2023 AMC 8 Problems/Problem 24
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Thorough)
- 4 Solution 3 (Faster)
- 5 Solution 4 (Slower)
- 6 Video Solution 1 by OmegaLearn (Using Similarity)
- 7 Video Solution 2 by SpreadTheMathLove(Using Area-Similarity Relaitionship)
- 8 Video Solution 3 by Magic Square (Using Similarity and Special Value)
- 9 Video Solution by Interstigation
- 10 Video Solution by WhyMath
- 11 Video Solution by harungurcan
- 12 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 ?
(note: diagrams are not necessarily drawn to scale)
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 ILoveMath31415926535
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
Solution 4 (Slower)
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
Video Solution 1 by OmegaLearn (Using Similarity)
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)
https://www.youtube.com/watch?v=-N46BeEKaCQ&t=1569s
Video Solution by Interstigation
https://youtu.be/1bA7fD7Lg54?t=2454
Video Solution by WhyMath
~savannahsolver
Video Solution by harungurcan
https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1593s
~harungurcan
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 |
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