anonymous
  • anonymous
If the circle with center O has area 9 pi, what is the area of equilateral triangle ABC?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1441314908068:dw|
misty1212
  • misty1212
HI!!
anonymous
  • anonymous
hey misty!

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misty1212
  • misty1212
area is \(9\pi \)r right?
anonymous
  • anonymous
yea so the diameter is 6
misty1212
  • misty1212
ok so you got that the height is 6
anonymous
  • anonymous
|dw:1441315064479:dw|
anonymous
  • anonymous
^is that for the height of the triangle? Like we're converting the 6 to an apropriate side that corresponds to the 30-60-90 triangle right?
misty1212
  • misty1212
you are way ahead of me
anonymous
  • anonymous
|dw:1441315150836:dw|
misty1212
  • misty1212
yeah each side of your triangle is \[\frac{2\times 6}{\sqrt3}\]
misty1212
  • misty1212
you lost me there height of the triangle is 6
misty1212
  • misty1212
|dw:1441315234092:dw|
misty1212
  • misty1212
the short leg of either right triangle on the left and right is \(\frac{6}{\sqrt{3}}\)
misty1212
  • misty1212
area of each right triangle is base times height
misty1212
  • misty1212
rather one half base times height but together it is base times height so \[6\times \frac{6}{\sqrt3}\]
misty1212
  • misty1212
|dw:1441315367905:dw|
anonymous
  • anonymous
wait what? so \[4\sqrt{3}\] is for the bottom left and right of the traingles?
anonymous
  • anonymous
how why?
misty1212
  • misty1212
|dw:1441315446382:dw|
misty1212
  • misty1212
the short leg is the long leg divided by root 3
misty1212
  • misty1212
so the area of the two right triangles taken together is base times height
misty1212
  • misty1212
ok i see \[\frac{6}{\sqrt{3}}=2\sqrt3\]
misty1212
  • misty1212
so total area is \[6\times 2\sqrt3 = 12\sqrt 3\]
anonymous
  • anonymous
fml i've been doin it all wrong this whole time
misty1212
  • misty1212
you got the height right away then i think you confused yourself and turned the height in to something else first you said it was 6 (correct) then you said it was \(4\sqrt3\)
anonymous
  • anonymous
So |dw:1441315715658:dw|
misty1212
  • misty1212
yeah that looks good to me
anonymous
  • anonymous
i dont get why it's 2 root 3 is the problem like why does it convert that way and the process
anonymous
  • anonymous
i thought in a 30-60-90 triangle the root three would replace the 6
misty1212
  • misty1212
k lets go slow you said you had a 30 - 60 -90 triangle right?
anonymous
  • anonymous
yea
misty1212
  • misty1212
the ratios are \[1:\sqrt3:2\] for short leg ; long leg ; hypotenuse
anonymous
  • anonymous
yea
misty1212
  • misty1212
the long leg here is 6
misty1212
  • misty1212
so the short leg is \[\frac{6}{\sqrt3}=2\sqrt3\]
anonymous
  • anonymous
how?
misty1212
  • misty1212
you multiply the short leg by root 3 to get the long leg right?
misty1212
  • misty1212
so to go the other way, from the long leg to the short leg, you divide
anonymous
  • anonymous
yea but like why is it that way? shouldn't the short leg not contain the root 3?
misty1212
  • misty1212
lol no it does here
anonymous
  • anonymous
that so weird
misty1212
  • misty1212
if the short leg was a whole number the long leg would contain root 3
misty1212
  • misty1212
but the long leg is a whole number, so the short leg as the root 3 in it
misty1212
  • misty1212
it is not weird, only seems weird because you usually see it with the short leg a whole number
anonymous
  • anonymous
hmmmm yea i guess so

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