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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>.
  
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*Note - If you don't know the diagnal 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.
+
*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==

Revision as of 19:09, 15 January 2024

Problem

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

[asy]/* AMC8 2003 #22 Problem */ size(3inch, 2inch); unitsize(1cm); pen outline = black+linewidth(1); filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle, mediumgrey, outline); filldraw(shift(3,0)*((0,0)--(2,0)--(2,2)--(0,2)--cycle), mediumgrey, outline); filldraw(Circle((7,1), 1), mediumgrey, black+linewidth(1)); filldraw(Circle((1,1), 1), white, outline); filldraw(Circle((3.5,.5), .5), white, outline); filldraw(Circle((4.5,.5), .5), white, outline); filldraw(Circle((3.5,1.5), .5), white, outline); filldraw(Circle((4.5,1.5), .5), white, outline); filldraw((6.3,.3)--(7.7,.3)--(7.7,1.7)--(6.3,1.7)--cycle, white, outline); label("A", (1, 2), N); label("B", (4, 2), N); label("C", (7, 2), N); draw((0,-.5)--(.5,-.5), BeginArrow); draw((1.5, -.5)--(2, -.5), EndArrow); label("2 cm", (1, -.5)); draw((3,-.5)--(3.5,-.5), BeginArrow); draw((4.5, -.5)--(5, -.5), EndArrow); label("2 cm", (4, -.5)); draw((6,-.5)--(6.5,-.5), BeginArrow); draw((7.5, -.5)--(8, -.5), EndArrow); label("2 cm", (7, -.5)); draw((6,1)--(6,-.5), linetype("4 4")); draw((8,1)--(8,-.5), linetype("4 4"));[/asy]

$\textbf{(A)}\ \text{A only}\qquad\textbf{(B)}\ \text{B only}\qquad\textbf{(C)}\ \text{C only}\qquad\textbf{(D)}\ \text{both A and B}\qquad\textbf{(E)}\ \text{all are equal}$


Solution

First we have to find the area of the shaded region in each of the figures. In figure $\textbf{A}$ the area of the shaded region is the area of the circle subtracted from the area of the square. That is $2^2-1^2 \pi=4-\pi$. In figure $\textbf{B}$ 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 $2^2-4 \left( \left( \frac{1}{2} \right)^2 \pi \right)=4-4 \left(\frac{\pi}{4} \right)=4-\pi$. In figure $\textbf{C}$ 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 $\frac{d_1 d_2}{2}$. So the area of the shaded region is $1^2 \pi-\frac{2\cdot{2}}{2}=\pi-2$. Clearly the largest area that we found among the three shaded regions is $\pi-2$. Thus the answer is $\boxed{C}$.

  • 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 $x$), which is $\sqrt{2}$, and get that the area is $2$, and continute as shown in the solution.

Video Solution

https://youtu.be/EESSjTQ87YU Soo, DRMS, NM

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

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