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AMC-8 Geometry Math Olympiad

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

Try this beautiful problem from AMC-8(2008),Problem-25,Geometry based on Area of a Circle. You may use sequential hints to solve the problem.

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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