Hsy there
#Geometric
edgy steeple
fervent nimbusBOT
Hsy there
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edgy steeple
Hi can someone teach me this question?
I just did h = √(3)x/2 so far
Am i in the right track?
And what to do next step ?
grave cipher
there are actually a lot of ways you can do this
edgy steeple
Hi... is you again glad you're here to help
grave cipher
if you label the vertices A, B, C and call your circumcenter O, you could find <AOC?
edgy steeple
if you label the vertices A, B, C and call your circumcenter O, you could find <...
120
edgy steeple
typo
grave cipher
now since AOC is isosceles, you can find <OAC too
edgy steeple
well i just let the length randomly "x" the midpoint AC is x/2
grave cipher
and then since AOC is isosceles, OM should be perpendicular to AC
OMA is similar to BMA, and BM is equal to the height
grave cipher
not quite
you have OA/AM = AB/BM
which means OA/(x/2) = x/(√(3)x/2)
edited for clarity
grave cipher
which means OA/(x/2) = x/(√(3)x/2)
you can find OA in terms of x
sorry i’m rushing; my phone is dying
edgy steeple
sorry i’m rushing; my phone is dying
√(3)x/2 no x with it ?
edgy steeple
sorry i’m rushing; my phone is dying
ahhh nevermind i also have to hurry gotta sleep
edgy steeple
which means OA/(x/2) = x/(√(3)x/2)
OA = x^2/√(3)