Difference between revisions of "2003 AMC 8 Problems/Problem 22"
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First we have to find the area of the shaded region in each of the figures. In figure <math>\textbf{A}</math> the area of the shaded region is the area of the circle subtracted from the area of the square. That is <math>2^2-1^2 \pi=4-\pi</math>. In figure <math>\textbf{B}</math> the area of the shaded region is the sum of the areas of the 4 circles subtracted from the area of the square. That is <math>2^2-4 \left( \left( \frac{1}{2} \right)^2 \pi \right)=4-4 \left(\frac{\pi}{4} \right)=4-\pi</math>. In figure <math>\textbf{C}</math> the area of the shaded region is the area of the square subtracted from the area of the circle. The diameter of the circle and the diagonal of the square are equal to 2. We can easily find the area of the square using the area formula <math>\frac{d_1 d_2}{2}</math>. So the area of the shaded region is <math>1^2 \pi-\frac{2\cdot{2}}{2}=\pi-2</math>. Clearly the largest area that we found among the three shaded regions is <math>\pi-2</math>. Thus the answer is <math>\boxed{C}</math>. | First we have to find the area of the shaded region in each of the figures. In figure <math>\textbf{A}</math> the area of the shaded region is the area of the circle subtracted from the area of the square. That is <math>2^2-1^2 \pi=4-\pi</math>. In figure <math>\textbf{B}</math> the area of the shaded region is the sum of the areas of the 4 circles subtracted from the area of the square. That is <math>2^2-4 \left( \left( \frac{1}{2} \right)^2 \pi \right)=4-4 \left(\frac{\pi}{4} \right)=4-\pi</math>. In figure <math>\textbf{C}</math> the area of the shaded region is the area of the square subtracted from the area of the circle. The diameter of the circle and the diagonal of the square are equal to 2. We can easily find the area of the square using the area formula <math>\frac{d_1 d_2}{2}</math>. So the area of the shaded region is <math>1^2 \pi-\frac{2\cdot{2}}{2}=\pi-2</math>. Clearly the largest area that we found among the three shaded regions is <math>\pi-2</math>. Thus the answer is <math>\boxed{C}</math>. | ||
− | *Note - If you don't know the | + | *Note - If you don't know the diagonal formula, you can also find the length of the square using the Pythagorean Theorem (set up an equation and solve for <math>x</math>), which is <math>\sqrt{2}</math>, and get that the area is <math>2</math>, and continute as shown in the solution. |
==Video Solution== | ==Video Solution== | ||
− | |||
https://www.youtube.com/watch?v=ei9blxnl9Gw ~David | https://www.youtube.com/watch?v=ei9blxnl9Gw ~David |
Latest revision as of 10:40, 14 June 2024
Contents
Problem
The following figures are composed of squares and circles. Which figure has a shaded region with largest area?
Solution
First we have to find the area of the shaded region in each of the figures. In figure the area of the shaded region is the area of the circle subtracted from the area of the square. That is
. In figure
the area of the shaded region is the sum of the areas of the 4 circles subtracted from the area of the square. That is
. In figure
the area of the shaded region is the area of the square subtracted from the area of the circle. The diameter of the circle and the diagonal of the square are equal to 2. We can easily find the area of the square using the area formula
. So the area of the shaded region is
. Clearly the largest area that we found among the three shaded regions is
. Thus the answer is
.
- Note - If you don't know the diagonal formula, you can also find the length of the square using the Pythagorean Theorem (set up an equation and solve for
), which is
, and get that the area is
, and continute as shown in the solution.
Video Solution
https://www.youtube.com/watch?v=ei9blxnl9Gw ~David
See Also
2003 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.