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

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==Solution==
 
==Solution==
  
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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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{{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 (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

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