Difference between revisions of "1986 AJHSME Problems/Problem 13"
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==Solution== | ==Solution== | ||
− | + | 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 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>. | |
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− | For the segments parallel to the side with side length | ||
− | |||
− | |||
So the perimeter of the polygon would be... | So the perimeter of the polygon would be... | ||
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<math>8 + 6 + a + b + c + d</math> | <math>8 + 6 + a + b + c + d</math> | ||
− | + | Note that <math>a + b = 8</math>, and <math>c + d = 6</math>. | |
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− | <math>8 + 6 + a + b + c + d = 8 + 6 + 8 + 6 = 14 \times 2 = 28< | + | Now we plug those in: |
+ | <math></math>\begin{align*} | ||
+ | 8 + 6 + a + b + c + d &= 8 + 6 + 8 + 6 \\ | ||
+ | &= 14 \times 2 \\ | ||
+ | &= 28 \\ | ||
+ | \end{align*}<math> | ||
− | 28 is C. | + | 28 is </math>\boxed{\text{C}}$. |
==See Also== | ==See Also== | ||
[[1986 AJHSME Problems]] | [[1986 AJHSME Problems]] |
Revision as of 19:24, 24 January 2009
Problem
The perimeter of the polygon shown is
Solution
For the segments parallel to the side with side length 8, let's call those two segments and
, the longer segment being
, the shorter one being
.
For the segments parallel to the side with side length 6, let's call those two segments and
, the longer segment being
, the shorter one being
.
So the perimeter of the polygon would be...
Note that , and
.
Now we plug those in:
$$ (Error compiling LaTeX. Unknown error_msg)\begin{align*}
8 + 6 + a + b + c + d &= 8 + 6 + 8 + 6 \\
&= 14 \times 2 \\
&= 28 \\
\end{align*}\boxed{\text{C}}$.