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

Revision as of 15:57, 19 November 2020

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

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! So all we have to do is find the perimeter of the shape as if it would be a rectangle. After that, we get $\boxed{\text{C 28}}$ .

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]$

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

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.

@@ Line 25: / Line 25: @@
 ===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>.
+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! 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>.
-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 <math>d</math>, the shorter one being <math>c</math>.
-So the perimeter of the polygon would be...
-<math>8 + 6 + a + b + c + d</math>
-Note that <math>a + b = 8</math>, and <math>c + d = 6</math>.
-Now we plug those in:
-<cmath>\begin{align*}
-+ 6 + a + b + c + d &= 8 + 6 + 8 + 6 \\
-&= 14 \times 2 \\
-&= 28 \\
-\end{align*}</cmath>
-is <math>\boxed{\text{C}}</math>.
 ===Solution 2===

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