2018 AMC 10A Problems/Problem 24
Problem
Triangle with
and
has area
. Let
be the midpoint of
, and let
be the midpoint of
. The angle bisector of
intersects
and
at
and
, respectively. What is the area of quadrilateral
?
Solution 1
Let ,
,
, and the length of the perpendicular to
through
be
. By angle bisector theorem, we have that
where
. Therefore substituting we have that
. By similar triangles, we have that
, and the height of this trapezoid is
. Then, we have that
. We wish to compute
, and we have that it is
by substituting.
(rachanamadhu)
Solution 2
is midway from
to
, and
. Therefore,
is a quarter of the area of
, which is
. Subsequently, we can compute the area of quadrilateral
to be
. Using the angle bisector theorem in the same fashion as the previous problem, we get that
is
times the length of
. We want the larger piece, as described by the problem. Because the heights are identical, one area is
times the other, and
.
Solution 3
The area of to the area of
is
by Law of Sines. So the area of
is
. Since
is the midsegment of
, so
is the midsegment of
. So the area of
to the area of
is
, so the area of
is
, by similar triangles. Therefore the area of quad
is
(steakfails)
Solution 4
The area of quadrilateral is the area of
minus the area of
. Notice,
, so
, and since
, the area of
. Given that the area of
is
, using
on side
yields
. Using the Angle Bisector Theorem,
, so the height of
. Therefore our answer is
-Solution by ktong
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
2018 AMC 12A (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 12 Problems and Solutions |
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