2000 AMC 12 Problems/Problem 19
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Problem
In triangle 
, 
, 
, 
. Let 
denote the midpoint of 
and let 
denote the intersection of with the bisector of angle 
. Which of the following is closest to the area of the triangle 
?
Solution
By the Angle Bisector Theorem, 
. Since 
, then 
.
By Heron's Formula, ![[ABC] = \sqrt{21(6)(7)(8)} = 84](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/2/7/d/27da688bdf54e79cdbcfe9fabd2fdc2dc6a946e0.gif)
, so the height of 
from 
is 
. Notice that the heights of and 
are the same, so ![[ADE] = \frac{1}{2}(12)(0.5) = 3\ \mathrm{(C)}](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/6/a/b/6abc70900499d87d2af7ff017bba1f7cf5d07419.gif)
.
See also
| 2000 AMC 12 (Problems • 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 | ||
