Difference between revisions of "2024 AMC 10B Problems/Problem 11"
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+ | ==Solution 4 (Similar Triangles)== | ||
+ | We are given <math>WX = 4</math>, <math>WZ = 8</math>. △ WXM and △ MYA have equal area, so let <math>XM = 2x</math> and <math>ZA = x</math>. <math>MY = 8-2x</math> and <math>AY = 4-x</math>. | ||
+ | From this, we can conclude that <math>\frac{MY}{AY} = \frac{8-2x}{4-x} = \frac{2}{1}</math> | ||
+ | |||
+ | Since <math>WM</math> intersects parallel lines <math>WZ</math> and <math>XY</math>, <math>\angle ZWM = \angle WMZ</math>. <math>\angle ZWM + \angle MWX = 90^\circ</math>, so <math>180^\circ - 90^\circ = \angle WMZ + \angle AMY</math>. Thus, <math>\angle MWX = \angle AMY</math> and △ WXM ~ △ MYA due to AA Similarity. | ||
+ | |||
+ | Corresponding sides of similar triangles are proportional, so <math>\frac{WX}{XM} = \frac{MY}{AY}</math> or <math>\frac{4}{2x} = \frac{2}{1}</math>. It is clear that <math>2x = 2</math>, and <math>x = 1</math>. Now, all we have to do is subtract the area of the rectangle by each of the three triangles. | ||
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+ | |||
+ | △ WMA = <math>8</math> · <math>4</math> - (<math>\frac{1}{2}</math> · <math>4</math> · <math>2</math>) - (<math>\frac{1}{2}</math> · <math>8</math> · <math>1</math>) - (<math>\frac{1}{2}</math> · <math>6</math> · <math>3</math>) | ||
+ | |||
+ | △ WMA = <math>32 - 4 - 4 - 9</math> | ||
+ | |||
+ | △ WMA = <math>\boxed{\textbf{(C) }15}</math> | ||
+ | |||
+ | ~peeghj | ||
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== |
Latest revision as of 21:39, 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:
~pepper2831
Solution 4 (Similar Triangles)
We are given , . △ WXM and △ MYA have equal area, so let and . and . From this, we can conclude that
Since intersects parallel lines and , . , so . Thus, and △ WXM ~ △ MYA due to AA Similarity.
Corresponding sides of similar triangles are proportional, so or . It is clear that , and . Now, all we have to do is subtract the area of the rectangle by each of the three triangles.
△ WMA = · - ( · · ) - ( · · ) - ( · · )
△ WMA =
△ WMA =
~peeghj
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.