Difference between revisions of "2017 AMC 8 Problems/Problem 16"
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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>. | 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>. | ||
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+ | ==Solution 2== | ||
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+ | Since <math>\overline{AC}</math> is <math>1</math> less than <math>\overline{BC}</math>, <math>\overline{CD}</math> must be <math>1</math> more than <math>\overline{BD}</math> to equate the perimeter. Hence, <math>\overline{BD}+\overline{BD}+1=5</math>, so <math>\overline{BD}=2</math>. Therefore, the area of <math>\triangle ABD</math> is <math>(2)(4)(\sin B)=8(\frac{3}{5})=\boxed{\textbf{(D) } \frac{12}{5}}</math> | ||
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+ | ~megaboy6679 | ||
==Video Solutions== | ==Video Solutions== |
Revision as of 13:56, 2 November 2024
Contents
[hide]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 .
Solution 2
Since is less than , must be more than to equate the perimeter. Hence, , so . Therefore, the area of is
~megaboy6679
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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