Difference between revisions of "2004 AMC 10B Problems/Problem 18"
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==Solution 4== | ==Solution 4== | ||
− | You can also place a point <math>X</math> on <math>CE</math> such that <math>CX</math> is <math> | + | You can also place a point <math>X</math> on <math>CE</math> such that <math>CX</math> is <math>12</math>, creating trapezoid <math>CBFX</math>. Then, you can find the area of the trapezoid, subtract the area of the two right triangles <math>DFX</math> and <math>BCD</math>, divide by the area of <math>ABC</math>, and get the ratio of <math>7/16</math>. |
== See also == | == See also == |
Revision as of 22:11, 2 January 2020
Problem
In the right triangle , we have
,
, and
. Points
,
, and
are located on
,
, and
, respectively, so that
,
, and
. What is the ratio of the area of
to that of
?
Solution 1
Let . Because
is divided into four triangles,
.
Because of triangle area,
.
and
, so
.
, so
.
Solution 2
First of all, note that , and therefore
.
Draw the height from onto
as in the picture below:
Now consider the area of . Clearly the triangles
and
are similar, as they have all angles equal. Their ratio is
, hence
.
Now the area
of
can be computed as
=
.
Similarly we can find that as well.
Hence , and the answer is
.
Solution 3
The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9), and let F be (12, 3). You can then use the shoelace theorem to find the area of DBF, which is 42.
Solution 4
You can also place a point on
such that
is
, creating trapezoid
. Then, you can find the area of the trapezoid, subtract the area of the two right triangles
and
, divide by the area of
, and get the ratio of
.
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 AMC 10 Problems and Solutions |
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