Difference between revisions of "1984 AHSME Problems/Problem 8"
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==See Also== | ==See Also== | ||
{{AHSME box|year=1984|num-b=7|num-a=9}} | {{AHSME box|year=1984|num-b=7|num-a=9}} | ||
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Latest revision as of 11:49, 5 July 2013
Problem
Figure 
is a trapezoid with 
, 
, 
, 
, and 
. The length of 
is
Solution
![[asy] unitsize(1.5cm); draw((0,0)--(5,0)--(8,-3)--(-sqrt(3),-3)--cycle); label("$A$",(0,0),NW); label("$B$",(5,0),NE); label("$C$",(8,-3),ENE); label("$D$",(-sqrt(3),-3),WNW); draw((0,0)--(0,-3)); draw((5,0)--(5,-3)); label("$E$",(5,-3),S); label("$F$",(0,-3),S); label("$5$",(2.5,0),N); label("$3\sqrt{2}$",(6.5,-1.5),ENE); label("$45^\circ$",(8,-3),WNW); label("$60^\circ$",(-sqrt(3),-3),ENE); draw((-.25,-2.75)--(-.25,-3)); draw((-.25,-2.75)--(0,-2.75)); draw((5.25,-3)--(5.25,-2.75)); draw((5.25,-2.75)--(5,-2.75));[/asy]](http://latex.artofproblemsolving.com/3/c/f/3cf6923a49746d60152a783c64711841538406a4.png)
Drop perpendiculars from 
and 
to 
, thus forming a 30-60-90 right triangle, a 45-45-90 right triangle, and a rectangle, as shown in the diagram. Since 
is a rectangle, we have 
. Also, from the properties of 45-45-90 right triangles, we have 
. Again since is a rectangle, 
, too. From the properties of 30-60-90 right triangles, we have 
. Therefore, 
.
See Also
| 1984 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
