Difference between revisions of "2024 AMC 12A Problems/Problem 3"

(Removed redirect to 2024 AMC 10A Problems/Problem 4)
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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]]}}
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#redirect[[2024 AMC 10A Problems/Problem 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 19:01, 21 March 2025