Difference between revisions of "2024 AMC 10B Problems/Problem 11"
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==Solution 2== | ==Solution 2== | ||
− | Let XM=b, ZA = a , 4 | + | Let <math>XM=b</math>, <math>ZA = a</math>, <math>4\cdot b= 8\cdot a</math>, <math>b = 2a</math>, |
<cmath>WM^2 + AM^2 = AW^2</cmath> | <cmath>WM^2 + AM^2 = AW^2</cmath> | ||
<cmath>(b^2+4^2) + (4-a)^2 + (8-b)^2 = (a^2 + 8^2)</cmath> | <cmath>(b^2+4^2) + (4-a)^2 + (8-b)^2 = (a^2 + 8^2)</cmath> | ||
Line 59: | Line 59: | ||
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
+ | ~minor edits by EaZ_Shadow | ||
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== |
Revision as of 10:14, 14 November 2024
- The following problem is from both the 2024 AMC 10B #11 and 2024 AMC 12B #7, so both problems redirect to this page.
Contents
[hide]Problem
In the figure below is a rectangle with and . Point lies , point lies on , and is a right angle. The areas of and are equal. What is the area of ?
Solution 1
We know that , , so and . Since , triangles and are similar. Therefore, , which gives . We also know that the areas of triangles and are equal, so , which implies . Substituting this into the previous equation, we get , yielding and . Thus,
Solution 2
Let , , , , a=1, b=2 ,
~luckuso ~minor edits by EaZ_Shadow
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/YqKmvSR1Ckk?feature=shared
~ Pi Academy
See also
2024 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 AMC 10 Problems and Solutions |
2024 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.