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Difference between revisions of "1995 AHSME Problems/Problem 22"

m (changed diagram from having sides of lengths 13, 19, 5, 25, 31 to 13, 19, 20, 25, 31 as stated in the problem)
 
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Since the pentagon is cut from a rectangle, the cut-off triangle must be right. Since all of the lengths given are integers, it follows that this triangle is a [[Pythagorean Triple]]. We know that <math>31</math> and either <math>25,\, 20</math> must be the dimensions of the rectangle, since they are the largest lengths. With some trial-and-error, if we assign the shortest side, <math>13</math>, to be the [[hypotenuse]] of the triangle, we see the <math>5-12-13</math> triple. Indeed this works, by placing the <math>31</math> side opposite from the <math>19</math> side and the <math>25</math> side opposite from the <math>20</math> side, leaving the cutaway side to be, as before, <math>13</math>.
 
Since the pentagon is cut from a rectangle, the cut-off triangle must be right. Since all of the lengths given are integers, it follows that this triangle is a [[Pythagorean Triple]]. We know that <math>31</math> and either <math>25,\, 20</math> must be the dimensions of the rectangle, since they are the largest lengths. With some trial-and-error, if we assign the shortest side, <math>13</math>, to be the [[hypotenuse]] of the triangle, we see the <math>5-12-13</math> triple. Indeed this works, by placing the <math>31</math> side opposite from the <math>19</math> side and the <math>25</math> side opposite from the <math>20</math> side, leaving the cutaway side to be, as before, <math>13</math>.
  
−
To find the area of the pentagon, we subtract the area of the triangle from that of the big rectangle: <math>31\cdot 25-\frac{12\cdot5}{2}=775-30=745\Longrightarrow \mathrm{(E)}</math>.
+
To find the area of the pentagon, we subtract the area of the triangle from that of the big rectangle: <math>31\cdot 25-\frac{12\cdot5}{2}=775-30=745\Longrightarrow \boxed{\mathrm{(E)}745}</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 15:16, 14 July 2021

Problem

A pentagon is formed by cutting a triangular corner from a rectangular piece of paper. The five sides of the pentagon have lengths $13, 19, 20, 25$ and $31$, although this is not necessarily their order around the pentagon. The area of the pentagon is

$\mathrm{(A) \ 459 } \qquad \mathrm{(B) \ 600 } \qquad \mathrm{(C) \ 680 } \qquad \mathrm{(D) \ 720 } \qquad \mathrm{(E) \ 745 }$

Solution

[asy] defaultpen(linewidth(0.7)); draw((0,0)--(31,0)--(31,25)--(12,25)--(0,20)--cycle); draw((0,20)--(0,25)--(12,25)--cycle,linetype("4 4")); [/asy]

Since the pentagon is cut from a rectangle, the cut-off triangle must be right. Since all of the lengths given are integers, it follows that this triangle is a Pythagorean Triple. We know that $31$ and either $25,\, 20$ must be the dimensions of the rectangle, since they are the largest lengths. With some trial-and-error, if we assign the shortest side, $13$, to be the hypotenuse of the triangle, we see the $5-12-13$ triple. Indeed this works, by placing the $31$ side opposite from the $19$ side and the $25$ side opposite from the $20$ side, leaving the cutaway side to be, as before, $13$.

To find the area of the pentagon, we subtract the area of the triangle from that of the big rectangle: $31\cdot 25-\frac{12\cdot5}{2}=775-30=745\Longrightarrow \boxed{\mathrm{(E)}745}$.

See also

1995 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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