Difference between revisions of "2018 AMC 8 Problems/Problem 20"
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==Solution 3== | ==Solution 3== | ||
− | <math>\triangle{ADE} \sim \triangle{ABC} \sim \triangle{EFB}</math>. We can substitute <math>\overline{DA}</math> as <math>\frac{1}{3}x</math> and <math>\overline{CD}</math> as <math>\frac{2}{3}x</math>, where <math>x</math> is <math>\overline{AC}</math>. Side <math>\overline{CB}</math> having, distance <math>y</math>, has <math>2</math> parts also. | + | <math>\triangle{ADE} \sim \triangle{ABC} \sim \triangle{EFB}</math>. We can substitute <math>\overline{DA}</math> as <math>\frac{1}{3}x</math> and <math>\overline{CD}</math> as <math>\frac{2}{3}x</math>, where <math>x</math> is <math>\overline{AC}</math>. Side <math>\overline{CB}</math> having, distance <math>y</math>, has <math>2</math> parts also. Anbe <math>4.5z-(0.5z+2z)=4.5z-2.5z=2z</math>. Parallelogram <math>CDEF</math> to <math>\triangle{ABC}= \frac{2z}{4.5z}=\frac{2}{4.5}=\frac{4}{9}</math>. The answer is <math>\boxed{(A) \frac{4}{9}}</math>. |
− | + | rfedswedfvgb by theultimatepooooooper | |
==See Also== | ==See Also== |
Revision as of 16:23, 1 October 2019
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
. The answer is
.
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.