Difference between revisions of "2018 AMC 8 Problems/Problem 20"
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==Solution== | ==Solution== | ||
− | Looking at this diagram, we notice some similar triangles. Since <math>\overline{DE} \parallel \overline{BC}</math>, <math>\angle{ACB}=\angle{ADE}</math>. Since<math>\triangle{ABC}</math> and triangle <math>\triangle{AED}</math> share <math>\angle{A}</math>, <math>\triangle{ABC}</math> is similar to <math>\triangle{AED}</math> by the AA similarity theorem. Using similar logic we can find <math>\triangle{ABC}</math> is similar to <math>\triangle{EBF}</math>. The ratio of the areas of two similar triangles is equivalent to the square of the ratio of the lenths, so the area of <math>\triangle{AED}</math> is <math>\frac{1}{9}</math> times the area of <math>\triangle{ABC}</math> and <math>\triangle{EBF}</math> is <math>\frac{4}{9}</math> times the area of <math>\triangle{ABC}</math>. This means that the area of quadrilateral <math>CDEF</math> is is <math>1-\right(\frac{1}{9}+\frac{4}{9}\left)=\frac{4}{9}</math> times the area of <math>\triangle{ABC}</math>, so our answer is <math>\boxed{\textbf{(A) }\frac 49}</math> | + | Looking at this diagram, we notice some similar triangles. Since <math>\overline{DE} \parallel \overline{BC}</math>, <math>\angle{ACB}=\angle{ADE}</math>. Since<math>\triangle{ABC}</math> and triangle <math>\triangle{AED}</math> share <math>\angle{A}</math>, <math>\triangle{ABC}</math> is similar to <math>\triangle{AED}</math> by the AA similarity theorem. Using similar logic we can find <math>\triangle{ABC}</math> is similar to <math>\triangle{EBF}</math>. The ratio of the areas of two similar triangles is equivalent to the square of the ratio of the lenths, so the area of <math>\triangle{AED}</math> is <math>\frac{1}{9}</math> times the area of <math>\triangle{ABC}</math> and <math>\triangle{EBF}</math> is <math>\frac{4}{9}</math> times the area of <math>\triangle{ABC}</math>. This means that the area of quadrilateral <math>CDEF</math> is is <math>1- \right(\frac{1}{9}+\frac{4}{9} \left)=\frac{4}{9}</math> times the area of <math>\triangle{ABC}</math>, so our answer is <math>\boxed{\textbf{(A) }\frac 49}</math> |
==See Also== | ==See Also== |
Revision as of 15:46, 23 November 2018
Problem 20
In a point is on with and Point is on so that and point is on so that What is the ratio of the area of to the area of
Solution
Looking at this diagram, we notice some similar triangles. Since , . Since and triangle share , is similar to by the AA similarity theorem. Using similar logic we can find is similar to . The ratio of the areas of two similar triangles is equivalent to the square of the ratio of the lenths, so the area of is times the area of and is times the area of . This means that the area of quadrilateral is is $1- \right(\frac{1}{9}+\frac{4}{9} \left)=\frac{4}{9}$ (Error compiling LaTeX. ! Extra \right.) times the area of , so our answer is
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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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