2017 AMC 10B Problems/Problem 15
Contents
Problem
Rectangle has
and
. Point
is the foot of the perpendicular from
to diagonal
. What is the area of
?
Solution
First, note that because
is a right triangle. In addition, we have
, so
. Using similar triangles within
, we get that
and
.
Let be the foot of the perpendicular from
to
. Since
and
are parallel,
is similar to
. Therefore, we have
. Since
,
. Note that
is an altitude of
from
, which has length
. Therefore, the area of
is
Solution 2
From similar triangles, we know that (see Solution 1). Furthermore, we also know that
from the rectangle. Using the sine formula for area, we have
But, note that
. Thus, we see that
~coolwiz
Solution 3
Alternatively, we can use coordinates. Denote as the origin. We find the equation for
as
, and
as
. Solving for
yields
. Our final answer then becomes
Solution 4
We note that the area of must equal area of
because they share the base and the height of both is the altitude of congruent triangles. Therefore, we find the area of
to be
Solution 5
We know all right triangles are 5-4-3, so the areas are proportional to the square of like sides. Area of is
of
. Using similar logic in Solution 4, Area of
is the same as
.
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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