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

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==Problem==
 
==Problem==
 
Rectangle <math>ABCD</math> has sides <math>CD=3</math> and <math>DA=5</math>. A circle of radius <math>1</math> is centered at <math>A</math>, a circle of radius <math>2</math> is centered at <math>B</math>, and a circle of radius <math>3</math> is centered at <math>C</math>. Which of the following is closest to the area of the region inside the rectangle but outside all three circles?
 
Rectangle <math>ABCD</math> has sides <math>CD=3</math> and <math>DA=5</math>. A circle of radius <math>1</math> is centered at <math>A</math>, a circle of radius <math>2</math> is centered at <math>B</math>, and a circle of radius <math>3</math> is centered at <math>C</math>. Which of the following is closest to the area of the region inside the rectangle but outside all three circles?
 +
 
<asy>
 
<asy>
 
draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0));
 
draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0));
Line 13: Line 14:
 
label("3",(5,1.5),E);
 
label("3",(5,1.5),E);
 
</asy>
 
</asy>
−
<math> \textbf{(A) }3.5\qquad\textbf{(B) }4.0\qquad\textbf{(C) }4.5\qquad\textbf{(D) }5.0\qquad\textbf{(E) }5.5 </math>
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 +
<math> \text{(A) }3.5\qquad\text{(B) }4.0\qquad\text{(C) }4.5\qquad\text{(D) }5.0\qquad\text{(E) }5.5 </math>
 
==Solution==
 
==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 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 <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==
 +
 +
https://youtu.be/e4gDE2uLQ6A ~DSA_Catachu
 +
 +
https://youtu.be/EPILwH5iA9Q ~savannahsolver
 +
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/j3QSD5eDpzU?t=2327
 +
 +
~ pi_is_3.14
  
 
==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}}

Latest revision as of 19:36, 2 January 2023

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]

$\text{(A) }3.5\qquad\text{(B) }4.0\qquad\text{(C) }4.5\qquad\text{(D) }5.0\qquad\text{(E) }5.5$

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

https://youtu.be/EPILwH5iA9Q ~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/j3QSD5eDpzU?t=2327

~ pi_is_3.14

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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