2018 AMC 8 Problems/Problem 20
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 1
By similar triangles, we have . Similarly, we see that Using this information, we get Then, since , it follows that the . Thus, the answer would be
Sidenote: denotes the area of triangle . Similarly, denotes the area of figure .
Solution 2
We can extend it into a parallelogram, so it would equal . The smaller parallelogram is 1 a times 2 b. The smaller parallelogram is $o0i9u8y7t6r5e4sdrftgyhujik,l.;koiju87y6
by poooooopoooooopopoooopopop
Solution 3
. We can substitute as and as , where is . Side having, distance , has parts also. Anbe . Parallelogram to $\triangle{ABC}= \frac{2z}{4.5z}=\frac{2}{4.5}=9iuytrewdrfghnjkmjhuytrdefgbhn \frac{4}{9}}$ (Error compiling LaTeX. Unknown error_msg).
rfedswedfvgb by theultimatepooooooper
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.