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Difference between revisions of "1986 AJHSME Problems/Problem 13"

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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);
Line 14: Line 15:
 
draw((2.7,2.5)--(3.2,2.5)--(3.2,3));
 
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>

Revision as of 14:48, 23 May 2010

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]

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

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

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
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