2017 AMC 8 Problems/Problem 16
Contents
Problem
In the figure below, choose point 
on 
so that 
and 
have equal perimeters. What is the area of ?
![[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]](http://latex.artofproblemsolving.com/5/3/e/53eb67d35ecca7ac6f68e8dff60b0eb5f3e4a602.png)
Solution 2
We know that the perimeters of the two small triangles are 
and 
. Setting both equal and using 
, we have 
and 
. Now, we simply have to find the area of . Since 
, we must have ![$\frac{[ABD]}{[ACD]} = 2/3$](http://latex.artofproblemsolving.com/a/d/b/adbd8883f0cd8632fcfbbf06047ae4d60076fef4.png)
. Combining this with the fact that ![$[ABC] = [ABD] + [ACD] = \frac{3\cdot4}{2} = 6$](http://latex.artofproblemsolving.com/7/d/8/7d8610994146054d34d9933863709ec1574f11f1.png)
, we get ![$[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} \cdot 6 = \boxed{\textbf{(D) } \frac{12}{5}}$](http://latex.artofproblemsolving.com/a/a/5/aa5e715c3874487c0abd4ed1151a562b3058d244.png)
.
Video Solutions
See Also
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
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