Difference between revisions of "1986 AJHSME Problems/Problem 13"
LuppleAOPS (talk | contribs) (→See Also) |
m (combined two solutions) |
||
| (6 intermediate revisions by 5 users not shown) | |||
| Line 3: | Line 3: | ||
The perimeter of the polygon shown is | The perimeter of the polygon shown is | ||
| + | <center> | ||
<asy> | <asy> | ||
draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); | draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); | ||
label("$6$",(0,3),W); | label("$6$",(0,3),W); | ||
label("$8$",(4,6),N); | label("$8$",(4,6),N); | ||
| + | draw((0.5,0)--(0.5,0.5)--(0,0.5)); | ||
| + | draw((0.5,6)--(0.5,5.5)--(0,5.5)); | ||
| + | draw((7.5,6)--(7.5,5.5)--(8,5.5)); | ||
| + | draw((7.5,3)--(7.5,3.5)--(8,3.5)); | ||
| + | draw((2.2,0)--(2.2,0.5)--(2.7,0.5)); | ||
| + | draw((2.7,2.5)--(3.2,2.5)--(3.2,3)); | ||
</asy> | </asy> | ||
| + | </center> | ||
<math>\text{(A)}\ 14 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 48</math> | <math>\text{(A)}\ 14 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 48</math> | ||
| Line 15: | Line 23: | ||
==Solution== | ==Solution== | ||
| − | + | You might have not seen this coming but there is a very simple way to do this. If we try to make a rectangle out of this, we have to take out both of the lines that are taking out part of the rectangle we want to make. but now we see that to finish the rectangle, we have to use those same irregular lines! | |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
<asy> | <asy> | ||
| Line 46: | Line 33: | ||
</asy> | </asy> | ||
| − | + | So all we have to do is find the perimeter of the shape as if it would be a rectangle. After that, we get <math>\boxed{\text{(C) 28}}</math>. | |
==See Also== | ==See Also== | ||
| Line 52: | Line 39: | ||
{{AJHSME box|year=1986|num-b=12|num-a=14}} | {{AJHSME box|year=1986|num-b=12|num-a=14}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 10:01, 28 June 2023
Problem
The perimeter of the polygon shown is
![[asy] draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); label("$6$",(0,3),W); label("$8$",(4,6),N); draw((0.5,0)--(0.5,0.5)--(0,0.5)); draw((0.5,6)--(0.5,5.5)--(0,5.5)); draw((7.5,6)--(7.5,5.5)--(8,5.5)); draw((7.5,3)--(7.5,3.5)--(8,3.5)); draw((2.2,0)--(2.2,0.5)--(2.7,0.5)); draw((2.7,2.5)--(3.2,2.5)--(3.2,3)); [/asy]](http://latex.artofproblemsolving.com/f/1/9/f19e2361261c1f97b4b8c8b6958ec7f30d342c8e.png)
Solution
You might have not seen this coming but there is a very simple way to do this. If we try to make a rectangle out of this, we have to take out both of the lines that are taking out part of the rectangle we want to make. but now we see that to finish the rectangle, we have to use those same irregular lines!
![[asy] unitsize(12); draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); label("$6$",(0,3),W); label("$8$",(4,6),N); draw((8,3)--(8,0)--(2.7,0),dashed); [/asy]](http://latex.artofproblemsolving.com/9/c/2/9c2c60ec3f042b22a74ebe21e5a64a6233dfa214.png)
So all we have to do is find the perimeter of the shape as if it would be a rectangle. After that, we get 
.
See Also
| 1986 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
