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
times
. The smaller parallelogram is
of the larger parallelogram, so the answer would be
, since the triangle is
of the parallelogram, so the answer is
.
By babyzombievillager with credits to many others who helped with the solution:)
Solution 3
. We can substitute
as
and
as
, where
is
. Side
having, distance
, has
parts also. And
and
are
and
respectfully. You can consider the height of
and
as
and
respectfully. The area of
is
because the area formula for a triangle is
or
. The area of
will be
. So the area of
will be
. The area of parallelogram
will be
. Parallelogram
to
. The answer is
Solution 4
If you have little time to calculate, divide DEFC into triangles that are equal to dae. Also cut triangle EFB into triangles similar to DAE. We see that there are 9 total triangles, and 4 of those are occupied by DEFC. Thus, 4/9.
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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