If the circle with center O has area 9 pi, what is the area of equilateral triangle ABC?

- anonymous

If the circle with center O has area 9 pi, what is the area of equilateral triangle ABC?

- chestercat

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- anonymous

|dw:1441314908068:dw|

- misty1212

HI!!

- anonymous

hey misty!

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## More answers

- misty1212

area is \(9\pi \)r right?

- anonymous

yea so the diameter is 6

- misty1212

ok so you got that the height is 6

- anonymous

|dw:1441315064479:dw|

- 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

you are way ahead of me

- anonymous

|dw:1441315150836:dw|

- misty1212

yeah each side of your triangle is \[\frac{2\times 6}{\sqrt3}\]

- misty1212

you lost me there
height of the triangle is 6

- misty1212

|dw:1441315234092:dw|

- misty1212

the short leg of either right triangle on the left and right is \(\frac{6}{\sqrt{3}}\)

- misty1212

area of each right triangle is base times height

- misty1212

rather one half base times height
but together it is base times height so
\[6\times \frac{6}{\sqrt3}\]

- misty1212

|dw:1441315367905:dw|

- anonymous

wait what? so \[4\sqrt{3}\] is for the bottom left and right of the traingles?

- anonymous

how why?

- misty1212

|dw:1441315446382:dw|

- misty1212

the short leg is the long leg divided by root 3

- misty1212

so the area of the two right triangles taken together is base times height

- misty1212

ok i see \[\frac{6}{\sqrt{3}}=2\sqrt3\]

- misty1212

so total area is
\[6\times 2\sqrt3 = 12\sqrt 3\]

- anonymous

fml i've been doin it all wrong this whole time

- 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

So |dw:1441315715658:dw|

- misty1212

yeah that looks good to me

- anonymous

i dont get why it's 2 root 3 is the problem like why does it convert that way and the process

- anonymous

i thought in a 30-60-90 triangle the root three would replace the 6

- misty1212

k lets go slow
you said you had a 30 - 60 -90 triangle right?

- anonymous

yea

- misty1212

the ratios are \[1:\sqrt3:2\] for short leg ; long leg ; hypotenuse

- anonymous

yea

- misty1212

the long leg here is 6

- misty1212

so the short leg is \[\frac{6}{\sqrt3}=2\sqrt3\]

- anonymous

how?

- misty1212

you multiply the short leg by root 3 to get the long leg right?

- misty1212

so to go the other way, from the long leg to the short leg, you divide

- anonymous

yea but like why is it that way? shouldn't the short leg not contain the root 3?

- misty1212

lol no it does here

- anonymous

that so weird

- misty1212

if the short leg was a whole number the long leg would contain root 3

- misty1212

but the long leg is a whole number, so the short leg as the root 3 in it

- misty1212

it is not weird, only seems weird because you usually see it with the short leg a whole number

- anonymous

hmmmm yea i guess so

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