Mathematics
8 Online
OpenStudy (leenathan):
helpppppp plz
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OpenStudy (mecharv):
Hey, do you know Pythagoras theorem?
OpenStudy (leenathan):
yus i think the answer is B
OpenStudy (mecharv):
|dw:1480695288154:dw|
here, \[(AC)^2=(AB)^2+(BC)^2\]
OpenStudy (mecharv):
How did you come up with B?
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OpenStudy (kaithang):
\[a ^{2 } + b ^{2} = c ^{2}\]
OpenStudy (kaithang):
Therefore 2 + ? = 4
OpenStudy (leenathan):
i asked my friend im trying to solve it now tho
OpenStudy (leenathan):
2+(2)=4
OpenStudy (leenathan):
oh so a
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OpenStudy (mecharv):
It is given \[AC=4units \]\[BC/AB=2units\]
OpenStudy (kaithang):
Idk if that is right but I' think it is
OpenStudy (mecharv):
So, \[4^2=2^2+(AB)^2\]
OpenStudy (leenathan):
hmmm i think its right but i want to make sure its correct
OpenStudy (mecharv):
\[16=4+(AB)^2\]
So now can you figure out what AB must be?
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OpenStudy (kaithang):
OMg that makes since @mecharv I prolly confused the heck out of @leenathan
OpenStudy (leenathan):
wat?
OpenStudy (mecharv):
@kaithang that is okay :)
You too got to learn something.
Now @leenathan what is the answer?
OpenStudy (leenathan):
a
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OpenStudy (mecharv):
No. How is it a?
I have almost given you the answer....
OpenStudy (leenathan):
@mecharv 12?
OpenStudy (leenathan):
A^2 + B^2 = C^2
we have A = 2 and C = 4
we need to find B
2^2 + B^2 = 4^2
4 + B^2 = 16
B^2 = 12
B = sqrt12
OpenStudy (leenathan):
c
OpenStudy (mecharv):
Yes that is correct :)
Always work out your problems mate. Cheers :)