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Difference between revisions of "2014 AMC 8 Problems/Problem 20"

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The area of the rectangle is <math>3\cdot5 =15</math>. The area of all 3 quarter circles is <math>\frac{\pi}{4}+\frac{\pi(2)^2}{4}+\frac{\pi(3)^2}{4} = \frac{14\pi}{4} = \frac{7\pi}{2}</math>. Therefore the area in the rectangle but outside the circles is <math>15-\frac{7\pi}{2}</math>.  <math>\pi</math> is approximately <math>\dfrac{22}{7},</math> and substituting that in will give <math>15-11=\boxed{\text{(B) }4.0}</math>
 
The area of the rectangle is <math>3\cdot5 =15</math>. The area of all 3 quarter circles is <math>\frac{\pi}{4}+\frac{\pi(2)^2}{4}+\frac{\pi(3)^2}{4} = \frac{14\pi}{4} = \frac{7\pi}{2}</math>. Therefore the area in the rectangle but outside the circles is <math>15-\frac{7\pi}{2}</math>.  <math>\pi</math> is approximately <math>\dfrac{22}{7},</math> and substituting that in will give <math>15-11=\boxed{\text{(B) }4.0}</math>
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==Video Solution==
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 +
https://youtu.be/e4gDE2uLQ6A ~DSA_Catachu
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2014|num-b=19|num-a=21}}
 
{{AMC8 box|year=2014|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:27, 6 July 2020

Problem

Rectangle $ABCD$ has sides $CD=3$ and $DA=5$. A circle of radius $1$ is centered at $A$, a circle of radius $2$ is centered at $B$, and a circle of radius $3$ is centered at $C$. Which of the following is closest to the area of the region inside the rectangle but outside all three circles? [asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw(Circle((0,0),1)); draw(Circle((0,3),2)); draw(Circle((5,3),3)); label("A",(0.2,0),W); label("B",(0.2,2.8),NW); label("C",(4.8,2.8),NE); label("D",(5,0),SE); label("5",(2.5,0),N); label("3",(5,1.5),E); [/asy] $\textbf{(A) }3.5\qquad\textbf{(B) }4.0\qquad\textbf{(C) }4.5\qquad\textbf{(D) }5.0\qquad\textbf{(E) }5.5$

Solution

The area in the rectangle but outside the circles is the area of the rectangle minus the area of all three of the quarter circles in the rectangle.

The area of the rectangle is $3\cdot5 =15$. The area of all 3 quarter circles is $\frac{\pi}{4}+\frac{\pi(2)^2}{4}+\frac{\pi(3)^2}{4} = \frac{14\pi}{4} = \frac{7\pi}{2}$. Therefore the area in the rectangle but outside the circles is $15-\frac{7\pi}{2}$. $\pi$ is approximately $\dfrac{22}{7},$ and substituting that in will give $15-11=\boxed{\text{(B) }4.0}$

Video Solution

https://youtu.be/e4gDE2uLQ6A ~DSA_Catachu

See Also

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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

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