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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

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

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

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