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

(New page: ==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> <math>\text{(A)}\ ...)
 
Line 15: Line 15:
 
==Solution==
 
==Solution==
  
−
{{Solution}}
+
Note: I'm assuming all concave angles are right angles.
 +
 
 +
At first, we may immediately put down E, because of course we don't know many of the segments. But let's give it a shot.
 +
 
 +
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 b, the shorter one being a.
 +
 
 +
For the segments parallel to the side with side length 6, let's call those two segments <math>c</math> and <math>d</math>, the longer segment being d, the shorter one being c.
 +
 
 +
So the perimeter of the polygon would be...
 +
 
 +
<math>8 + 6 + a + b + c + d</math>
 +
 
 +
And it looks like we're stuck. But wait!
 +
 
 +
Notice that <math>a + b = 8</math>, and <math>c + d = 6</math>.
 +
 
 +
Now we plug those in...
 +
 
 +
<math>8 + 6 + a + b + c + d = 8 + 6 + 8 + 6 = 14 \times 2 = 28</math>
 +
 
 +
28 is C.
  
 
==See Also==
 
==See Also==
  
 
[[1986 AJHSME Problems]]
 
[[1986 AJHSME Problems]]

Revision as of 18:16, 24 January 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

Note: I'm assuming all concave angles are right angles.

At first, we may immediately put down E, because of course we don't know many of the segments. But let's give it a shot.

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...

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

And it looks like we're stuck. But wait!

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

Now we plug those in...

$8 + 6 + a + b + c + d = 8 + 6 + 8 + 6 = 14 \times 2 = 28$

28 is C.

See Also

1986 AJHSME Problems

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