Difference between revisions of "2002 AMC 10B Problems/Problem 5"
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| − | A line going through the centers of the two smaller circles also | + | A line going through the centers of the two smaller circles also goes through the diameter. The length of this line within the circle is <math>3+3+2+2=10.</math> Because this is the length of the larger circle's diameter, the length of its radius is <math>5.</math> |
The area of the large circle is <math>25\pi</math>, and the area of the two smaller circles is <math>9\pi + 4\pi = 13\pi.</math> To find the area of the shaded region, subtract the area of the two smaller circles from the area of the large circle. <math>\longrightarrow 25\pi - 13\pi = \boxed{\mathrm{(E) \ } 12\pi}</math> | The area of the large circle is <math>25\pi</math>, and the area of the two smaller circles is <math>9\pi + 4\pi = 13\pi.</math> To find the area of the shaded region, subtract the area of the two smaller circles from the area of the large circle. <math>\longrightarrow 25\pi - 13\pi = \boxed{\mathrm{(E) \ } 12\pi}</math> | ||
Latest revision as of 16:30, 5 January 2021
Problem
Circles of radius 
and 
are externally tangent and are circumscribed by a third circle, as shown in the figure. Find the area of the shaded region.
![[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real r1=3; real r2=2; real r3=5; pair A=(-2,0), B=(3,0), C=(0,0); pair X=(1,0), Y=(5,0); path circleA=Circle(A,r1); path circleB=Circle(B,r2); path circleC=Circle(C,r3); fill(circleC,gray); fill(circleA,white); fill(circleB,white); draw(circleA); draw(circleB); draw(circleC); draw(A--X); draw(B--Y); pair[] ps={A,B}; dot(ps); label("$3$",midpoint(A--X),N); label("$2$",midpoint(B--Y),N); [/asy]](http://latex.artofproblemsolving.com/2/c/5/2c5a7bbafa30af5c988825a58d564ad2037aaa7b.png)
Solution
A line going through the centers of the two smaller circles also goes through the diameter. The length of this line within the circle is 
Because this is the length of the larger circle's diameter, the length of its radius is 
The area of the large circle is 
, and the area of the two smaller circles is 
To find the area of the shaded region, subtract the area of the two smaller circles from the area of the large circle. 
See Also
| 2002 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
