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Geometry 21 Online
OpenStudy (anonymous):

help plz What is the value of x to the nearest tenth? https://fisd3939-acp-ccl.gradpoint.com/Resource/3688351,8BD,0,0,0,0,0/Assets/testitemimages/geometry_b/circles/mc034-1.jpg

OpenStudy (anonymous):

First of all, Welcome to Open Study!

OpenStudy (anonymous):

thanks i have no idea how to do this

OpenStudy (anonymous):

Sorry don't know how to do this...but @Luigi0210 might

OpenStudy (luigi0210):

@ganeshie8

OpenStudy (luigi0210):

I think you have to use \(\Large a^2+b^2=c^2 \) right?

OpenStudy (anonymous):

i think so

OpenStudy (luigi0210):

Hmm, okay, I'mma try and see if I can solve it since Gane isn't here yet >.< So what I would do is use \(a^2+b^2=c^2 \) Since we have the diameter as 16, we can get the radius by cutting it in half and use that as the hypotenuse: \( a^2+b^2=8^2 \) That 6.5 could be used as one of the sides of the triangle: \( 6.5^2+b^2=8^2 \) So we have this triangle: |dw:1423514990365:dw| So just solve this equation for b and you should get x: \(\Large \color{green}{6.5^2+b^2=8^2} \) Do you have any answer choices?

OpenStudy (anonymous):

yea 1 sec

OpenStudy (anonymous):

yeah i think that was right

OpenStudy (anonymous):

there is 4.7

OpenStudy (luigi0210):

Awesome :D Did the explanation make sense tho?

OpenStudy (anonymous):

yea thanks!

OpenStudy (luigi0210):

You're welcome :)

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