Difference between revisions of "2000 AMC 12 Problems/Problem 19"

(Solution)
Line 5: Line 5:
  
 
== Solution ==
 
== Solution ==
The area of <math>ADE = 3</math> which rounded equals = <math>2\ \mathrm{(C)}</math>.
+
The area of <math>ADE = 3</math>\ \mathrm{(C)}$.
  
 
== See also ==
 
== See also ==

Revision as of 20:08, 10 January 2015

Problem

In triangle $ABC$, $AB = 13$, $BC = 14$, $AC = 15$. Let $D$ denote the midpoint of $\overline{BC}$ and let $E$ denote the intersection of $\overline{BC}$ with the bisector of angle $BAC$. Which of the following is closest to the area of the triangle $ADE$?

$\text {(A)}\ 2 \qquad \text {(B)}\ 2.5 \qquad \text {(C)}\ 3 \qquad \text {(D)}\ 3.5 \qquad \text {(E)}\ 4$

Solution

The area of $ADE = 3$\ \mathrm{(C)}$.

See also

2000 AMC 12 (ProblemsAnswer KeyResources)
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 AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png