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
By angle bisector theorem, . By similar triangles,
, and the height of this trapezoid is
, where
is the length of the altitude to
. Then
and we wish to compute
.
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)
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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