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Geometry 7 Online
OpenStudy (anonymous):

can someone answer this for me !! what is the value of x? (A) 6 (B) 16/3 (C) 27 (D) sqaure root over 27 (E) 27/16 (F) square root over 324

OpenStudy (anonymous):

what is the value of x?

OpenStudy (anonymous):

using pythagora's theorem

OpenStudy (anonymous):

find AC.

OpenStudy (anonymous):

wads AC? did u get \[\sqrt{585}\] ?

OpenStudy (anonymous):

no ..?

OpenStudy (anonymous):

the answer choices are up there

OpenStudy (anonymous):

\[\sqrt{27^2 -12^2} \] ?

OpenStudy (anonymous):

i need to guide u because givin u the answer is against policy of open study

OpenStudy (anonymous):

ok thats better

OpenStudy (anonymous):

ok so using AC^2-(27-x)^2 = 12^2-x^2

OpenStudy (anonymous):

can u find x now?

OpenStudy (anonymous):

i dont understand how to do thaty ! :(

OpenStudy (anonymous):

because by pythagora's theorem, AC^2-DC^2=AD^2=AB^2-BD^2 , which give u the above equation

OpenStudy (anonymous):

As AC^2=585, x is 16/3

OpenStudy (anonymous):

try to digest it

OpenStudy (anonymous):

hm thank u

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