Difference between revisions of "2002 AMC 10B Problems/Problem 5"

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== Problem 5 ==
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== Problem==
  
 
Circles of radius <math>2</math> and <math>3</math> are externally tangent and are circumscribed by a third circle, as shown in the figure. Find the area of the shaded region.
 
Circles of radius <math>2</math> and <math>3</math> are externally tangent and are circumscribed by a third circle, as shown in the figure. Find the area of the shaded region.
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==Solution==
 
==Solution==
  
A line going through the centers of the two smaller circles also go 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>
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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>
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==See Also==
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{{AMC10 box|year=2002|ab=B|num-b=4|num-a=6}}
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[[Category:Introductory Geometry Problems]]
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[[Category:Area Problems]]
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{{MAA Notice}}

Latest revision as of 16:30, 5 January 2021

Problem

Circles of radius $2$ and $3$ 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]

$\mathrm{(A) \ } 3\pi\qquad \mathrm{(B) \ } 4\pi\qquad \mathrm{(C) \ } 6\pi\qquad \mathrm{(D) \ } 9\pi\qquad \mathrm{(E) \ } 12\pi$

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 $3+3+2+2=10.$ Because this is the length of the larger circle's diameter, the length of its radius is $5.$

The area of the large circle is $25\pi$, and the area of the two smaller circles is $9\pi + 4\pi = 13\pi.$ To find the area of the shaded region, subtract the area of the two smaller circles from the area of the large circle. $\longrightarrow 25\pi - 13\pi = \boxed{\mathrm{(E) \ } 12\pi}$

See Also

2002 AMC 10B (ProblemsAnswer KeyResources)
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

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