Difference between revisions of "2017 AMC 8 Problems/Problem 16"
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− | ==Problem | + | ==Problem== |
In the figure below, choose point <math>D</math> on <math>\overline{BC}</math> so that <math>\triangle ACD</math> and <math>\triangle ABD</math> have equal perimeters. What is the area of <math>\triangle ABD</math>? | In the figure below, choose point <math>D</math> on <math>\overline{BC}</math> so that <math>\triangle ACD</math> and <math>\triangle ABD</math> have equal perimeters. What is the area of <math>\triangle ABD</math>? | ||
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<asy>draw((0,0)--(4,0)--(0,3)--(0,0)); | <asy>draw((0,0)--(4,0)--(0,3)--(0,0)); | ||
label("$A$", (0,0), SW); | label("$A$", (0,0), SW); | ||
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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> | ||
− | ==Solution== | + | ==Solution 1== |
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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</math>. Combining this with the fact that <math>[ABC] = [ABD] + [ACD] = \frac{3\cdot4}{2} = 6</math>, we get <math>[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} \cdot 6 = \boxed{\textbf{(D) } \frac{12}{5}}</math>. | |
− | + | ==Video Solutions== | |
+ | https://youtu.be/itz3JyoZQYg | ||
==See Also== | ==See Also== |
Latest revision as of 19:14, 5 July 2024
Contents
Problem
In the figure below, choose point on
so that
and
have equal perimeters. What is the area of
?
Solution 1
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
. Combining this with the fact that
, we get
.
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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