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

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== Solution ==
 
== Solution ==
The answer is exactly 3, choice \mathrm{(C)}$.
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The answer is exactly 3, choice <math>\mathrm{(C)}</math>.
 
We can use either Heron's on the larger triangle to calculate the height of ADE or we can decompose ABC into two right triangles of 5-12-13 and 9-12-15 to notice that height is 12. Then angle bisector theorem leads directly to the base of ADE, which leads to an area of 3.
 
We can use either Heron's on the larger triangle to calculate the height of ADE or we can decompose ABC into two right triangles of 5-12-13 and 9-12-15 to notice that height is 12. Then angle bisector theorem leads directly to the base of ADE, which leads to an area of 3.
  

Revision as of 05:11, 4 February 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 answer is exactly 3, choice $\mathrm{(C)}$. We can use either Heron's on the larger triangle to calculate the height of ADE or we can decompose ABC into two right triangles of 5-12-13 and 9-12-15 to notice that height is 12. Then angle bisector theorem leads directly to the base of ADE, which leads to an area of 3.

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

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