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Geometry 16 Online
OpenStudy (anonymous):

if the lengths of the segments are 5 inches and 20 inches, find the length of the altitude

jigglypuff314 (jigglypuff314):

could you try providing a diagram? :)

OpenStudy (anonymous):

the problem doesn't provide a diagram

jigglypuff314 (jigglypuff314):

then what topic is this covering? :) triangles? parallelograms?

OpenStudy (anonymous):

Ratio, proportion, and similarity

jigglypuff314 (jigglypuff314):

are you sure there isn't a picture of the shape or something that goes alone with it? :/

OpenStudy (anonymous):

no its just a problem to solve

jigglypuff314 (jigglypuff314):

then I'm not sure... sorry... perhaps I can call for some help from some others that might know... @jdoe0001 @Loser66 @agent0smith

OpenStudy (anonymous):

yes could you please

OpenStudy (jdoe0001):

looks pretty ambiguous

OpenStudy (inkyvoyd):

@sharona ,either you have failed to provide us a diagram, or you are missing information. Check the problem again.

OpenStudy (jdoe0001):

a segment it's just a line could be altitude or any part of any polygon really

OpenStudy (jdoe0001):

got picture?

OpenStudy (anonymous):

no no picture just applying skills

OpenStudy (jdoe0001):

hehe, clairvoyance skills? =)

OpenStudy (inkyvoyd):

I can't help you, and I doubt anyone else can help you with the information you are given. Could you please provide a screenshot of the problem?

OpenStudy (jdoe0001):

well, to get an altitude, you'd need more info than that

OpenStudy (anonymous):

the altitude to the hypotenuse of a right triangle divides the hypotenuse into two segments

OpenStudy (anonymous):

thats all it says her and then they give me the problem

OpenStudy (inkyvoyd):

|dw:1389139789748:dw|

OpenStudy (anonymous):

here*

OpenStudy (inkyvoyd):

NOW the problem is solvable.

OpenStudy (anonymous):

yes this looks right can you please help me solve it

OpenStudy (jdoe0001):

hehehe, so there's a picture you didn't provide =)

OpenStudy (inkyvoyd):

|dw:1389139837208:dw| the three triangles are similar by AA similarity. That is, the two inner ones and the big one. Therfore take 5/x=x/20 multiply both sides by x: 5=x^2/20 x^2-100=0 Solve that equation and take the positive solution, which will be your side length.

OpenStudy (anonymous):

no there is no picture i just learned this today in class and i am just so confused

OpenStudy (jdoe0001):

ok.... so.... what you're after is the so-called "geometric mean" so following the picture by inkyvoyd it'd be like -> http://www.mathwarehouse.com/geometry/similar/triangles/geometric-mean.php notice in the picture below here running a line likewise from the top right-angle vertex, making right angle at the bottom gives 3 SIMILAR triangles the 2 smaller ones contained and the bigger one containing the smaller 2 so you use their ratio

OpenStudy (jdoe0001):

and solve for "x"

OpenStudy (anonymous):

okay thanks a lot

OpenStudy (jdoe0001):

yw

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