Difference between revisions of "2003 AMC 8 Problems/Problem 22"
Mathisfun7 (talk | contribs) (Created page with "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 f...") |
Mathisfun7 (talk | contribs) |
||
| Line 1: | Line 1: | ||
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}} | ||
Revision as of 11:25, 24 August 2012
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 | ||
