Difference between revisions of "2003 AMC 8 Problems/Problem 22"
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+ | ==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> | ||
+ | |||
+ | <math> \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}</math> | ||
+ | |||
+ | |||
+ | ==Solution== | ||
+ | |||
First we have to find the area of the shaded region in each of the figures. In figure <math>\bolded{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>\bolded{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((\frac{1}{2})^2 \pi)=4-4(\frac{\pi}{4})=4-\pi</math>. In figure <math>\bolded{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>4-\pi</math>. Both figure <math>\bolded{A}</math> and figure <math>\bolded{B}</math> had that area so the answer is <math>{(D)}\ \text{both A and B}</math> | First we have to find the area of the shaded region in each of the figures. In figure <math>\bolded{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>\bolded{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((\frac{1}{2})^2 \pi)=4-4(\frac{\pi}{4})=4-\pi</math>. In figure <math>\bolded{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>4-\pi</math>. Both figure <math>\bolded{A}</math> and figure <math>\bolded{B}</math> had that area so the answer is <math>{(D)}\ \text{both A and B}</math> | ||
{{AMC8 box|year=2003|num-b=21|num-a=23}} | {{AMC8 box|year=2003|num-b=21|num-a=23}} |
Revision as of 11:28, 24 August 2012
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 $\bolded{A}$ (Error compiling LaTeX. Unknown error_msg) the area of the shaded region is the area of the circle subtracted from the area of the square. That is . In figure $\bolded{B}$ (Error compiling LaTeX. Unknown error_msg) 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 $\bolded{C}$ (Error compiling LaTeX. Unknown error_msg) 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 . Both figure $\bolded{A}$ (Error compiling LaTeX. Unknown error_msg) and figure $\bolded{B}$ (Error compiling LaTeX. Unknown error_msg) had that area so the answer is
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 |