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

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<math>\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}</math>
 
<math>\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}</math>
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==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> 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>
 
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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==See Also==
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{{AMC8 box|year=2017|num-b=15|num-a=17}}
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{{MAA Notice}}

Revision as of 14:44, 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 $\triangle ABD.$ 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
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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