Difference between revisions of "2024 AMC 8 Problems/Problem 3"
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-Benedict T (countmath1) | -Benedict T (countmath1) | ||
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| + | ==Solution 2== | ||
| + | We can calculate it as the sum of the areas of <math>2</math> smaller trapezoids and <math>2</math> larger trapezoids. | ||
| + | <cmath>2\left(\cfrac{(7+4)(7-4)}{2}\right)+2\left(\cfrac{(10+9)(10-9)}{2}\right)=10^2 - 9^2 + 7^2 - 4^2 = 19 + 33 = \boxed{\textbf{(E)}\ 52}</cmath> | ||
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| + | -SahanWijetunga | ||
==Video Solution 1 (easy to digest) by Power Solve== | ==Video Solution 1 (easy to digest) by Power Solve== | ||
Revision as of 20:38, 26 January 2024
Contents
Problem 3
Four squares of side length 
and 
are arranged in increasing size order so that their left edges and bottom edges align. The squares alternate in color white-gray-white-gray, respectively, as shown in the figure. What is the area of the visible gray region in square units?
![[asy] size(150); filldraw((0,0)--(10,0)--(10,10)--(0,10)--cycle,gray(0.7),linewidth(1)); filldraw((0,0)--(9,0)--(9,9)--(0,9)--cycle,white,linewidth(1)); filldraw((0,0)--(7,0)--(7,7)--(0,7)--cycle,gray(0.7),linewidth(1)); filldraw((0,0)--(4,0)--(4,4)--(0,4)--cycle,white,linewidth(1)); draw((11,0)--(11,4),linewidth(1)); draw((11,6)--(11,10),linewidth(1)); label("$10$",(11,5),fontsize(14pt)); draw((10.75,0)--(11.25,0),linewidth(1)); draw((10.75,10)--(11.25,10),linewidth(1)); draw((0,11)--(4,11),linewidth(1)); draw((6,11)--(10,11),linewidth(1)); draw((0,11.25)--(0,10.75),linewidth(1)); draw((10,11.25)--(10,10.75),linewidth(1)); label("$9$",(5,11),fontsize(14pt)); draw((-1,0)--(-1,1),linewidth(1)); draw((-1,3)--(-1,7),linewidth(1)); draw((-1.25,0)--(-0.75,0),linewidth(1)); draw((-1.25,7)--(-0.75,7),linewidth(1)); label("$7$",(-1,2),fontsize(14pt)); draw((0,-1)--(1,-1),linewidth(1)); draw((3,-1)--(4,-1),linewidth(1)); draw((0,-1.25)--(0,-.75),linewidth(1)); draw((4,-1.25)--(4,-.75),linewidth(1)); label("$4$",(2,-1),fontsize(14pt)); [/asy]](http://latex.artofproblemsolving.com/8/8/7/887a43d8afeeeef5afbec8561bfde36d524b34ab.png)
Solution 1
We work inwards. The area of the outer shaded square is the area of the whole square minus the area of the second largest square. The area of the inner shaded region is the area of the third largest square minus the area of the smallest square. The sum of these areas is
![\[10^2 - 9^2 + 7^2 - 4^2 = 19 + 33 = \boxed{\textbf{(E)}\ 52}\]](http://latex.artofproblemsolving.com/6/0/e/60e3d82bd69bcfc64e4084908e40474b3db9af9b.png)
-Benedict T (countmath1)
Solution 2
We can calculate it as the sum of the areas of 
smaller trapezoids and larger trapezoids.
![\[2\left(\cfrac{(7+4)(7-4)}{2}\right)+2\left(\cfrac{(10+9)(10-9)}{2}\right)=10^2 - 9^2 + 7^2 - 4^2 = 19 + 33 = \boxed{\textbf{(E)}\ 52}\]](http://latex.artofproblemsolving.com/7/0/0/700260a89bb96e26af11b7f5bfb448d4fa1499c4.png)
-SahanWijetunga
Video Solution 1 (easy to digest) by Power Solve
https://youtu.be/HE7JjZQ6xCk?si=39xd5CKI9nx-7lyV&t=118
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=Ylw-kJkSpq8
~NiuniuMaths
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=L83DxusGkSY
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://youtu.be/sPzsce8FQtY?si=IS1hVC0cMd-0CYj5
See Also
| 2024 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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. 
