_{1}$ and the height of the second cylinder be $h_{2}$.

Can you now finish the problem ……….

According to the problem,

$r_{2}=\frac{11 r_{1}}{10}$
$\pi r_{1}^{2} h_{1}=\pi r_{2}^{2} h_{2}$

can you finish the problem……..

$r_{1}^{2} h_{1}=\frac{121 r_{1}^{2}}{100} h_{2} \Rightarrow h_{1}=\frac{121 h_{2}}{100}$

Therefore the Possible answer will be (D) The first height is $21 \%$ more than the second.

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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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AMC-8 USA Math Olympiad

Mensuration : Area of Triangle , AMC 8 2015 Problem 21

What is Area of a triangle?


Area of a triangle $=\frac{1}{2}\times BASE \times HEIGHT $

Try the problem


In the given figure hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas $18$ and $32$ respectively, $\triangle JBK$ is equilateral and $FE=BC$. What is the area of $\triangle KBC$?

$\textbf{(A) }6\sqrt{2}\quad\textbf{(B) }9\quad\textbf{(C) }12\quad\textbf{(D) }9\sqrt{2}\quad\textbf{(E) }32$.

AMC 8 2015 Problem 21

Area of Squares and triangles.

6 out of 10

Challenges and Thrills of Pre college Mathematics

Knowledge Graph


area of triangle- knowledge graph

Use some hints


Can you find the lengths of one side of the squares $ABJI$ and $FEHG$ ??

If you can, then try to use the given conditions to find out the lengths of two sides of $\triangle KBC$

Area of a square =$(\textbf{Side of the square})^2$

Then Side of a square =$\sqrt{(\textbf{Area of the square})}$

Then you can easily find the lengths of a side of the squares $ABJI$ and $FEHG$ which is $3\sqrt{2}$ and $4\sqrt{2}$ respectively.

Then by the given condition $\overline{FE}=\overline{BC}=4\sqrt{2}$

Since $\triangle JBK$ is an equilateral triangle then all of its sides are equal.

and $\overline{JB}=\overline{BK}=3\sqrt{2}$

Now as you know that $\triangle JBK$ is equilateral and the hexagon $ABCDEF$ is equiangular. Can you find out the measure of $\angle KBC$ ??

From the figure we can clearly see $\angle JBA + \angle ABC + \angle KBC + \angle KBJ = 360^{\circ}$

$\angle KBC = 360^{\circ}-90^{\circ}-120^{\circ}-60^{\circ}$ [Since $\angle JBA=90^{\circ}$ (an angle of a square) $\angle ABC=120^{\circ}$ (an angle of an equiangular hexagon) and $\angle JBK= 60^{\circ} $(an angle of an equilateral triangle)]

i.e., $\angle KBC= 90^{\circ}$

Then $\triangle KBC$ is a right angle triangle

Then the base and height of $\triangle KBC$ are $BC$ and $KB$

So the area of $\triangle KBC=\frac12 \times KB \times BC = \frac12 \times 4\sqrt{2} \times 3\sqrt{2} = 12$

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