Difference between revisions of "2007 AMC 10A Problems/Problem 15"
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− | + | == Problem == | |
− | of | + | Four circles of radius <math>1</math> are each tangent to two sides of a square and externally tangent to a circle of radius <math>2</math>, as shown. What is the area of the square? |
− | + | <center> | |
− | of | + | <asy> |
− | + | unitsize(5mm); | |
+ | defaultpen(linewidth(.8pt)+fontsize(8pt)); | ||
+ | real h=3*sqrt(2)/2; | ||
+ | pair O0=(0,0), O1=(h,h), O2=(-h,h), O3=(-h,-h), O4=(h,-h); | ||
+ | pair X=O0+2*dir(30), Y=O2+dir(45); | ||
+ | draw((-h-1,-h-1)--(-h-1,h+1)--(h+1,h+1)--(h+1,-h-1)--cycle); | ||
+ | draw(Circle(O0,2)); | ||
+ | draw(Circle(O1,1)); | ||
+ | draw(Circle(O2,1)); | ||
+ | draw(Circle(O3,1)); | ||
+ | draw(Circle(O4,1)); | ||
+ | draw(O0--X); | ||
+ | draw(O2--Y); | ||
+ | label("$2$",midpoint(O0--X),NW); | ||
+ | label("$1$",midpoint(O2--Y),SE); | ||
+ | </asy> | ||
+ | </center> | ||
+ | <math>\text{(A)}\ 32 \qquad \text{(B)}\ 22 + 12\sqrt {2}\qquad \text{(C)}\ 16 + 16\sqrt {3}\qquad \text{(D)}\ 48 \qquad \text{(E)}\ 36 + 16\sqrt {2}</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | Draw a square connecting the centers of the four small circles of radius <math>1</math>. This square has a diagonal of length <math>6</math>, as it includes the diameter of the big circle of radius <math>2</math> and two radii of the small circles of radius <math>1</math>. Therefore, the side length of this square is <cmath>\frac{6}{\sqrt{2}} = 3\sqrt{2}</cmath>. | + | Draw a square connecting the centers of the four small circles of radius <math>1</math>. This square has a diagonal of length <math>6</math>, as it includes the diameter of the big circle of radius <math>2</math> and two radii of the small circles of radius <math>1</math>. Therefore, the side length of this square is <cmath>\frac{6}{\sqrt{2}} = 3\sqrt{2}.</cmath> The side length of the larger square is <math>2</math> units greater than the one found by connecting the midpoints, so its side length is <cmath>2 + 3\sqrt{2}.</cmath> The area of the larger square is <math>(2+3\sqrt{2})^2 = 22 + 12\sqrt{2}</math> <math>(B).</math> |
− | + | == Solution 2 == | |
+ | |||
+ | We draw the diagonal of the square. This diagonal yields <math>2\sqrt{2}+1+1+2+2=2\sqrt{2}+6</math>. We know that the side length <math>s</math> in terms of the diagonal <math>d</math> is <math>s=\frac{d}{\sqrt{2}}</math>, so our side length is <math>\frac{2\sqrt{2}+6}{\sqrt{2}}</math>. However, we are trying to look for the area, so squaring <math>\frac{2\sqrt{2}+6}{\sqrt{2}}</math> yields <math>\frac{44+24\sqrt{2}}{2}=\boxed{\text{(B)}22+12\sqrt{2}}</math> | ||
==See Also== | ==See Also== |
Latest revision as of 10:29, 4 June 2021
Contents
Problem
Four circles of radius are each tangent to two sides of a square and externally tangent to a circle of radius , as shown. What is the area of the square?
Solution 1
Draw a square connecting the centers of the four small circles of radius . This square has a diagonal of length , as it includes the diameter of the big circle of radius and two radii of the small circles of radius . Therefore, the side length of this square is The side length of the larger square is units greater than the one found by connecting the midpoints, so its side length is The area of the larger square is
Solution 2
We draw the diagonal of the square. This diagonal yields . We know that the side length in terms of the diagonal is , so our side length is . However, we are trying to look for the area, so squaring yields
See Also
2007 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 AMC 10 Problems and Solutions |
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