Difference between revisions of "2024 AMC 12A Problems/Problem 3"
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| − | # | + | {{duplicate|[[2024 AMC 12A Problems/Problem 3|2024 AMC 12A #3]] and [[2024 AMC 10A Problems/Problem 4|2024 AMC 10A #4]]}} |
| + | |||
| + | ==Problem== | ||
| + | A square and an isosceles triangle are joined along an edge to form a pentagon <math>10</math> inches tall and <math>22</math> 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> | ||
| + | |||
| + | <math>\textbf{(A)}~54\qquad \textbf{(B)}~56 \qquad \textbf{(C)}~62 \qquad \textbf{(D)}~64 \qquad \textbf{(E)}~66</math> | ||
| + | |||
| + | ==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 <math>\tfrac{10}{2} = 5</math> and <math>22 - 10 = 12</math>. It follows that the two congruent sides have length <math>13</math>, hence, the perimeter of the pentagon is <math>3 \cdot 10 + 2 \cdot 13 = 30 + 26 = \boxed{\textbf{(B)}~56}</math>. | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC12 box|year=2024|ab=A|num-b=2|num-a=4}} | ||
| + | {{AMC10 box|year=2024|ab=A|num-b=3|num-a=5}} | ||
| + | {{MAA Notice}} | ||
Revision as of 19:02, 21 March 2025
- 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 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
