2024 AMC 12A Problems/Problem 3
- The following problem is from both the 2024 AMC 12A #3 and 2024 AMC 10A #4, so both problems redirect to this page.
Problem
A square and an isosceles triangle are joined along an edge to form a pentagon 
inches tall and 
inches wide, as shown below. What is the perimeter of the pentagon, in inches?
![[asy] import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ pen GGG = grey; draw((10, 0)--(0, 0)--(0, 10)--(10, 10)); draw((10, 0)--(10, 10), dashed); draw((10, 0)--(22, 5)--(10, 10)); draw((-1.5, 0)--(-1.5, 10), arrow = ArcArrow(SimpleHead), GGG); draw((-1.5, 10)--(-1.5, 0), arrow = ArcArrow(SimpleHead), GGG); draw((0, 11.5)--(22, 11.5), arrow = ArcArrow(SimpleHead), GGG); draw((22, 11.5)--(0, 11.5), arrow = ArcArrow(SimpleHead), GGG); label("$10$ in.", (-3.5, 5), GGG); label("$22$ in.", (11, 12.75), GGG); dot((0, 0)); dot((0, 10)); dot((10, 10)); dot((10, 0)); dot((22, 5)); [/asy]](http://latex.artofproblemsolving.com/e/b/1/eb1888f5cc06dc259137a47d906ed488b3dce141.png)
Solution
Drop an altitude from the vertex of the isosceles triangle to the midpoint of the base, thereby creating two right triangles whose legs are 
and 
. It follows that the two congruent sides have length 
, hence, the perimeter of the pentagon is 
.
Video Solution by TheBeautyofMath
https://youtu.be/zaswZfIEibA?t=900
~IceMatrix
See also
| 2024 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
| 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 | |
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
