Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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)
 
(One intermediate revision by one other user not shown)
Line 10: Line 10:
 
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 ==
Line 16: Line 16:
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 16:16, 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1995_AHSME_Problems/Problem_22&oldid=158495"