Difference between revisions of "1995 AHSME Problems/Problem 22"
(wik) |
|||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 6: | Line 6: | ||
<center><asy> | <center><asy> | ||
defaultpen(linewidth(0.7)); | defaultpen(linewidth(0.7)); | ||
| − | draw((0,0)--(31,0)--(31,25)--(12,25)--(0, | + | draw((0,0)--(31,0)--(31,25)--(12,25)--(0,20)--cycle); draw((0,20)--(0,25)--(12,25)--cycle,linetype("4 4")); |
</asy></center> | </asy></center> | ||
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 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 
and 
, although this is not necessarily their order around the pentagon. The area of the pentagon is
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]](http://latex.artofproblemsolving.com/3/c/e/3ce3d34a3db7974dc170215a2b6628242fe21d12.png)
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 and either 
must be the dimensions of the rectangle, since they are the largest lengths. With some trial-and-error, if we assign the shortest side, 
, to be the hypotenuse of the triangle, we see the 
triple. Indeed this works, by placing the side opposite from the 
side and the 
side opposite from the 
side, leaving the cutaway side to be, as before, .
To find the area of the pentagon, we subtract the area of the triangle from that of the big rectangle: 
.
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. 
