Difference between revisions of "2018 AMC 8 Problems/Problem 20"
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<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== | + | ==Solution 1== |
By similar triangles, we have <math>[ADE] = \frac{1}{9}[ABC]</math>. Similarly, we see that <math>[BEF] = \frac{4}{9}[ABC].</math> Using this information, we get <cmath>[ACFE] = \frac{5}{9}[ABC].</cmath> Then, since <math>[ADE] = \frac{1}{9}[ABC]</math>, it follows that the <math>[CDEF] = \frac{4}{9}[ABC]</math>. Thus, the answer would be <math>\boxed {A}.</math> | By similar triangles, we have <math>[ADE] = \frac{1}{9}[ABC]</math>. Similarly, we see that <math>[BEF] = \frac{4}{9}[ABC].</math> Using this information, we get <cmath>[ACFE] = \frac{5}{9}[ABC].</cmath> Then, since <math>[ADE] = \frac{1}{9}[ABC]</math>, it follows that the <math>[CDEF] = \frac{4}{9}[ABC]</math>. Thus, the answer would be <math>\boxed {A}.</math> |
Revision as of 20:16, 30 September 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
of the larger parallelogram, so the answer would be
, since the triangle is
of the parallelogram, so the answer is
By babyzombievillager
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
.
By: sap2018
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.