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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 Geometry Math Olympiad USA Math Olympiad

Area of Rectangle Problem | AMC 8, 2004 | Problem 24

Try this beautiful problem from Geometry from AMC-8, 2004 ,Problem-24, based on area of Rectangle.

Rectangle | AMC-8, 2004 | Problem 24


In the figure ABCD is a rectangle and EFGH  is a parallelogram. Using the measurements given in the figure, what is the length  d  of the segment that is perpendicular to  HE and FG?

Area of Rectangle Problem
  • $7.1$
  • $7.6$
  • $7.8$

Key Concepts


Geometry

Rectangle

Parallelogram

Check the Answer


Answer:$7.6$

AMC-8, 2004 problem 24

Pre College Mathematics

Try with Hints


Find Area of the Rectangle and area of the Triangles i.e \((\triangle AHE ,\triangle EBF , \triangle FCG , \triangle DHG) \)

Can you now finish the problem ……….

Area of the Parallelogram EFGH=Area Of Rectangle ABCD-Area of\((\triangle AHE +\triangle EBF + \triangle FCG + \triangle DHG) \)

can you finish the problem……..

Area of Rectangle Problem

Area of the Rectangle =\(CD \times AD \)=\(10 \times 8\)=80 sq.unit

Area of the \(\triangle AHE\) =\(\frac{1}{2} \times AH \times AE \)= \(\frac{1}{2} \times 4 \times 3\) =6 sq.unit

Area of the \(\triangle EBF\) =\(\frac{1}{2} \times EB \times BE \)= \(\frac{1}{2} \times 6 \times 5\) =15 sq.unit

Area of the \(\triangle FCG\) =\(\frac{1}{2} \times GC \times FC\)= \(\frac{1}{2} \times 4\times 3\) =6 sq.unit

Area of the \(\triangle DHG\) =\(\frac{1}{2} \times DG \times DH \)= \(\frac{1}{2} \times 6 \times 5\) =15 sq.unit

Area of the Parallelogram EFGH=Area Of Rectangle ABCD-Area of\((\triangle AHE +\triangle EBF + \triangle FCG + \triangle DHG) \)=\(80-(6+15+6+15)=80-42=38\)

As ABCD is a Rectangle ,\(\triangle GCF\) is a Right-angle triangle,

Therefore GF=\(\sqrt{4^2 + 3^2}\)=5 sq.unit

Now Area of the parallelogram EFGH=\( GF \times d\)=38

\(\Rightarrow 5 \times d\)=38

\(\Rightarrow d=7.6\)

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AMC-8 Geometry Math Olympiad

Area of Circle Problem | AMC 8, 2008 | Problem 25

Try this beautiful problem from Geometry based on the Area of a Circle.

Area of Circle | AMC-8, 2008 | Problem 25


Margie’s winning art design is shown. The smallest circle has radius  2 inches, with each successive circle’s radius increasing by 2 inches. Approximately what percent of the design is black?

Area of circle problem
  • $44$
  • $42$
  • $45$

Key Concepts


Geometry

Area

Circle

Check the Answer


Answer:$42$

AMC-8, 2008 problem 25

Pre College Mathematics

Try with Hints


Area of the square is \(\pi (r)^2\),where \(r\)=radius of the circle

Can you now finish the problem ……….

Find the total area of the black region……..

can you finish the problem……..

Area of circle problem

Given that The smallest circle has radius  2 inches, with each successive circle’s radius increasing by 2 inches .

The radius of the 1st circle is 2, So the area is \(\pi(2)^2\)=4\(\pi\) sq.unit

The radius of the 2nd circle is 4, So the area is \(\pi(4)^2\)=16\(\pi\) sq.unit

The radius of the 3rd circle is 6 So the area is \(\pi(6)^2\)=36\(\pi\) sq.unit

The radius of the 4th circle is 8, So the area is \(\pi(8)^2\)=64\(\pi\) sq.unit

The radius of the 5th circle is 10, So the area is \(\pi(10)^2\)=100\(\pi\) sq.unit

The radius of the 6th circle is 12, So the area is \(\pi(12)^2\)=144\(\pi\) sq.unit

Therefore The entire circle’s area is 144\(\pi\)

The area of the black regions is \((100\pi-64\pi)+(36\pi-16\pi)+4\pi=60\pi \)sq.unit

The percentage of the design that is black is  \((\frac{60\pi}{144\pi} \times 100)\%=(\frac{5}{12} \times 100) \% \approx 42\%\)

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