Difference between revisions of "1999 AMC 8 Problems/Problem 14"
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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== | ||
Revision as of 12:50, 20 February 2022
Contents
Problem
In trapezoid 
, the sides 
and 
are equal. The perimeter of is
![[asy] draw((0,0)--(4,3)--(12,3)--(16,0)--cycle); draw((4,3)--(4,0),dashed); draw((3.2,0)--(3.2,.8)--(4,.8)); label("$A$",(0,0),SW); label("$B$",(4,3),NW); label("$C$",(12,3),NE); label("$D$",(16,0),SE); label("$8$",(8,3),N); label("$16$",(8,0),S); label("$3$",(4,1.5),E); [/asy]](http://latex.artofproblemsolving.com/b/e/b/beb4d75c7db4c3afcc5a11cb678b437390df2029.png)
Solution
Im on my way
![[asy] draw((0,0)--(4,3)--(12,3)--(16,0)--cycle); draw((4,3)--(4,0),dashed); draw((12,3)--(12,0),dashed); draw((3.2,0)--(3.2,.8)--(4,.8)); label("$A$",(0,0),SW); label("$B$",(4,3),NW); label("$C$",(12,3),NE); label("$D$",(16,0),SE); label("$8$",(8,3),N); label("$8$",(8,0),S); label("$3$",(4,1.5),E); label("$4$",(2,0),S); label("$4$",(14,0),S); label("$5$",(0,0)--(4,3),NW); label("$5$",(12,3)--(16,0),NE); [/asy]](http://latex.artofproblemsolving.com/2/a/a/2aa5b1156ba4ecb2097fc5b495f4be7e51665123.png)
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. 
