Difference between revisions of "2004 AMC 12B Problems/Problem 7"

(New page: == Problem 7 == A square has sides of length 10, and a circle centered at one of its vertices has radius 10. What is the area of the union of the regions enclosed by the square and the cir...)
 
(Solution)
 
(7 intermediate revisions by 6 users not shown)
Line 1: Line 1:
== Problem 7 ==
+
{{duplicate|[[2004 AMC 12B Problems|2004 AMC 12B #7]] and [[2004 AMC 10B Problems|2004 AMC 10B #9]]}}
A square has sides of length 10, and a circle centered at one of its vertices has radius 10. What is the area of the union of the regions enclosed by the square and the circle?
+
 
 +
== Problem ==
 +
A square has sides of length <math>10</math>, and a circle centered at one of its vertices has radius <math>10</math>. What is the area of the union of the regions enclosed by the square and the circle?
 +
 
 +
<math>\mathrm{(A)\ }200+25\pi\quad\mathrm{(B)\ }100+75\pi\quad\mathrm{(C)\ }75+100\pi\quad\mathrm{(D)\ }100+100\pi\quad\mathrm{(E)\ }100+125\pi</math>
 +
 
 +
== Video Solution 1==
 +
https://youtu.be/IGN4XxJIbE0
 +
 
 +
~Education, the Study of Everything
 +
 
  
<math>(\mathrm {A}) 200+25\pi \quad (\mathrm {B}) 100+75\pi \quad (\mathrm {C}) 75+100\pi \quad (\mathrm {D}) 100+100\pi \quad (\mathrm {E}) 100+125\pi</math>
 
  
 
== Solution ==
 
== Solution ==
  
The area of the circle is <math>S_{\bigcirc}=100\pi</math>, the area of the square is <math>S_{\square}=100</math>.
+
The area of the circle is <math>S_{\bigcirc}=100\pi</math>; the area of the square is <math>S_{\square}=100</math>.
  
Exactly <math>1/4</math> of the circle lies inside the square. Thus the total area is <math>\dfrac34 S_{\bigcirc} + S_{\square} = \boxed{100+75\pi} \Longrightarrow \mathrm{(B)}</math>.
+
Exactly <math>\frac{1}{4}</math> of the circle lies inside the square. Thus the total area is <math>\dfrac34 S_{\bigcirc}+S_{\square}=\boxed{\mathrm{(B)\ }100+75\pi}</math>.
 +
 
 +
<asy>
 +
Draw(Circle((0,0),10));
 +
Draw((0,0)--(10,0)--(10,10)--(0,10)--(0,0));
 +
label("$10$",(5,0),S);
 +
label("$10$",(0,5),W);
 +
dot((0,0));
 +
</asy>
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2004|ab=B|num-b=6|num-a=8}}
 
{{AMC12 box|year=2004|ab=B|num-b=6|num-a=8}}
 +
{{AMC10 box|year=2004|ab=B|num-b=8|num-a=10}}
 +
 +
[[Category:Introductory Geometry Problems]]
 +
[[Category:Area Problems]]
 +
{{MAA Notice}}

Latest revision as of 19:23, 22 October 2022

The following problem is from both the 2004 AMC 12B #7 and 2004 AMC 10B #9, so both problems redirect to this page.

Problem

A square has sides of length $10$, and a circle centered at one of its vertices has radius $10$. What is the area of the union of the regions enclosed by the square and the circle?

$\mathrm{(A)\ }200+25\pi\quad\mathrm{(B)\ }100+75\pi\quad\mathrm{(C)\ }75+100\pi\quad\mathrm{(D)\ }100+100\pi\quad\mathrm{(E)\ }100+125\pi$

Video Solution 1

https://youtu.be/IGN4XxJIbE0

~Education, the Study of Everything


Solution

The area of the circle is $S_{\bigcirc}=100\pi$; the area of the square is $S_{\square}=100$.

Exactly $\frac{1}{4}$ of the circle lies inside the square. Thus the total area is $\dfrac34 S_{\bigcirc}+S_{\square}=\boxed{\mathrm{(B)\ }100+75\pi}$.

[asy] Draw(Circle((0,0),10)); Draw((0,0)--(10,0)--(10,10)--(0,10)--(0,0)); label("$10$",(5,0),S); label("$10$",(0,5),W); dot((0,0)); [/asy]

See Also

2004 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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 12 Problems and Solutions
2004 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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. AMC logo.png