In circle O, CD = 44, OM = 20, ON = 19,CD l OM, EF l ON (The figure is not drawn to scale.)

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In circle O, CD = 44, OM = 20, ON = 19,CD l OM, EF l ON (The figure is not drawn to scale.)

Mathematics
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a.Find the radius. If your answer is not an integer, express it in radical form. b.Find FN. If your answer is not an integer, express it in radical form. c.Find EF. Express it as a decimal rounded to the nearest tenth.

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|dw:1360903051401:dw|
|dw:1360903107022:dw|
i hate geometry..lol
|dw:1360903151843:dw| solve for x to find the radius a^2 + b^2 = c^2 20^2 + 22^2 = x^2
is this the answer
tell me what you get for x
42?
oops 22
20^2 + 22^2 = x^2 400 + 484 = x^2 x^2 = 884 x = ???
52
i don't know
x^2 = 884 x = sqrt(884) x = 2*sqrt(221)
so thats your radius
now you can use this radius to find EF and FN
202 + 222
solve for y y^2 + 19^2 = (2*sqrt(221))^2 y^2 + 381 = 884 keep going
What do you do after you solve for y? @jim_thompson5910
I let y = EN so to find EF, you need to find 2y this is because EF = 2*EN EF = 2y
also, FN = EN, so once you get EN, you have found FN
y^2 + 361=884?? y^2=523 right?
oh somehow I had 381...hmm lol yeah it's 361
y^2=523 y=sqrt(523)
How did you get 884 up there?
(2*sqrt(221))^2 = 4*221 = 884
alright so y=29.73?
leave it in radical form
unless it asks about the approximation and rounding it
to leave it in radical form do I just leave it as y=sqrt(523)?
yes square roots are considered radicals
Solving for y solves for both FN and EF..?
FN yes since FN = EN
but you have to double it to find EF
So the one we just solved (y) is FN right? and I add FN and EN together.. but how do I do that since its in radical form?
or you can double FN to get EF
1,046? right
is EF
no EF = 2*sqrt(523)
How do I double Fn to get Ef
I just did it
-___- if 2*sqrt(523) is EF then half of that is Fn and En right..
correct
Alright lol sorry I was confused on which was which. How do I find out what half of the radical is? just divide 523 by 2?
it's just 2*sqrt(523) and sqrt(523)
EF=2*sqrt(523) and EN and FN = sqrt(523) .. wow haha I feel stupid. Thank you so much though!
it's fine, yw
to find 2*sqrt(523) as a decimal rounded to the nearest tenth I multiply 523*2 and then take the square root of that number right?
you first take the square root of 523, then you double that result
then you round to whatever they want
so the answer would be 45.74?
yep 2*sqrt(523) = 45.74 roughly
Thank you again
sure thing

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