High school geometry
Course: High school geometry > Unit 4Lesson 7: Solving modeling problems with similar & congruent triangles
Geometry word problem: a perfect pool shot
Sal uses triangle similarity to plan the perfect shot in a pool game. Created by Sal Khan.
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- At1:55Sal says that the two green angles are congruent because each is complementary to the black angle. How do we know that they are complementary to the black angle? It seems to me that we're just assuming that the line of reflection (the black line between the two blue segmented lines) is perpendicular to right (east) side of the pool, which is not necessarily true.(5 votes)
- The perpendicular is a perpendicular because we have defined it that way. We have just taken the point where the ball hits and purposefully constructed a perpendicular there. In terms of measuring the angles of the path that the ball follows, the only way to "know" how real physical objects behave is to take measurements.
You may have a math theory as to how they behave, but you must test the theory against a real experiment. For instance, in this case, you could take a pool ball, and cover it in wet paint. You could then shoot the ball.
The path that the ball leaves in paint on the table would give clues. You may have a math theory beforehand (a theory based on similar triangles), or you may just perform the experiment, and guess at the math might be afterwards, based on what you see --
You may say to yourself : "the path the ball leaves in paint on the table appears to describe similar triangles : Maybe pool balls bouncing off walls follow a path based on similar triangles" -- and from there , you may draw a perpendicular on the pool table at the point the ball bounced on the wall to see if you seem correct.
Then, you may repeat the experiment numerous times, and test the math against the reality : Test many many many times to feel more confident that the math is really describing what is happening -- and try your best to consider variables which may conflict with your math :
Such as : irregularities in the pool table, the shape of the ball, whether you hit the ball directly in the centre of the ball, etc.) -- but all things considered, if the math seems to generally describe what is occurring physically, over continual / "endless" experiments, there is probably a true relationship between the "reality" and the math theory you are trying to apply to it.
This is one way we can sort of "know", or strongly suspect and feel confident if there is some truth in an idea.
I only know a little tiny bit of high-school entry level physics, but this pool table problem makes me think of "Total internal reflection" in optics, which is to do with similar triangles and a ray being reflected off a surface, and seems quite similar.
See "Snell's Law".(11 votes)
- Hello. Can someone please explain to me @6:10, how does Sal go from 7/4x to 11/4x?(1 vote)
- He started with 7/4x + x, got a common denominator 7/4 x + 4/4 x then added 7+4 to get 11/4 x.(5 votes)
- I need help on pool problems.(2 votes)
- If you elaborate, I may be able to help!(1 vote)
- Sometimes saul does the correspondence for sides as (7/4)/1 = ((3/4)-x)/x , while other times he does it as x/1 = ((3/4)-x)/(7/4). When do i use one or another?(1 vote)
- It doesn't matter. The two are equivalent. Take the first equality and see what happens when you multiply by x/(7/4).(3 votes)
- If two angles of a triangle are equal to each other, does it mean their opposite sites must be equal? And vice versa, please explain it to me.(2 votes)
- At1:52, how are the angles complementary to each other? How do you know that mirror line is perpendicular to the rectangle?(2 votes)
- Because the tricky part of the question,
"the angles form as the ball approaches and deflects form a mirror image of each other "
Imagine if we change the ball's position from the original and place it on the north wall center pocket .
We hit the ball and it approached the East wall,
because the question says it form a image,
when the ball leaves the East wall toward South wall, it always creates the exact same angle .
So, the bisector that between these 2 angles will always perpendicular to the East wall .(1 vote)
- will we get such questions in the GRE examination?(1 vote)
- They'll not give you these types of problems but the problems that you have to know the logic.(2 votes)
- I still don't understand what the problem means by "the angles formed as the ball approaches and deflects." Isn't there one angle formed between the two rays that mark the ball's travel? Otherwise, shouldn't it be referring to the other two supplementary angles?(2 votes)
- Would geometry be used in other sports? Such as baseball or basketball?(1 vote)
- i don't think that when playing pool you would actually use geometry but i could be wrong.(1 vote)
- 1/4 from the north wall or the west wall? shouldn't it be 3/4 from the north wall?(1 vote)
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.