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Mathematics 14 Online
OpenStudy (christina166):

The diagonal of a square is 8 units. Find the length of the side.

OpenStudy (sshayer):

use pythagoras's theorm to find the side.

OpenStudy (christina166):

what is the pythagoras theorm

OpenStudy (sshayer):

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OpenStudy (christina166):

lol sorry

OpenStudy (christina166):

the diagonal is 8 but in order to use that theorm dont you need 2 diagonals

OpenStudy (sshayer):

i have shown one diagonal is sufficient. Diagonals of a square are equal.

OpenStudy (christina166):

okay

OpenStudy (christina166):

now what do i do :/ im sooo sorry i really am dumb

OpenStudy (sshayer):

x^2+x^2=8^2 find x

OpenStudy (christina166):

okay

OpenStudy (christina166):

im sorry it would be x^2=32

OpenStudy (christina166):

hello?

OpenStudy (mathstudent55):

correct keep going and solve for x

OpenStudy (christina166):

how do i eliminate the ^2

OpenStudy (mathstudent55):

Since you have \(x^2 = 32\), and you want just x, you take the square root of each side.

OpenStudy (christina166):

okay but how do i cancel it out

OpenStudy (christina166):

thanks for taking your time for this by the way

OpenStudy (mathstudent55):

\(x^2 = 32\) \(\sqrt{x^2} = \sqrt{32} \) \(x = \sqrt {32} \)

OpenStudy (mathstudent55):

You're welcome.

OpenStudy (christina166):

it equals to 5.66

OpenStudy (christina166):

THANK YOU SO MUCH!!!!

OpenStudy (mathstudent55):

That is an approximation. If you are asked for the exact length, then it is \(\sqrt{32} = \sqrt{16 \times 2} = 4 \sqrt 2\)

OpenStudy (mathstudent55):

You're welcome.

OpenStudy (christina166):

OHH OKAY

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