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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]]}}
| | #redirect[[2024 AMC 10A Problems/Problem 4]] |
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| ==Problem==
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| 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?
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| <asy>
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| import graph; size(7cm);
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| real labelscalefactor = 0.5; /* changes label-to-point distance */
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| pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
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| pen dotstyle = black; /* point style */
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| pen GGG = grey;
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| draw((10, 0)--(0, 0)--(0, 10)--(10, 10));
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| draw((10, 0)--(10, 10), dashed);
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| draw((10, 0)--(22, 5)--(10, 10));
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| draw((-1.5, 0)--(-1.5, 10), arrow = ArcArrow(SimpleHead), GGG);
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| draw((-1.5, 10)--(-1.5, 0), arrow = ArcArrow(SimpleHead), GGG);
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| draw((0, 11.5)--(22, 11.5), arrow = ArcArrow(SimpleHead), GGG);
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| draw((22, 11.5)--(0, 11.5), arrow = ArcArrow(SimpleHead), GGG);
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| label("$10$ in.", (-3.5, 5), GGG);
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| label("$22$ in.", (11, 12.75), GGG);
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| dot((0, 0));
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| dot((0, 10));
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| dot((10, 10));
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| dot((10, 0));
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| dot((22, 5));
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| </asy>
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| <math>\textbf{(A)}~54\qquad \textbf{(B)}~56 \qquad \textbf{(C)}~62 \qquad \textbf{(D)}~64 \qquad \textbf{(E)}~66</math>
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| ==Solution==
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| 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>.
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| ==Video Solution by TheBeautyofMath==
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| https://youtu.be/zaswZfIEibA?t=900
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| ~IceMatrix
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| ==See also==
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| {{AMC12 box|year=2024|ab=A|num-b=2|num-a=4}}
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| {{AMC10 box|year=2024|ab=A|num-b=3|num-a=5}}
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| {{MAA Notice}}
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