2009 AMC 10A Problems/Problem 17
Contents
Problem
Rectangle has
and
. Segment
is constructed through
so that
is perpendicular to
, and
and
lie on
and
, respectively. What is
?
Solutions
Solution 1
The situation is shown in the picture below.
From the Pythagorean theorem we have .
Triangle is similar to
, as they have the same angles. Segment
is perpendicular to
, meaning that angle
and
are right angles and congruent. Also, angle
is a right angle. Because it is a rectangle, angle
is congruent to
and angle
is also a right angle. By the transitive property:
Next, because every triangle has a degree measure of , angle
and angle
are similar.
Hence , and therefore
.
Also triangle is similar to
. Hence
, and therefore
.
We then have .
Solution 2
Since is the altitude from
to
, we can use the equation
.
Looking at the angles, we see that triangle is similar to
. Because of this,
. From the given information and the Pythagorean theorem,
,
, and
. Solving gives
.
We can use the above formula to solve for .
. Solve to obtain
.
We now know and
.
.
Solution 3
There is a better solution where we find directly instead of in parts (use similarity). The strategy is similar.
Solution 4(Coordinate Bash)
To keep things simple, we will use coordinates in only the first quadrant. The picture will look like the diagram above reflected over the -axis.It is also worth noting the
will lie on the
axis and
on the
. Let
be the origin,
,
, and
. We can express segment
as the line
.
Since
is perpendicular to
, and we know that
lies on it, we can use this information to find that segment
is on the line
. Since
and
are on the
and
axis, respectively, we plug in
for
and
, we find that point
is at
, and point
is at
. Applying the distance formula,
we obtain that
=
.
See Also
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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All AMC 10 Problems and Solutions |
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