I am Stuck. Help. MEDAL!!!
What is the length of stack H D with bar on top when BH = 20 in., BC = 12 in., and DC = 14 in.? Round your answer to the nearest tenth of an inch.
in.
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OpenStudy (anonymous):
Image
OpenStudy (anonymous):
anyone
OpenStudy (anonymous):
@IrishBoy123
OpenStudy (anonymous):
@Loser66 can you help me.
OpenStudy (xapproachesinfinity):
just the use of Pythatgorean theorem couple of times
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OpenStudy (loser66):
@xapproachesinfinity is a perfect guy for this
OpenStudy (anonymous):
ok
OpenStudy (xapproachesinfinity):
ok let come from a clearer picture!
you recognize HDC is right triangle yes
OpenStudy (anonymous):
yes
OpenStudy (xapproachesinfinity):
we are given DC
but we need HC
to do the so called Pythagoras yes
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OpenStudy (anonymous):
yes
OpenStudy (xapproachesinfinity):
now what we do is look at the triangle HCB
it is also a right triangle
and we are given BC and BH
so we can find HC
OpenStudy (anonymous):
ok how?
OpenStudy (xapproachesinfinity):
so then we have the following:\
HC^2+BC^2=BH^2
HC^2=BH^2-BC^2
HC^2=20^2-12^2
OpenStudy (xapproachesinfinity):
now tell me what is after you take the square root
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OpenStudy (xapproachesinfinity):
you have a calculator with right
OpenStudy (xapproachesinfinity):
i got HC=16
OpenStudy (anonymous):
me too
OpenStudy (xapproachesinfinity):
now we go back to our first triangle HDC
now we have HC=16 and DC=14
we use again that same theorem to find HD
HD^2+DC^2=HC^2
HD^2=HC^2-DC^2
HD^2=16^2-14^2
OpenStudy (xapproachesinfinity):
let see what you got
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OpenStudy (anonymous):
60?
OpenStudy (xapproachesinfinity):
sqrt of 60 not just 60
because i told you to take square root after that too
OpenStudy (xapproachesinfinity):
so then HD=7.75 inch
OpenStudy (anonymous):
is that the answer?
OpenStudy (xapproachesinfinity):
if we wanted the decimal, well in a way we want the decimal since we measuring length
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