Difference between revisions of "2020 AMC 8 Problems/Problem 25"
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− | Assuming that the problem is well-posed, it should be true in the particular case where <math>S_1 \cong S_3</math> and <math>R_1 \cong R_2</math>. Let the sum of the side lengths of <math>S_1</math> and <math>S_3</math> be <math>x</math>, and let the length of square <math>S_2</math> be <math>y</math>. We then have the system <cmath>\begin{dcases}x+y =3322 \\x-y=2020\end{dcases}</cmath> which we solve to determine <math>y=\boxed{\textbf{(A) }651}</math>. | + | Assuming that the problem is well-posed, it should be true in the particular case where <math>S_1 \cong S_3</math> and <math>R_1 \cong R_2</math>. why? Let the sum of the side lengths of <math>S_1</math> and <math>S_3</math> be <math>x</math>, and let the length of square <math>S_2</math> be <math>y</math>. We then have the system <cmath>\begin{dcases}x+y =3322 \\x-y=2020\end{dcases}</cmath> which we solve to determine <math>y=\boxed{\textbf{(A) }651}</math>. |
==Solution 3 (faster version of Solution 1)== | ==Solution 3 (faster version of Solution 1)== |
Revision as of 16:09, 2 December 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (faster version of Solution 1)
- 5 Solution 4
- 6 Video Solution by Math-X (First understand the problem!!!)
- 7 Video Solution(🚀Just 1 min🚀)
- 8 Video Solution
- 9 Video Solution by OmegaLearn
- 10 Video Solution by WhyMath
- 11 Video Solution by The Learning Royal
- 12 Video Solution by Interstigation
- 13 Video Solution by STEMbreezy
- 14 See also
Problem
Rectangles and and squares and shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of in units?
Solution 1
Let the side length of each square be . Then, from the diagram, we can line up the top horizontal lengths of , , and to cover the top side of the large rectangle, so . Similarly, the short side of will be , and lining this up with the left side of to cover the vertical side of the large rectangle gives . We subtract the second equation from the first to obtain , and thus .
Solution 2
Assuming that the problem is well-posed, it should be true in the particular case where and . why? Let the sum of the side lengths of and be , and let the length of square be . We then have the system which we solve to determine .
Solution 3 (faster version of Solution 1)
Since, for each pair of rectangles, the side lengths have a sum of or and a difference of , the answer must be .
Solution 4
Let the side length of be s, and the shorter side length of and be . We have
From this diagram, it is evident that . Also, the side length of and is . Then, . Now, we have 2 systems of equations.
We can see an in the 2nd equation, so substituting that in gives us .
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/UnVo6jZ3Wnk?si=ryNc6kwFiy7YkEbc&t=6127
~Math-X
Video Solution(🚀Just 1 min🚀)
~Education, the Study of Everything
Video Solution
Please like and subscribe!
https://www.youtube.com/watch?v=gJXMZq2Rbwg ~David
Video Solution by OmegaLearn
https://youtu.be/jhJifWaoUI8?t=441
~ pi_is_3.14
Video Solution by WhyMath
~savannahsolver
Video Solution by The Learning Royal
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=1639
~Interstigation
Video Solution by STEMbreezy
https://youtu.be/wq8EUCe5oQU?t=588
~STEMbreezy
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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.