Difference between revisions of "2024 AMC 10B Problems/Problem 11"
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~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
~minor edits by EaZ_Shadow | ~minor edits by EaZ_Shadow | ||
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+ | ==Solution 3 (Pythagorean Theorem) == | ||
+ | Assign ZA as <math>x</math>, then AY as <math>4 - x</math>. Assign XM as <math>y</math> and MY as <math>8 - y</math>. Since triangles WXM and WZA are together, we can say <math>4x = 8y</math>, so <math>y = 2x</math>. Then therefore, XM is <math>2x</math> and MY has length <math>8 - 2x</math>. We can use the Pythagorean theorem to find WM, which is actually <math>\sqrt{(2x)^2 + 4^2)} = \sqrt{4x^2 + 16}</math>. We don't factor it yet - we are going to find <math>x</math> again using the Pythagorean Theorem. Similarly, finding MA is just the square root of the squares of AY and MY individually, or <math>\sqrt{(8 - 2x)^2 + (4 - x)^2} = \sqrt{64 - 32x + 4x^2 + 16 - 8x + x^2} = \sqrt{5x^2 - 40x + 80}</math>. Then simply, WA is really <math>\sqrt{x^2 + 64}</math>. | ||
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+ | Now we have the three sides of the right triangle: <math>\sqrt{4x^2 + 16}</math>, <math>\sqrt{5x^2 - 40x + 80}</math>, and <math>\sqrt{x^2 + 64}</math>. Per the Pythagorean theorem again, we can see <math>(4x^2 + 16) + (5x^2 - 40x + 80) = (x^2 + 64)</math>. Combining like terms gives us <math>8x^2 - 40x + 32 = 0</math>, then dividing by 8 gives <math>x^2 - 5x + 4 = 0</math>. As this elementary and well-known quadratic gives us the roots of <math>1</math> and <math>4</math>, we can see it is a bit weird to have <math>x = 4</math>, as then point Z is point A. So we'll assume <math>x = 1</math>. We have two legs of the triangle by plugging in the sides with x in them, given that <math>x = 1</math>: <math>\sqrt{20}</math> and <math>\sqrt{45}</math>. We should know that <math>20 \cdot 45 = 900</math>, and <math>\sqrt{900} = 30.</math> Dividing by 2 reveals us our answer: \boxed{\textbf{(C) }15} | ||
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+ | ~pepper2831 | ||
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== |
Revision as of 20:46, 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
Solution 3 (Pythagorean Theorem)
Assign ZA as , then AY as . Assign XM as and MY as . Since triangles WXM and WZA are together, we can say , so . Then therefore, XM is and MY has length . We can use the Pythagorean theorem to find WM, which is actually . We don't factor it yet - we are going to find again using the Pythagorean Theorem. Similarly, finding MA is just the square root of the squares of AY and MY individually, or . Then simply, WA is really .
Now we have the three sides of the right triangle: , , and . Per the Pythagorean theorem again, we can see . Combining like terms gives us , then dividing by 8 gives . As this elementary and well-known quadratic gives us the roots of and , we can see it is a bit weird to have , as then point Z is point A. So we'll assume . We have two legs of the triangle by plugging in the sides with x in them, given that : and . We should know that , and Dividing by 2 reveals us our answer: \boxed{\textbf{(C) }15}
~pepper2831
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/YqKmvSR1Ckk?feature=shared
~ Pi Academy
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=24EZaeAThuE
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.