Difference between revisions of "2018 AMC 8 Problems/Problem 20"
m (/* Sol7trdszxdcfvgbhnjmk,l. pooop poop poop) |
m (poop poop poop) |
||
Line 17: | Line 17: | ||
<math>\textbf{(A) } \frac{4}{9} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{5}{9} \qquad \textbf{(D) } \frac{3}{5} \qquad \textbf{(E) } \frac{2}{3}</math> | <math>\textbf{(A) } \frac{4}{9} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{5}{9} \qquad \textbf{(D) } \frac{3}{5} \qquad \textbf{(E) } \frac{2}{3}</math> | ||
− | ==Solution 1== | + | ==Solution 1==fdsdfghjk./.lkjhyt5r4e3w2qqasxdcfvgbhnjm,./ pooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooop |
− | |||
− | |||
− | |||
− | |||
==Solution 2== | ==Solution 2== |
Revision as of 15:25, 1 October 2019
Contents
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==fdsdfghjk./.lkjhyt5r4e3w2qqasxdcfvgbhnjm,./ pooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooop
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.