2023 AMC 8 Problems/Problem 24
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Thorough)
- 4 Solution 3 (Faster)
- 5 Solution 4(Based on solution 3) (VERY SLOW AND SHOULD ONLY BE USED AS LAST RESORT)
- 6 Video Solution (Solve under 60 seconds!!!)
- 7 Video Solution by Math-X (First understand the problem!!!)
- 8 Video Solution (THINKING CREATIVELY!!!)
- 9 Video Solution 1 by OmegaLearn (Using Similar Triangles)
- 10 Video Solution 2 by SpreadTheMathLove(Using Area-Similarity Relaitionship)
- 11 Video Solution 3 by Magic Square (Using Similarity and Special Value)(best solution)
- 12 Video Solution by Interstigation
- 13 Video Solution by WhyMath
- 14 Video Solution by harungurcan
- 15 Video Solution by Dr. David
- 16 See Also
Problem
Isosceles has equal side lengths and . In the figure below, segments are drawn parallel to so that the shaded portions of have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of of ?
Solution 1
First, we notice that the smaller isosceles triangles are similar to the larger isosceles triangles. We can find that the area of the gray area in the first triangle is . Similarly, we can find that the area of the gray part in the second triangle is . These areas are equal, so . Simplifying yields so .
~MathFun1000 (~edits apex304)
Solution 2 (Thorough)
We can call the length of AC as . Therefore, the length of the base of the triangle with height is . Therefore, the base of the smaller triangle is . We find that the area of the trapezoid is .
Using similar triangles once again, we find that the base of the shaded triangle is . Therefore, the area is .
Since the areas are the same, we find that . Multiplying each side by , we get . Therefore, we can subtract from both sides, and get . Finally, we divide both sides by and get . is .
Solution by CHECKMATE2021
Solution 3 (Faster)
Since the length of AC does not matter, we can assume the base of triangle ABC is . Therefore, the area of the trapezoid in the first diagram is .
The area of the triangle in the second diagram is now .
Therefore, . Multiplying both sides by , we get . Subtracting from both sides, we get and is .
Solution by ILoveMath31415926535, and CHECKMATE2021
Solution 4(Based on solution 3) (VERY SLOW AND SHOULD ONLY BE USED AS LAST RESORT)
The answers are there on the bottom, so start with the middle one, . After calculating, we find that we need a shorter length, so try . Still, we need a shorter answer, so we simply choose without trying it out.
~SaxStreak (~edits by KMSONI)
Video Solution (Solve under 60 seconds!!!)
https://youtu.be/6O5UXi-Jwv4?si=LIt75cZMdNbGeWMB&t=1116
~hsnacademy
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/Ku_c1YHnLt0?si=NHwA1x9STOJLG8IP&t=5349
~Math-X
~please like and subscribe
Video Solution (THINKING CREATIVELY!!!)
~Education, the Study of Everything
Video Solution 1 by OmegaLearn (Using Similar Triangles)
Video Solution 2 by SpreadTheMathLove(Using Area-Similarity Relaitionship)
https://www.youtube.com/watch?v=GTlkTwxSxgo
Video Solution 3 by Magic Square (Using Similarity and Special Value)(best solution)
https://www.youtube.com/watch?v=-N46BeEKaCQ&t=1569s
Video Solution by Interstigation
https://youtu.be/DBqko2xATxs&t=3270
Video Solution by WhyMath
~savannahsolver
Video Solution by harungurcan
https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1593s
~harungurcan
Video Solution by Dr. David
See Also
2023 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.