2017 AMC 8 Problems/Problem 16

Revision as of 15:27, 22 November 2017 by Doingwhatcounts (talk | contribs) (Solution)

Problem 16

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 $\triangle ABD$? [asy]draw((0,0)--(4,0)--(0,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,0), ESE); label("$C$", (0, 3), N); label("$3$", (0, 1.5), W); label("$4$", (2, 0), S); label("$5$", (2, 1.5), NE);[/asy]

$\textbf{(A) }\frac{3}{4}\qquad\textbf{(B) }\frac{3}{2}\qquad\textbf{(C) }2\qquad\textbf{(D) }\frac{12}{5}\qquad\textbf{(E) }\frac{5}{2}$

Solution

We know that the perimeters of the two small triangles are $3+CD+AD$ and $4+BD+AD.$ Setting both equal and using $BD+CD = 5,$ we have $BD = 2$ and $CD = 3.$ Now, we simply have to find the area of $\triangle ABD$. We can use $AB$ as the base and the altitude from $D$. Let's call the foot of the altitude $E.$ We have $\triangle BDE$ similar to $BAC.$

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AJHSME/AMC 8 Problems and Solutions

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