Difference between revisions of "2005 AMC 8 Problems/Problem 19"
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==Video Solution== | ==Video Solution== | ||
https://youtu.be/h0V8qcEo0U8 Soo, DRMS, NM | https://youtu.be/h0V8qcEo0U8 Soo, DRMS, NM | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/abSgjn4Qs34?t=2362 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2005|num-b=18|num-a=20}} | {{AMC8 box|year=2005|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 20:49, 2 January 2023
Problem
What is the perimeter of trapezoid 
?
![[asy]size(3inch, 1.5inch); pair a=(0,0), b=(18,24), c=(68,24), d=(75,0), f=(68,0), e=(18,0); draw(a--b--c--d--cycle); draw(b--e); draw(shift(0,2)*e--shift(2,2)*e--shift(2,0)*e); label("30", (9,12), W); label("50", (43,24), N); label("25", (71.5, 12), E); label("24", (18, 12), E); label("$A$", a, SW); label("$B$", b, N); label("$C$", c, N); label("$D$", d, SE); label("$E$", e, S);[/asy]](http://latex.artofproblemsolving.com/0/9/6/096a58ba6264c3736c928365300c0d13f53fcf0b.png)
Solution
Draw altitudes from 
and 
to base 
to create a rectangle and two right triangles. The side opposite 
is equal to 
. The bases of the right triangles can be found using Pythagorean or special triangles to be 
and 
. Add it together to get 
. The perimeter is 
.
Video Solution
https://youtu.be/h0V8qcEo0U8 Soo, DRMS, NM
Video Solution by OmegaLearn
https://youtu.be/abSgjn4Qs34?t=2362
~ pi_is_3.14
See Also
| 2005 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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. 
