Difference between revisions of "2023 AMC 8 Problems/Problem 24"
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==Solution 1== | ==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 grey part in the first triangle is <math>[\text{ABC}]\cdot\left(1-\left(\frac{11}{h}\right)^2\right)</math>. IN PROGRESS | |
~MathFun1000 | ~MathFun1000 | ||
Revision as of 22:19, 24 January 2023
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 ?
- Add asymptote diagram*
(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 grey part in the first triangle is ![$[\text{ABC}]\cdot\left(1-\left(\frac{11}{h}\right)^2\right)$](http://latex.artofproblemsolving.com/5/1/4/51419272f759f77aec5ffe7b8846262198b1d432.png)
. IN PROGRESS
~MathFun1000
Video Solution 1 by OmegaLearn (Using Similarity)
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. 
