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## Solving modeling problems with similar & congruent triangles

Current time:0:00Total duration:6:49

# Geometry word problem: a perfect pool shot

CCSS Math: HSG.SRT.B.5

## Video transcript

A pool table is 1
meter by 2 meters. So this is 1 meter-- let me
label that-- so this distance right over here is 1 meter. This distance right
over here is 2 meters. That's 2 meters
right over there. There are 6 pockets total, 1,
2, 3, 4, 5, 6, 4 in the corners and 2 at the midpoints of
each of the 2-meter sides, or the midpoints of
each 2-meter side. A cue ball is placed
0.25, or a quarter meter, from the north wall and
a quarter meter away from the west wall, so
this right over here, this distance right over here. So it's a quarter meter
away from the north wall. So this is 1/4 meter. And this distance right over
here is also 1/4 of a meter. So that's that distance,
that's that distance. From the west wall. The angles formed as the
ball approaches and deflects form a mirror image
of each other. So this is as we approach,
and then we deflect. And they're mirror images. So if we were to draw-- if
we could imagine a mirror right over there--
we see that they are mirror images of each other. At what distance
x-- so then they labeled x right over here--
from the southeast corner should the cue ball hit
the east wall-- so this is the distance from
the southeast corner-- so the cue ball
sinks into the pocket at the midpoint
of the south wall? And I encourage you
to pause the video. And I'll give you one hint. It might involve
similar triangles. So let's try to work it through. So a big clue is that the
approach and the deflection, that they're going to be
mirror images of each other. So if they're mirror
images, this angle is going to be
congruent to that angle. If those two angles
are congruent, then this angle,
which is complementary to that black angle, must
be equal to this angle. Each of these are going to be 90
degrees minus that black angle right over there. So you have this angle is
congruent to that angle. And now we can construct
two right triangles. So you can imagine one up
here, this is the larger one. So imagine this is our approach
right triangle right over here. So the top of it is parallel
to the side of the pool table. And then this is our deflection
right triangle right over here. And the whole reason why I show
that these two green angles are going to be congruent is to show
that these two triangles are similar to each other. How do we know they're similar? Well, if you have
two angles, they both have a 90 degree angle, and
they both have this green angle, so that means that the third
angle must also be the same. If you know two angles, you know
what the third angle has to be. If two corresponding angles
of two different triangles are congruent,
then the triangles are going to be similar. So this top triangle is similar
to this bottom triangle. And what that helps
us is that means that the ratio of the lengths
of corresponding parts of those triangles are
going to be the same. So for example, we've already
said that this distance-- let's figure out what we know
about these triangles-- so that distance is x. Now, what is this
distance right over here? What is this distance right
over here going to be? Well, let's think
about it a little bit. We know that this distance
is 1/4 of a meter. We know that this entire
distance is 1 meter. So this distance
right-- let me do this in a color you can see--
this distance right over here is going to be 3/4 of a meter. And so if this distance
is 3/4 of a meter, then this part
right over here is going to be 3/4 minus x meters. So let me write that down. 3/4 minus x is this
magenta length. And what else do we know? Well, we definitely
know the length of this segment right over here. We know that the pockets
are 1 meter apart, so that is 1 meter. And we also know the length of
this segment right over here. We know that this is 1 meter
and that this is another 3/4 of a meter. This whole distance right
over here is 1 and 3/4 meters. Or we could write
that as 7/4 meters. So let me write it like this. This is 7/4. I like to write everything
as an improper fraction because I have a
feeling that I'm going to have to do
some ratios in a second. So corresponding parts of these
triangles-- these two triangles are similar, so
corresponding parts are going to have
the same ratio. So for example, this green
segment right over here, this is the longer
side that's not the hypotenuse of this
top right triangle. That's going to correspond
to the longer side that's not the hypotenuse
of this triangle. The sides that are
opposite this green angle, they correspond to each other. So we could say that
the ratio of 7/4 to 1, the ratio of 7/4
meters to 1 meter, is going to be equal to
the ratio of the sides that are opposite the magenta angles. So it's going to be equal
to 3/4 minus x to x. I'm just showing that the
ratio of corresponding parts are the same. So let's now just solve for x. So let's see, if we
multiply both sides of this by x, on the left hand
side, we end up with 7/4 x. And on the right hand side,
we're left with 3/4 minus x. Well, now we can add
an x to both sides. And we're going to get
7/4 x plus another 4/4 x is going to give us
11/4 x is equal to 3/4. And now to solve
for x, we can just multiply both sides times the
reciprocal of its coefficient, so times 4/11. This is going to give us that
x is equal to 3/11 of a meter. So if we hit 3/11 of a meter
above the southeast corner of this wall, then we should
hit this pocket right over here.