Question

In triangle $ABC,$ $AB = 13,$ $BC = 14,$ and $AC = 15.$ 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 triangle $ADE?$

$2$

$2.5$

$3$

$3.5$

$4$

Solution:

By Heron's formula, the area of $\triangle ABC$ is $\sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84,$ so the altitude from $A$ to $BC$ is $\dfrac{2 \cdot 84}{14} = 12.$

The midpoint $D$ is $7$ from $B.$ The bisector from $A$ meets $BC$ at $E$ with $BE : EC = AB : AC = 13 : 15,$ so $BE = 14 \cdot \dfrac{13}{28} = 6.5.$

Both $D$ and $E$ lie on $BC,$ so $\triangle ADE$ has base $DE = 7 - 6.5 = 0.5$ and altitude $12,$ giving area $\tfrac12 \cdot 0.5 \cdot 12 = 3.$

Thus, the correct answer is C.

2000 AMC 12 Problem 19

Solution:

Problem 19 in Other Years