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

Revision as of 14:47, 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 $2(6+8)=28\rightarrow \boxed{\text{C}}$

1986 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AJHSME/AMC 8 Problems and Solutions

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