Difference between revisions of "2024 AMC 8 Problems/Problem 24"
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<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 4\sqrt{2} \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 5\sqrt{2}</math> | <math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 4\sqrt{2} \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 5\sqrt{2}</math> | ||
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| + | ==Video Solution- (Extremely Simple Triangle Theories!!)== | ||
| + | https://www.youtube.com/watch?v=3UC1rwf6vfo | ||
| + | ~TheMathGeek | ||
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==Solution1== | ==Solution1== | ||
Revision as of 16:36, 20 December 2024
Contents
- 1 Problem
- 2 Video Solution- (Extremely Simple Triangle Theories!!)
- 3 Solution1
- 4 Solution 2 (Using a Ruler)
- 5 Video Solution 1 by Math-X (First understand the problem!!!)
- 6 Video Solution (A Clever Explanation You’ll Get Instantly)
- 7 Video Solution (Quick solution) by Parshwa
- 8 Video Solution by Power Solve (easy to understand)
- 9 Video Solution by OmegaLearn.org
- 10 Video Solution by SpreadTheMathLove
- 11 Video Solution by NiuniuMaths (Easy to understand!)
- 12 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 13 Video Solution by Interstigation
- 14 2 minute solve (fast and elegant) by MegaMath
- 15 Video solution by Dr. David
- 16 Video Solution by WhyMath
- 17 See Also
Problem
Jean has made a piece of stained glass art in the shape of two mountains, as shown in the figure below. One mountain peak is 
feet high while the other peak is 
feet high. Each peak forms a 
angle, and the straight sides form a 
angle with the ground. The artwork has an area of 
square feet. The sides of the mountain meet at an intersection point near the center of the artwork, 
feet above the ground. What is the value of 
![[asy] unitsize(.3cm); filldraw((0,0)--(8,8)--(11,5)--(18,12)--(30,0)--cycle,gray(0.7),linewidth(1)); draw((-1,0)--(-1,8),linewidth(.75)); draw((-1.4,0)--(-.6,0),linewidth(.75)); draw((-1.4,8)--(-.6,8),linewidth(.75)); label("$8$",(-1,4),W); label("$12$",(31,6),E); draw((-1,8)--(8,8),dashed); draw((31,0)--(31,12),linewidth(.75)); draw((30.6,0)--(31.4,0),linewidth(.75)); draw((30.6,12)--(31.4,12),linewidth(.75)); draw((31,12)--(18,12),dashed); label("$45^{\circ}$",(.75,0),NE,fontsize(10pt)); label("$45^{\circ}$",(29.25,0),NW,fontsize(10pt)); draw((8,8)--(7.5,7.5)--(8,7)--(8.5,7.5)--cycle); draw((18,12)--(17.5,11.5)--(18,11)--(18.5,11.5)--cycle); draw((11,5)--(11,0),dashed); label("$h$",(11,2.5),E); [/asy]](http://latex.artofproblemsolving.com/0/6/2/062be4008fc995e68ebbe56afa0a90b07160a7d6.png)
Video Solution- (Extremely Simple Triangle Theories!!)
https://www.youtube.com/watch?v=3UC1rwf6vfo ~TheMathGeek
Solution1
Extend the "inner part" of the mountain so that the image is two right triangles that overlap in a third right triangle as shown.
![[asy] unitsize(.2cm); draw((0,0)--(8,8)--(11,5)--(18,12)--(30,0)--cycle,linewidth(1)); draw((11,5)--(6, 0)--(16, 0)--cycle,linewidth(0.5)); label("$8\sqrt{2}$",(4,4),NW); label("$12\sqrt{2}$",(24,6),NE); draw((8,8)--(7.5,7.5)--(8,7)--(8.5,7.5)--cycle); draw((18,12)--(17.5,11.5)--(18,11)--(18.5,11.5)--cycle); draw((11,5)--(11,0),dashed); label("$h$",(11,2.5),E); [/asy]](http://latex.artofproblemsolving.com/8/8/b/88bf4e27a470961cc074edc3adc9d53475fc71e8.png)
The side length of the largest right triangle is 
which means its area is 
Similarly, the area of the second largest right triangle is 
(the side length is 
), and the area of the overlap is 
(the side length is 
). Thus,
![\[144+64-h^2=183,\]](http://latex.artofproblemsolving.com/d/a/3/da33465ebae23a898b76506cd66992f6ff48089f.png)
which means that the answer is 
~BS201
Solution 2 (Using a Ruler)
You can measure with a ruler(rulers are allowed on the AMC 8), and see that is closest to
~SID-DARTH-VATER
(Please note: This is not the best way, as diagrams are NOT DRAWN TO SCALE) ~Anonymous ~Continuous_Pi
Video Solution 1 by Math-X (First understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=793ml6R_qT79h-aI&t=7651
~Math-X
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/5ZIFnqymdDQ?si=D8pGHQYbzJ_eMsEy&t=3678 ~hsnacademy
Video Solution (Quick solution) by Parshwa
Video Solution by Power Solve (easy to understand)
https://www.youtube.com/watch?v=PlMGNmWIkBg
Video Solution by OmegaLearn.org
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=RiSt6_WLfrM
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=8353ngt05N0
~NiuniuMaths
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=H5Pq8mf-OVk
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=3126
2 minute solve (fast and elegant) by MegaMath
https://www.youtube.com/watch?v=hUh0hux3xuU
Video solution by Dr. David
Video Solution by WhyMath
See Also
| 2024 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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. 
