## anonymous one year ago The figure shows three right triangles. Triangles JKM, KLM, and JLK are similar.

1. anonymous

Theorem: If two triangles are similar, the corresponding sides are in proportion.

2. anonymous

3. anonymous

Using the given theorem, which two statements help to prove that if segment JL is x, then x2 = 100? A.Segment JL • segment JM = 64 Segment JL • segment LM = 48 B. Segment JL • segment JM = 48 Segment JL • segment LM = 36 C. Segment JL • segment JM = 64 Segment JL • segment LM = 36 D.Segment JL • segment JM = 36 Segment JL • segment LM = 64

4. anonymous

anyone know how to do this?

5. anonymous

do you know? @Calcmathlete

6. anonymous

I know how to do the problem, but it's worded very strangely to me. It would make much more sense for x to be JK, KL, or KM. Give me a second, and I'll help you.

7. anonymous

Ok. Have you done work with geometric means in these types of problems before?

8. anonymous

no its actually a pretest so i havent done this before

9. anonymous

but if x^2 is 100 then x must be 10

10. anonymous

JL is 10 i mean

11. anonymous

i thought it was x2 not x^2

12. anonymous

Whenever you copy and paste, an exponent, it usually pastes it as x2 or something. @Venice-Gribbin

13. anonymous

sorry hehe its x squared

14. anonymous

i see.

15. anonymous

Alright. Before we start, you may have noticed that you could've used Pythagorean Theorem to verify that JL = 10. Did you notice that?

16. anonymous

yeah 36 plus 64=100

17. anonymous

Ok. That's something that happens a lot in geometry. The key is that you need to use the given, which is the theorem they gave you. So, looking at similarity. The way the triangles are written are very significant. The order in which the letters appear tell you which sides are proportionate to what. So, if they were written as (JKM, KLM, and JLK) as you said, JL is proportionate to KL and JK. So, you could say that:$\frac{JL}{KL}=\frac{JK}{KM}$Are you following me so far?

18. anonymous

yes so can we plug in their values into this proportion?

19. anonymous

$\frac{ 10 }{ 8 }=\frac{ 6 }{ ? }$

20. anonymous

If you have them, yes. Now, with this type of problem, you can only work with the values they gave you, even though it's obvious that JL does indeed equal 10. So, the only values you can actually plug in are JK and KL. The above proportion won't help you here. That was just for example. Hold on, and I'll help you get the right equation.

21. anonymous

Lol @Gjallerhorn your name is perfect man

22. anonymous

fellow destiny player?

23. anonymous

Yeah I play it on xbox one. But before we continue this convo I want you to succeed in your math lol so ill wait till you get the answer

24. anonymous

anyone who is familiar with Feizel crux's masterpiece knows their "Destiny..."

25. anonymous

lolwut

26. anonymous

I'll tell you how I got this specific proportion afterwards, but the one that will help you is$\frac{JM}{JK}=\frac{JK}{JL}$This can be simplified to $JM\times JL=JK^2$Now, plug in the values you have.

27. anonymous

into the proportion or the equation under it?

28. anonymous

The equation under it. I got the bottom equation by cross multiplying. The only values they gave you were JK and KL.

29. anonymous

JMxJL=6^2 or 36?

30. anonymous

Yes. That gives you the answer you need because none of the other answers have that as part of the answer. If you wanted to completely do the question, the other proportion you would use is:$\frac{JL}{KL}=\frac{KL}{LM}$which simplifies to $JL\times LM=KL^2$

31. anonymous

ok lemme plug in

32. anonymous

JLxLM=64 or 8^2

33. anonymous

with those two values my answer would be D. correct?

34. anonymous

Yes. That's correct. I'll write how I got those specific proportions in a second.

35. anonymous

Now, the way I got those proportions is something called geometric means. A geometric mean between two numbers a and b is defined as $Geometric ~Mean=\sqrt{ab}$or you can also say $(Geometric~Mean)^2=a\times b$Whenever you have a situation like this where you have the three similar right triangles with 2 of them within the bigger one, there are three relationships of "geometric means" that are extremely helpful (they are just similar triangles, so you can derive them later too). |dw:1436552274512:dw|

36. anonymous

One of the geometric mean relationships is that BD is the geometric mean of AD and DC. In other words, $\frac{AD}{BD}=\frac{BD}{DC}$or$AD\times DC=BD^2$which is a geometric mean. The other two are that BC is the geometric mean of CD and AC and that AB is the geometric mean of AD and AC.

37. anonymous

I remember it with the pneumonic "ALL".|dw:1436552524856:dw| That's the 'A' part of it.

38. anonymous

These are the other two 'L' |dw:1436552558243:dw| |dw:1436552572987:dw| Sorry that the explanation is so long, just thought it'd be good if you knew where I got the information from.

39. anonymous

wow thanks so much!!!!

40. anonymous

No problem. If you need me to explain anything else, let me know.

41. anonymous

Also, keep in mind that they need to be right triangles for these things to work because otherwise, they wouldn't be similar triangles.

42. anonymous

im definitely fanning you.

43. anonymous

thanks for the guidance and explanation!

44. anonymous

No problem. I'm not online very often tbh, but when I am, I try to help as many people as possible.