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Difference between revisions of "2024 AMC 8 Problems/Problem 3"

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Solve for the areas for each square inside the larger one , and then substract the areas of the white regions from the areas of each square.
 
Solve for the areas for each square inside the larger one , and then substract the areas of the white regions from the areas of each square.
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-MrTechguyPCMT
 
-MrTechguyPCMT
  

Revision as of 19:54, 8 February 2024

Problem 3

Four squares of side length $4, 7, 9,$ and $10$ 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)--(3,11),linewidth(1)); draw((5,11)--(9,11),linewidth(1)); draw((0,11.25)--(0,10.75),linewidth(1)); draw((9,11.25)--(9,10.75),linewidth(1)); label("$9$",(4,11),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] $\textbf{(A)}\ 42 \qquad \textbf{(B)}\ 45\qquad \textbf{(C)}\ 49\qquad \textbf{(D)}\ 50\qquad \textbf{(E)}\ 52$

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 \[10^2 - 9^2 + 7^2 - 4^2 = 19 + 33 = \boxed{\textbf{(E)}\ 52}\]

This problem appears multiple times in various math competitions including the AMC and MATHCOUNTS.

-Benedict (countmath1) ~Nivaar


Solution 2

Solve for the areas for each square inside the larger one , and then substract the areas of the white regions from the areas of each square.

-MrTechguyPCMT

Solution 2

We can calculate it as the sum of the areas of $2$ smaller trapezoids and $2$ larger trapezoids. \[2\left(\cfrac{(7+4)(7-4)}{2}\right)+2\left(\cfrac{(10+9)(10-9)}{2}\right)=10^2 - 9^2 + 7^2 - 4^2 = 19 + 33 = \boxed{\textbf{(E)}\ 52}\]

-SahanWijetunga

Video Solution 1 (easy to digest) by Power Solve

https://youtu.be/HE7JjZQ6xCk?si=39xd5CKI9nx-7lyV&t=118

Video Solution 2 by Parshwa

https://www.youtube.com/watch?v=DQbe_BNfPYw

Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=F37qz7vyfRgZm-1u&t=469

~Math-X

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

Video Solution by Interstigation

https://youtu.be/ktzijuZtDas&t=158

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

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