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

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==Problem 3 ==
 
==Problem 3 ==
 
Four squares of side length <math>4, 7, 9,</math> and <math>10</math> 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?
 
Four squares of side length <math>4, 7, 9,</math> and <math>10</math> 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>
+
<asy>  
−
        size(150);
+
size(150);
−
        filldraw((0,0)--(10,0)--(10,10)--(0,10)--cycle,gray(0.7),linewidth(1));
+
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)--(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>
 +
<math>\textbf{(A)}\ 42 \qquad \textbf{(B)}\ 45\qquad \textbf{(C)}\ 49\qquad \textbf{(D)}\ 50\qquad \textbf{(E)}\ 52</math>
  
−
        filldraw((0,0)--(7,0)--(7,7)--(0,7)--cycle,gray(0.7),linewidth(1));
+
==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
 +
<cmath>10^2 - 9^2 + 7^2 - 4^2 = 19 + 33 = \boxed{\textbf{(E)}\ 52}</cmath>
  
−
        filldraw((0,0)--(4,0)--(4,4)--(0,4)--cycle,white,linewidth(1));
+
This problem appears multiple times in various math competitions including the AMC and MATHCOUNTS.
  
−
        draw((11,0)--(11,4),linewidth(1));
+
-Benedict (countmath1) ~Nivaar
−
        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));
+
==Solution 2==
−
        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));
+
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.  
−
        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>
 
−
    <math>\textbf{(A)}\ 42 \qquad \textbf{(B)}\ 45\qquad \textbf{(C)}\ 49\qquad \textbf{(D)}\ 50\qquad \textbf{(E)}\ 52</math>
 
  
−
==Solution 1==  
+
-MrTechguyPCMT
 +
 
 +
==Solution 3==  
 +
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>
 +
 
 +
-SahanWijetunga
 +
 
 +
==Video Solution (A Clever Explanation You’ll Get Instantly)==
 +
https://youtu.be/5ZIFnqymdDQ?si=vSpf507m-7QMFkbZ&t=238
 +
 
 +
~hsnacademy
  
−
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
+
==Video Solution by Math-X (First fully understand the problem!!!)==
−
    <cmath>10^2 - 9^2 + 7^2 - 4^2 = 19 + 33 = \boxed{\textbf{(E)}\ 52}</cmath>
+
https://youtu.be/BaE00H2SHQM?si=F37qz7vyfRgZm-1u&t=469
  
−
-Benedict T (countmath1)
+
~Math-X
  
−
==Video Solution 1 (easy to digest) by Power Solve==
+
==Video Solution (easy to digest) by Power Solve==
 
https://youtu.be/HE7JjZQ6xCk?si=39xd5CKI9nx-7lyV&t=118
 
https://youtu.be/HE7JjZQ6xCk?si=39xd5CKI9nx-7lyV&t=118
 +
 +
==Video Solution  by Parshwa==
 +
https://www.youtube.com/watch?v=DQbe_BNfPYw
 +
 +
==Video Solution by NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=Ylw-kJkSpq8
 +
 +
~NiuniuMaths
 +
 +
==Video Solution  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
 +
 +
==Video Solution by Daily Dose of Math (Understandable, Quick, and Speedy)==
 +
 +
https://youtu.be/bSPWqeNO11M?si=HIzlxPjMfvGM5lxR
 +
 +
~Thesmartgreekmathdude
 +
 +
==See Also==
 +
{{AMC8 box|year=2024|num-b=2|num-a=4}}
 +
{{MAA Notice}}

Revision as of 19:33, 28 August 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 3

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 (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=vSpf507m-7QMFkbZ&t=238

~hsnacademy

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

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

~Math-X

Video Solution (easy to digest) by Power Solve

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

Video Solution by Parshwa

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

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=Ylw-kJkSpq8

~NiuniuMaths

Video Solution 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

Video Solution by Daily Dose of Math (Understandable, Quick, and Speedy)

https://youtu.be/bSPWqeNO11M?si=HIzlxPjMfvGM5lxR

~Thesmartgreekmathdude

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. AMC logo.png

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