Difference between revisions of "2018 AMC 8 Problems/Problem 20"
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==Solution 2== | ==Solution 2== | ||
− | We can extend it into a parallelogram, so it would equal <math>3a \cdot 3b</math>. The smaller parallelogram is 1 a times 2 b. The smaller parallelogram is <math>\frac{2}{9}</math> of the larger parallelogram, so the answer would be <math>\frac{2}{9} \cdot 2</math>, since the triangle is <math>\frac{1}{2}</math> of the parallelogram, so the answer is <math>\boxed{(A) \frac{4}{9}}</math> | + | We can extend it into a parallelogram, so it would equal <math>3a \cdot 3b</math>. The smaller parallelogram is 1 a times 2 b. The smaller parallelogram is <math>\frac{2}{9}</math> of the larger parallelogram, so the answer would be <math>\frac{2}{9} \cdot 2</math>, since the triangle is <math>\frac{1}{2}</math> of the parallelogram, so the answer is <math>\boxed{(A) \frac{4}{9}}</math> |
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==Solution 3== | ==Solution 3== |
Revision as of 21:08, 8 November 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
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
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 |
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