Difference between revisions of "2017 AMC 8 Problems/Problem 16"
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==Solution== | ==Solution== | ||
| − | We know that the perimeters of the two small triangles are <math>3+CD+AD</math> and <math>4+BD+AD</math>. Setting both equal and using <math>BD+CD = 5</math>, we have <math>BD = 2</math> and <math>CD = 3</math>. Now, we simply have to find the area of <math>\triangle ABD</math>. Since <math>\frac{BD}{CD} = \frac{2}{3}</math>, we must have <math>\frac{[ABD]}{ACD]} = 2/3</math>. Combining this with the fact that <math>[ABC] = [ABD] + [ACD] = \frac{3*4}{2} = 6</math>, we get <math>[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} * 6 = \boxed{\textbf{(D) } | + | We know that the perimeters of the two small triangles are <math>3+CD+AD</math> and <math>4+BD+AD</math>. Setting both equal and using <math>BD+CD = 5</math>, we have <math>BD = 2</math> and <math>CD = 3</math>. Now, we simply have to find the area of <math>\triangle ABD</math>. Since <math>\frac{BD}{CD} = \frac{2}{3}</math>, we must have <math>\frac{[ABD]}{ACD]} = 2/3</math>. Combining this with the fact that <math>[ABC] = [ABD] + [ACD] = \frac{3*4}{2} = 6</math>, we get <math>[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} * 6 = \boxed{\textbf{(D) } \frac{12}{5}}</math> |
==See Also== | ==See Also== | ||
Revision as of 16:35, 22 November 2017
Problem 16
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
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/f/e/9/fe9792dcd5ec9e17365c91523549220808939768.png)
. Combining this with the fact that ![$[ABC] = [ABD] + [ACD] = \frac{3*4}{2} = 6$](http://latex.artofproblemsolving.com/6/4/e/64e2d85b6138ea15949b7362bd0d9b2204e8a359.png)
, we get ![$[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} * 6 = \boxed{\textbf{(D) } \frac{12}{5}}$](http://latex.artofproblemsolving.com/5/8/3/583f355c81572d0b7d5075d6e2e3a9726fed0f35.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
