Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2004 AMC 8 Problems/Problem 25"
(→‎Solution 2)
 
(2 intermediate revisions by 2 users not shown)
Line 18: Line 18:
 
The area of the shaded region is the area of the two squares minus the area of the circle which is <math>\boxed{\textbf{(D)}\ 28-2\pi}</math>.
 
The area of the shaded region is the area of the two squares minus the area of the circle which is <math>\boxed{\textbf{(D)}\ 28-2\pi}</math>.
  
−
==Solution 2==
+
==Video Solution by OmegaLearn==
 +
https://youtu.be/j3QSD5eDpzU?t=2824
 +
 
 +
~ pi_is_3.14
  
−
Find the area of the overlapping squares. <math>2 \cdot 4^2 - 2^2=28</math>. Now, we find the chord of the circle to be 2, so then the radius of the circle would be <math>\sqrt{2}</math>. Now, the area of the circle is <math>\pi r^2</math>, so we put in the values to get <math>2\pi</math>. Now, we can subtract the two values. Now, we can find the answer to be <math>\boxed{\textbf{(D)}\ 28-2\pi}</math>.
 
 
==See Also:==
 
==See Also:==
−
{{AMC8 box|year=2004|num-b=24|after=Last <br /> Question of 2004}}
+
{{AMC8 box|year=2004|num-b=24|after=Last <br /> Question of 2004 AMC8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 08:28, 17 June 2024

Problem

Two $4 \times 4$ squares intersect at right angles, bisecting their intersecting sides, as shown. The circle's diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the squares?

[asy] unitsize(6mm); draw(unitcircle); filldraw((0,1)--(1,2)--(3,0)--(1,-2)--(0,-1)--(-1,-2)--(-3,0)--(-1,2)--cycle,lightgray,black); filldraw(unitcircle,white,black); [/asy]

$\text{(A)}\ 16-4\pi\qquad \text{(B)}\ 16-2\pi \qquad \text{(C)}\ 28-4\pi \qquad \text{(D)}\ 28-2\pi \qquad \text{(E)}\ 32-2\pi$

Solution 1

If the circle was shaded in, the intersection of the two squares would be a smaller square with half the sidelength, $2$. The area of this region would be the two larger squares minus the area of the intersection, the smaller square. This is $4^2 + 4^2 - 2^2 = 28$.

The diagonal of this smaller square created by connecting the two points of intersection of the squares is the diameter of the circle. This value can be found with Pythagorean or a $45^\circ - 45^\circ - 90^\circ$ circle to be $2\sqrt{2}$. The radius is half the diameter, $\sqrt{2}$. The area of the circle is $\pi r^2 = \pi (\sqrt{2})^2 = 2\pi$.

The area of the shaded region is the area of the two squares minus the area of the circle which is $\boxed{\textbf{(D)}\ 28-2\pi}$.

Video Solution by OmegaLearn

https://youtu.be/j3QSD5eDpzU?t=2824

~ pi_is_3.14

See Also:

2004 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last
Question of 2004 AMC8
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_8_Problems/Problem_25&oldid=220649"