anonymous
  • anonymous
how do you get this length...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
i feel that your questions might be lacking some vital components :/
anonymous
  • anonymous
|dw:1327247673332:dw|
amistre64
  • amistre64
thats a scaled version of the base

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More answers

anonymous
  • anonymous
That's a secret length.
amistre64
  • amistre64
if you knew some angles itd be doable as well
anonymous
  • anonymous
I think we can use Euler straight edge for that.
AravindG
  • AravindG
wt kind of triangle is it???
amistre64
  • amistre64
did Fool just suggest a ruler? lol
anonymous
  • anonymous
i don't need a numerical answer. length in question is c, height of c from base b is y
anonymous
  • anonymous
not a right trianle
amistre64
  • amistre64
|dw:1327247884124:dw|
anonymous
  • anonymous
just need to solve in terms of y. no angle given
anonymous
  • anonymous
^^^^yes
amistre64
  • amistre64
\[\frac{c}{h}=\frac{b}{h-y}\] \[c=\frac{bh}{h-y}\]is my only idea without knowing any fancy triangle thrms
amistre64
  • amistre64
where h = height of the altitude
anonymous
  • anonymous
hmmm i hate answering "none of the above"
amistre64
  • amistre64
we have options?
anonymous
  • anonymous
yes
amistre64
  • amistre64
way to make life harder for us by not posting them, kudos :)
anonymous
  • anonymous
Yes amstre that's a ruler :D
anonymous
  • anonymous
a) (h-y)(b/h) b)(h-y)(h/b) c)(b/h)y d)(h/b)y
anonymous
  • anonymous
sorry, i wasn't trying to get a straight answer in the first place, i want to solve :P
anonymous
  • anonymous
You need to use basic proportionality theorem.
amistre64
  • amistre64
i think my solution is adequate then, if we can rearrange it into one of those formats
amistre64
  • amistre64
\[\frac{bh}{h-y}\] \[b\frac{h}{h-y}\] \[b\frac{1}{1-y/h}\] maybe the wolf can give us options?
amistre64
  • amistre64
it was worth a shot, check my proportion; is it ok?
anonymous
  • anonymous
about to just plug in numbers and see
amistre64
  • amistre64
|dw:1327248716441:dw|
amistre64
  • amistre64
this is the same type of set up but with something i can see better
anonymous
  • 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
  • anonymous
from your expressions, c ends up being longer than b
amistre64
  • 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
  • amistre64
c is to b as h-y is to h \[c=\frac{b}{h}(h-y)\]might be better :)
amistre64
  • amistre64
i think that matches your "a" option
anonymous
  • 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
  • amistre64
no, i had to dust out the cobwebs and make sure I was putting things in their proper places.
amistre64
  • amistre64
similar triangle differ by some scalar multiple; they keep the same ratios from one tri to the next
amistre64
  • amistre64
so if we equate the proper ratios we can solve for them
amistre64
  • amistre64
bases with repect to heights will math in silimlar tris. b/h = c/(h-y) are the respective parts to match up
amistre64
  • amistre64
then we solve for c
anonymous
  • anonymous
i think i need to study "basic proportiionality theorem". don't recall learning about it
anonymous
  • anonymous
thx for your help
amistre64
  • amistre64
your welcome, and good luck :)

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