how do you get this length...

- anonymous

how do you get this length...

- chestercat

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- amistre64

i feel that your questions might be lacking some vital components :/

- anonymous

|dw:1327247673332:dw|

- amistre64

thats a scaled version of the base

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## More answers

- anonymous

That's a secret length.

- amistre64

if you knew some angles itd be doable as well

- anonymous

I think we can use Euler straight edge for that.

- AravindG

wt kind of triangle is it???

- amistre64

did Fool just suggest a ruler? lol

- anonymous

i don't need a numerical answer. length in question is c, height of c from base b is y

- anonymous

not a right trianle

- amistre64

|dw:1327247884124:dw|

- anonymous

just need to solve in terms of y. no angle given

- anonymous

^^^^yes

- amistre64

\[\frac{c}{h}=\frac{b}{h-y}\]
\[c=\frac{bh}{h-y}\]is my only idea without knowing any fancy triangle thrms

- amistre64

where h = height of the altitude

- anonymous

hmmm i hate answering "none of the above"

- amistre64

we have options?

- anonymous

yes

- amistre64

way to make life harder for us by not posting them, kudos :)

- anonymous

Yes amstre that's a ruler :D

- anonymous

a) (h-y)(b/h)
b)(h-y)(h/b)
c)(b/h)y
d)(h/b)y

- anonymous

sorry, i wasn't trying to get a straight answer in the first place, i want to solve :P

- anonymous

You need to use basic proportionality theorem.

- amistre64

i think my solution is adequate then, if we can rearrange it into one of those formats

- amistre64

\[\frac{bh}{h-y}\]
\[b\frac{h}{h-y}\]
\[b\frac{1}{1-y/h}\]
maybe the wolf can give us options?

- amistre64

it was worth a shot, check my proportion; is it ok?

- anonymous

about to just plug in numbers and see

- amistre64

|dw:1327248716441:dw|

- amistre64

this is the same type of set up but with something i can see better

- anonymous

your proportions aren't right but i also don't know if this pic is drawn to scale. h is longer than b, c is shorter than b, of coarse. and y is shorter than b and c

- anonymous

from your expressions, c ends up being longer than b

- amistre64

3/4 = 2/c
c = 8/3 = 2 2/3?
sqrt(2^2 + 8/3^2)= 10/3
5/3 = d/2
10/3 = d
thats a match in my book

- amistre64

c is to b
as
h-y is to h
\[c=\frac{b}{h}(h-y)\]might be better :)

- amistre64

i think that matches your "a" option

- anonymous

yes, it's exact. i don't get why though. so did you derive "a" from your earilier expression c=(bh)/(h-y)?

- amistre64

no, i had to dust out the cobwebs and make sure I was putting things in their proper places.

- amistre64

similar triangle differ by some scalar multiple; they keep the same ratios from one tri to the next

- amistre64

so if we equate the proper ratios we can solve for them

- amistre64

bases with repect to heights will math in silimlar tris.
b/h = c/(h-y) are the respective parts to match up

- amistre64

then we solve for c

- anonymous

i think i need to study "basic proportiionality theorem". don't recall learning about it

- anonymous

thx for your help

- amistre64

your welcome, and good luck :)

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