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Difference between revisions of "2017 AMC 8 Problems/Problem 16"

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==Solution==
 
==Solution==
  
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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>. We can use <math>AB</math> as the base and the altitude from <math>D</math>. Let's call the foot of the altitude <math>E.</math> We have <math>\triangle BDE</math> similar to <math>BAC.</math>
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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. 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}}$
  
 
==See Also==
 
==See Also==

Revision as of 15:33, 22 November 2017

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 $\triangle ABD$. Since $\frac{BD}{CD} = \frac{2}{3}$, we must have $\frac{[ABD]}{ACD]} = 2/3. Combining this with the fact that$[ABC] = [ABD] + [ACD] = \frac{3*4}{2} = 6$, we get$[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} * 6 = \boxed{\textbf{(D) }$ \frac{12}{5}}$

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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