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 "1986 AJHSME Problems/Problem 13"

m (→‎Solution)
m
Line 14: Line 14:
  
 
==Solution==
 
==Solution==
 +
 +
===Solution 1===
  
 
For the segments parallel to the side with side length 8, let's call those two segments <math>a</math> and <math>b</math>, the longer segment being <math>b</math>, the shorter one being <math>a</math>.
 
For the segments parallel to the side with side length 8, let's call those two segments <math>a</math> and <math>b</math>, the longer segment being <math>b</math>, the shorter one being <math>a</math>.
Line 33: Line 35:
  
 
28 is <math>\boxed{\text{C}}</math>.
 
28 is <math>\boxed{\text{C}}</math>.
 +
 +
===Solution 2===
 +
 +
<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>
 +
 +
Clearly the perimeter of the requested region is the same as the perimeter of the rectangle with the dashed portion.  This makes the answer obviously <math>2(6+8)=28\rightarrow \boxed{\text{C}}</math>
  
 
==See Also==
 
==See Also==
  
−
[[1986 AJHSME Problems]]
+
{{AJHSME box|year=1986|num-b=12|num-a=14}}
 +
[[Category:Introductory Geometry Problems]]

Revision as of 20:07, 22 May 2009

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); [/asy]

$\text{(A)}\ 14 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 48$

$\text{(E)}\ \text{cannot be determined from the information given}$

Solution

Solution 1

For the segments parallel to the side with side length 8, let's call those two segments $a$ and $b$, the longer segment being $b$, the shorter one being $a$.

For the segments parallel to the side with side length 6, let's call those two segments $c$ and $d$, the longer segment being $d$, the shorter one being $c$.

So the perimeter of the polygon would be...

$8 + 6 + a + b + c + d$

Note that $a + b = 8$, and $c + d = 6$.

Now we plug those in: \begin{align*} 8 + 6 + a + b + c + d &= 8 + 6 + 8 + 6 \\ &= 14 \times 2 \\ &= 28 \\ \end{align*}

28 is $\boxed{\text{C}}$.

Solution 2

[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]

Clearly the perimeter of the requested region is the same as the perimeter of the rectangle with the dashed portion. This makes the answer obviously $2(6+8)=28\rightarrow \boxed{\text{C}}$

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1986_AJHSME_Problems/Problem_13&oldid=31804"