2000 AMC 12 Problems/Problem 19
Problem
In triangle ,
,
,
. Let
denote the midpoint of
and let
denote the intersection of
with the bisector of angle
. Which of the following is closest to the area of the triangle
?
Solution 1
The answer is exactly , choice
.
We can find the area of triangle
by using the simple formula
. Dropping an altitude from
, we see that it has length
( we can split the large triangle into a
and a
triangle). Then we can apply the Angle Bisector Theorem on triangle
to solve for
. Solving
, we get that
.
is the midpoint of
so
. Thus we get the base of triangle
, to be
units long. Applying the formula
, we get
.
Solution 2
The area of is
where
is the height of triangle
. Using Angle Bisector Theorem, we find
, which we solve to get
.
is the midpoint of
so
. Thus we get the base of triangle
, to be
units long. We can now use Heron's Formula on
.
Therefore, the answer is
.
Solution 3
Let's find the area of by Heron,
Then,
Knowing that D is the midpoint of BC, then .
By Angle Bisector Theorem we know that:
Also, we know that:
And, we can easily see that , so,
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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