Difference between revisions of "1999 AMC 8 Problems/Problem 14"
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==Solution== | ==Solution== | ||
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<asy> | <asy> | ||
draw((0,0)--(4,3)--(12,3)--(16,0)--cycle); | draw((0,0)--(4,3)--(12,3)--(16,0)--cycle); | ||
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There is a rectangle present, with both horizontal bases being <math>8</math> units in length. The excess units on the bottom base must then be <math>16-8=8</math>. The fact that <math>AB</math> and <math>CD</math> are equal in length indicate, by the Pythagorean Theorem, that these excess lengths are equal. There are two with a total length of <math>8</math> units, so each is <math>4</math> units. The triangle has a hypotenuse of <math>5</math>, because the triangles are <math>3-4-5</math> right triangles. So, the sides of the trapezoid are <math>8</math>, <math>5</math>, <math>16</math>, and <math>5</math>. Adding those up gives us the perimeter, <math>8 + 5 + 16 + 5 = 13 + 21 = \boxed{\text{(D)}\ 34}</math> units. | There is a rectangle present, with both horizontal bases being <math>8</math> units in length. The excess units on the bottom base must then be <math>16-8=8</math>. The fact that <math>AB</math> and <math>CD</math> are equal in length indicate, by the Pythagorean Theorem, that these excess lengths are equal. There are two with a total length of <math>8</math> units, so each is <math>4</math> units. The triangle has a hypotenuse of <math>5</math>, because the triangles are <math>3-4-5</math> right triangles. So, the sides of the trapezoid are <math>8</math>, <math>5</math>, <math>16</math>, and <math>5</math>. Adding those up gives us the perimeter, <math>8 + 5 + 16 + 5 = 13 + 21 = \boxed{\text{(D)}\ 34}</math> units. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/DKoTd9S4y-Q Soo, DRMS, NM | ||
==See Also== | ==See Also== |
Latest revision as of 21:16, 24 March 2022
Contents
Problem
In trapezoid , the sides
and
are equal. The perimeter of
is
Solution
There is a rectangle present, with both horizontal bases being units in length. The excess units on the bottom base must then be
. The fact that
and
are equal in length indicate, by the Pythagorean Theorem, that these excess lengths are equal. There are two with a total length of
units, so each is
units. The triangle has a hypotenuse of
, because the triangles are
right triangles. So, the sides of the trapezoid are
,
,
, and
. Adding those up gives us the perimeter,
units.
Video Solution
https://youtu.be/DKoTd9S4y-Q Soo, DRMS, NM
See Also
1999 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.