Question

In the figure below, choose point $D$ on $\overline{BC}$ so that $\triangle ACD$ and $\triangle ABD$ have equal perimeters. What is the area of $\triangle ABD?$

$\dfrac{3}{4}$

$\dfrac{3}{2}$

$2$

$\dfrac{12}{5}$

$\dfrac{5}{2}$

Video solution:

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Written solution:

The only way to split $\overline{BC}$ into two parts such that the two triangles have the same perimeter is if $\overline{CD} = 3$ and $\overline{BD} = 2.$

$\triangle ACD$ and $\triangle ABD$ have the same altitudes, so their areas are proportional to their bases. This means that the area of $\triangle ABD$ is $\dfrac{2}{5}$ the area of $\triangle ABC,$ which is $\dfrac{2}{5} \cdot 3 \cdot \dfrac{4}{2} = \dfrac{12}{5}.$

Thus, D is the correct answer.

2017 AMC 8 Problem 16

Video solution:

Written solution:

Problem 16 in Other Years