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AMC-8 Math Olympiad USA Math Olympiad

Largest area Problem | AMC 8, 2003 | Problem 22

Try this beautiful problem from AMC 8, 2003, problem no-22 based on Largest area. You may use sequential hints to solve the problem.

Try this beautiful problem from Geometry based Largest area.

Largest area – AMC-8, 2003 – Problem 22


The following figures are composed of squares and circles. Which figure has a shaded region with largest area?

Problem figure
  • $A$
  • $B$
  • $C$

Key Concepts


Geometry

Circle

Square

Check the Answer


But try the problem first…

Answer:$C$

Source
Suggested Reading

AMC-8 (2003) Problem 22

Pre College Mathematics

Try with Hints


First hint

To find out the largest area at first we have to find out the radius of the circles . all the circles are inscribed ito the squares .now there is a relation between the radius and the side length of the squares….

Can you now finish the problem ……….

Second Hint

area of circle =\(\pi r^2\)

can you finish the problem……..

Final Step

Largest  area Problem

In A:

Total area of the square =\(2^2=4\)

Now the radius of the inscribed be 1(as the diameter of circle = side length of the side =2)

Area of the inscribed circle is \(\pi (1)^2=\pi\)

Therefore the shaded area =\(4- \pi\)

In B:

Largest  area Problem - figure

Total area of the square =\(2^2=4\)

There are 4 circle and radius of one circle be \(\frac{1}{2}\)

Total area pf 4 circles be \(4 \times \pi \times (\frac{1}{2})^2=\pi\)

Therefore the shaded area =\(4- \pi\)

In C:

finding the largest area

Total area of the square =\(2^2=4\)

Now the length of the diameter = length of the diagonal of the square=2

Therefore radius of the circle=\(\pi\) and lengthe of the side of the square=\(\sqrt 2\)

Thertefore area of the shaded region=Area of the square-Area of the circle=\(\pi (1)^2-(\sqrt 2)^2\)=\(\pi – 2\)

Therefore the answer is  C

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