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

(See Also)
Line 24: Line 24:
 
{{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}}
 
{{AMC10 box|year=2004|ab=B|num-b=8|num-a=10}}
 +
 +
[[Category:Introductory Geometry Problems]]
 +
[[Category:Area Problems]]

Revision as of 22:40, 11 April 2013

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?

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

Solution

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

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

[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