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Geometry 22 Online
OpenStudy (anonymous):

Please help me with this problem! I will attach the problem. Thank you very much! :)

OpenStudy (anonymous):

OpenStudy (anonymous):

I tried 4+3 = 7 7/x = x/3 x^2 = 21 x = sq root of 36 But that's not right!

OpenStudy (anonymous):

I meant sq root of 21

OpenStudy (e.mccormick):

Unless it is supposed to be some ratio, I am not sure... but I do not see proper information to assume it is a ratio. =/

OpenStudy (anonymous):

I also tried 5+4+3 = 12 12/x = x/3 x=36 squared but that is 6 and not an answer! anyone??????

OpenStudy (e.mccormick):

It is not a right triangle as far as I can tell....

OpenStudy (anonymous):

so where do I start?

OpenStudy (e.mccormick):

Have you been studying ratios recently or just stuff with triangles?

OpenStudy (anonymous):

ratios and geometric means so it's a ratio??

OpenStudy (e.mccormick):

I do not see how a geometric mean would apply... I think they are trying to say these are similar triangles... but doing it badly. That is a guess on my part, but if you assume that and do a ratio, it does give an answer that rounds to one of those.

OpenStudy (anonymous):

not getting it

OpenStudy (anonymous):

can you get me started??

OpenStudy (e.mccormick):

The bat that has been haning around pointed out Angle Bisector Theorem... which looks like it may show it would be a ratio. Have you done anything with that? http://www.cut-the-knot.org/triangle/AngleBisectorTheorem.shtml

OpenStudy (anonymous):

Well the line is a bisector of the angle, so we can set it up as follow \[\frac{ x }{ 5 } = \frac{ 3}{ 4 }\]

OpenStudy (anonymous):

x/5 = 3/4 x4/4 = 15/4 x=3.75 or 3.8

OpenStudy (anonymous):

Rounded to the nearest tenth, 3.8 sounds good to me

OpenStudy (anonymous):

Thank you!!

OpenStudy (anonymous):

Np know if you have a triangle as such |dw:1425426068694:dw| we can see we have the following \[\frac{ CA }{ CD } = \frac{ BA }{ DB }\] this is the angle bisector theorem. It has been quite some time I've used it myself, but you can see it's pretty useful :)

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