2009 AMC 10B Problems/Problem 18
Contents
Problem
Rectangle has
and
. Point
is the midpoint of diagonal
, and
is on
with
. What is the area of
?
Solution 1 (Coordinate Geo)
Set to
. Since
is the midpoint of the diagonal, it would be
. The diagonal
would be the line
. Since
is perpendicular to
, its line would be in the form
. Plugging in
and
for
and
would give
. To find the x-intercept of
we plug in
for
and get
. Then, using the Shoelace Formula for
,
, and
, we find the area is
.
Solution 2
By the Pythagorean theorem we have , hence
.
The triangles and
have the same angle at
and a right angle, thus all their angles are equal, and therefore these two triangles are similar.
The ratio of their sides is , hence the ratio of their areas is
.
And as the area of triangle is
, the area of triangle
is
.
Solution 3 (Only Pythagorean Theorem)
Draw as shown from the diagram. Since
is of length
, we have that
is of length
, because of the midpoint
. Through the Pythagorean theorem, we know that
, which means
. Define
to be
for the sake of clarity. We know that
. From here, we know that
. From here, we can write the expression
. Now, remember
.
, since we set
in the start of the solution. Now to find the area
Solution 4
We know by the Pythagorean theorem, and furthermore,
is similar to
. Therefore,
, and the area of the triangle is
.
See Also
2009 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.