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Challenging similarity problem

Interesting similarity problem where we don't have a lot of information to work with. Created by Sal Khan.

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  • duskpin sapling style avatar for user Gigi
    I am reeeally struggling with the exercises on 'Solving problems with similar and congruent triangles'. I feel like I'm missing some algebra knowledge or something, I can't even understand the Hints properly. Especially the exercise with Michelangelo's Doni Madonna pentagram, for example, how do you go from CD= x - x/phi to CD = (phi-1/phi)x??
    Is anyone else having problems with these exercises, or are there some areas/videos I should refresh myself on? I've gone though this playlist, but I'm still struggling :/
    (19 votes)
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    • piceratops ultimate style avatar for user ILoveToLearn
      Hi Gabrielle! (Beautiful name, I have a dear friend named Gabriella!)
      I was having quite some difficulty with that too, because I wasn't sure how to write the proportions correctly even though I understand how to do the math behind it, so I was pretty stuck! If you could detail what your confusion was about with the hints and alike, I'd be happy to help you.
      This link helped me a lot (especially the practice at the end, which, I noticed, was esthetically a take-off of the question-answer-hints button from Khan Academy, all the way down to the smiley face when you get it right! Same colors, everything!)
      I hope this helps you!
      Your fellow learner,
      (12 votes)
  • blobby green style avatar for user Janno Clay
    I ask if high school students are expected to solve the problem because I tried to and got stuck seemingly because I did not have the knowledge to apply algebra in this way. I suspect many are like me.

    What can I do about this lack of ability to solve such problems that are challenging and new to me?
    (16 votes)
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  • piceratops ultimate style avatar for user Hushus/حسين
    I understood the whole problem except the last thing he said at , when he found the answer of 36/7, what did he mean when he said "regardless of how far and wide it is, it will always be 36/7"? Does this mean that for every problem like above the answer will always be 36/7, even if their angles or vertical length change? I find this hard to believe, and if somebody could explain it, I would be indebted to them.
    (7 votes)
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    • piceratops ultimate style avatar for user greybeakteddy
      i recommend you get some squared paper and draw this problem to scale to prove it to yourself. make the left side 9cm and the right 12cm and use a ruler to connect the lines as in sal's diagram. now repeat but move both the lines closer to the middle , keeping them 9cm and 12cm. you'll notice when you connect the lines they pass through the same point. the length of y is unimportant. now repeat the diagram but make one side 4cm. you'll notice that the new CF has definately changed. the conclusions are: if the sides remain the same length you can move them as far apart or as close and CF will be the same. change the lengths and you'll get a new length CF
      (11 votes)
  • blobby green style avatar for user Janno Clay
    Are high school students expected to solve this problem?
    (11 votes)
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  • blobby green style avatar for user Isuru  Herath
    couldn't you also have done CF/9+CF/12=[(y-x)/y]+(x/y)? Then just simplifying that would get you the same answer.
    (7 votes)
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  • mr pink orange style avatar for user orange velociraptor
    I see many comments in the section that say, just use 1/AB + 1/DE = 1/CF.
    (I tried it, it works).
    I've asked all of them why it works but haven't gotten an answer for months.
    So could someone answer it here and help me understand the reason?
    I would really appreciate it since this has been bugging me for a long time.
    (6 votes)
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    • blobby green style avatar for user guruling2004cat
      ok here it goes.
      assume instead of 9 it was x
      and instead of 12 it was y
      i am following diagram for proof
      see triangles ABC and DEC
      now you can prove triangles BCF and DEB are similar easily
      so CF/DE=CB/DB
      we can CB and DC as xh and yh(h is a proportionality
      now CF/y=xh/(xh+yh)
      Now you can rewrite it as
      (1 vote)
  • hopper cool style avatar for user ALYSIQUE UNIQUE = )
    I'm confused for when the number 1 suddenly pops into the problem when it says cf/9=1-x/y
    (5 votes)
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  • leaf grey style avatar for user adsterr
    In this diagram, isn't triangle ABE similar to triangle DEB?
    (1 vote)
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  • duskpin ultimate style avatar for user Pierre Dob
    I encountered a strange problem;
    Since there are two right angles between ab and ed, they are parallel. That means that angle bac is congruent to angle dec. Angles acb and dce are congruent because they are vertical angles. From this, we can conclude that angle cbf and angle cef are congruent, because abc and dec are congruent and angles abf and def are right angles. That means that triangles dbf and cfe are congruent, by aas. But this also gives the congruence of acb and dce, also by aas. But we know this is not true, as we have corresponding sides are unequal. Is there a fault in my logic? If not, is this situation impossible? What can be adjusted to make it possible(while keeping the side lengths)?
    (3 votes)
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  • piceratops sapling style avatar for user seamus s.
    At x over y is circled in purple pointing to ce over de = x over y how come
    (3 votes)
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Video transcript

So given this diagram, we need to figure out what the length of CF right over here is. And you might already guess that this will have to do something with similar triangles, because at least it looks that triangle CFE is similar to ABE. And the intuition there is it's kind of embedded inside of it, and we're going to prove that to ourselves. And it also looks like triangle CFB is going to be similar to triangle DEB, but once again, we're going to prove that to ourselves. And then maybe we can deal with all the ratios of the different sides to CF right over here and then actually figure out what CF is going to be. So first, let's prove to ourselves that these definitely are similar triangles. So you have this 90-degree angle in ABE, and then you have this 90-degree angle in CFE. If we can prove just one other angle or one other set of corresponding angles is congruent in both, then we've proved that they're similar. And there we can either show that, look, they both share this angle right over here. Angle CEF is the same as angle AEB. So we've shown two corresponding angles in these triangles. This is an angle in both triangles. They are congruent, so the triangles are going to be similar. You can also show that this line is parallel to this line, because obviously these two angles are the same. And so these angles will also be the same. So they're definitely similar triangles, so let's just write that down. Get that out of the way. We know that triangle ABE is similar to triangle CFE. And you want to make sure you get it in the right order. F is where the 90-degree angle is. B is where the 90-degree angle is, and then E is where this orange angle is. So CFE. It's similar to triangle CFE. Now let's see if we can figure out that same statement going the other way, looking at triangle DEB. So once again, you have a 90-degree angle here. If this is 90, then this is definitely going to be 90, as well. You have a 90-degree angle here at CFB. You have a 90-degree angle at DEF or DEB, however you want to call it. So they have one set of corresponding angles that are congruent, and then you'll also see that they both share this angle right over here on the smaller triangle. So I'm now looking at this triangle right over here, as opposed to the one on the right. So they both share this angle right over here. Angle DBE is the same as angle CBF. So I've shown you already that we have this angle is congruent to this angle, and we have this angle is a part of both. So it's obviously congruent to itself. So we have two corresponding angles that are congruent to each other. So we know that this larger triangle over here is similar to this smaller triangle over there. So let me write this down. Scroll over to the right a little bit. We also know that triangle DEB is similar to triangle CFB. Now, what can we do from here? Well, we know that the ratios of corresponding sides, for each of those similar triangles, they're going to have to be the same. But we only have one side of one of the triangles. So in the case of ABE and CFE, we've only been given one side. In the case of DEB and CFB, we've only been given one side right over here, so there doesn't seem to be a lot to work with. And this is why this is a slightly more challenging problem here. Let's just go ahead and see if we can assume one of the sides, and actually, maybe a side that's shared by both of these larger triangles. And then maybe things will work out from there. So let's just assume that this length right over here-- let's just assume that BE is equal to y. So let me just write this down. This whole length is going to be equal to y, because this at least gives us something to work with. And y is shared by both ABE and DEB, so that seems useful. And then we're going to have to think about the smaller triangles right over there. So maybe we'll call BF x. Let's call BF x. And then let's FE-- well, if this is x, then this is y minus x. So we've introduced a bunch of variables here. But maybe with all the proportionalities and things maybe, just maybe, things will work out. Or at least we'll have a little bit more sense of where we can go with this actual problem. But now we can start dealing with the similar triangles. For example, so we want to figure out what CF is. We now know that for these two triangles right here, the ratio of the corresponding sides are going to be constant. So for example, the ratio between CF and 9, their corresponding sides, has got to be equal to the ratio between y minus x-- that's that side right there-- and the corresponding side of the larger triangle. Well, the corresponding side of the larger triangle is this entire length. And that entire length right over there is y. So it's equal to y minus x over y. So we could simplify this a little bit. Well, I'll hold off for a second. Let's see if we can do something similar with this thing on the right. So once again, we have CF, its corresponding side on DEB. So now we're looking at the triangle CFB, not looking at triangle CFE anymore. So now when we're looking at this triangle, CF corresponds to DE. So we have CF over DE is going to be equal to x-- let me do that in a different color. I'm using all my colors. It's going to be equal to x over this entire base right over here, so this entire BE, which once again, we know is y. So over y. And now this looks interesting, because it looks like we have three unknowns. Sorry, we know what DE is already. This is 12. I could have written CF over 12. The ratio between CF and 12 is going to be the ratio between x and y. So we have three unknowns and only two equations, so it seems hard to solve at first, because there's one unknown, another unknown, another unknown, another unknown, another unknown, and another unknown. But it looks like I can write this right here, this expression, in terms of x over y, and then we could do a substitution. So that's why this was a little tricky. So this one right here-- let me do it in that same green color. We can rewrite it as CF over 9 is equal to y minus x over y. It's the same thing as y over y minus x over y, or 1 minus x over y. All I did is, I essentially, I guess you could say, distributed the 1 over y times both of these terms. So y over y minus x over y, or 1 minus x minus y. And this is useful, because we already know what x over y is equal to. We already know that x over y is equal to CF over 12. So this right over here, I can replace with this, CF over 12. So then we get-- this is the home stretch here-- CF, which is what we care about, CF over 9 is equal to 1 minus CF over 12. And now we have one equation with one unknown, and we should be able to solve this right over here. So we could add CF over 12 to both sides, so you have CF over 9 plus CF over 12 is equal to 1. We just have to find a common denominator here, and I think 36 will do the trick. So 9 times 4 is 36, so if you have to multiply 9 times 4, you have to multiply CF times 4. So you have 4CF. 4CF over 36 is the same thing as CF over 9, and then plus CF over 12 is the same thing as 3CF over 36. And this is going to be equal to 1. And then we are left with 4CF plus 3CF is 7CF over 36 is equal to 1. And then to solve for CF, we can multiply both sides times the reciprocal of 7 over 36. So 36 over 7. Multiply both sides times that, 36 over 7. This side, things cancel out. And we are left with-- we get our drum roll now. So all of this stuff cancels out. CF is equal to 1 times 36 over 7, or just 36 over 7. And this was a pretty cool problem, because what it shows you is if you have two things-- let's say this thing is some type of a pole or a stick or maybe the wall of a building, or who knows what it is-- if this is 9 feet tall or 9 yards tall or 9 meters tall, and this over here, this other one, is 12 meters tall or 12 yards or whatever units you want to use, if you were to drape a string from either of them to the base of the other, from the top of one of them to the base of the other-- regardless of how far apart these two things are going to be-- we just decided they're y apart. Regardless of how far apart they are, the place where those two strings would intersect are going to be 36/7 high, or 5 and 1/7 high, regardless of how far they are. So I think that was a pretty cool problem.