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Mathematics 9 Online
OpenStudy (highschoolmom2010):

Find the value of x to the nearest tenth.

OpenStudy (highschoolmom2010):

OpenStudy (anonymous):

The Diameter is given to you as 20, so the radius is 10 because r = d/2 r=20/2 r=10 It's half of the circle. Wait sorry I thought I knew where I was going with this problem XD Is there a formula to find the answer?

OpenStudy (highschoolmom2010):

not that im aware

OpenStudy (anonymous):

Lol there has got to be or else you can't solve the problem. Unless you are only supposed to expropriate . . . I believe what you are trying to find is known as a "Chord" I'm doing some research on it right now.

OpenStudy (highschoolmom2010):

im lost on this whole topic tbh

OpenStudy (anonymous):

Chord length = 2rsin(c/2) A chord is when it touches the center line of the circle, and then juts out in a different direction.

OpenStudy (anonymous):

r = 10 20sin(c/2)

OpenStudy (anonymous):

C represents the angle The angle in this question is 90 degrees

OpenStudy (anonymous):

20sin(45) = 20(sqrt2/2) = 20sqrt(2)/2 =10sqrt(2) Does that look at all familiar to you?

OpenStudy (highschoolmom2010):

eh kinda

OpenStudy (anonymous):

Okay good. I'm hoping that i'm on the right track too Xd Does it make sense to you that sin(45) = sqrt(2)/2?

OpenStudy (anonymous):

What I might be wrong about is the angle, but I feel confident with that. Let me know if you need any more help.

OpenStudy (highschoolmom2010):

i just dont understand how i am supposed to find the value for x here :/

OpenStudy (anonymous):

Okay, now remember, I'm iffy about this problem too Xd If you take the radius of 10sqrt(2), you get 5sqrt(2) You now have two sides of a triangle a^2 + b^2 = c^2 5sqrt(2)^2 + b^2 = 10^2 Solve for b, and that is your x. Sorry I have to go.

OpenStudy (disco619):

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OpenStudy (disco619):

Get the idea?

OpenStudy (highschoolmom2010):

kinda

OpenStudy (highschoolmom2010):

yes ok

OpenStudy (disco619):

\[a ^{2} + b ^{2} = c ^{2}\] \[x ^{2} + 8 = 10\] Can you solve this now?

OpenStudy (highschoolmom2010):

x=6 8^2+B^2=10^2 64+B^2=100 -64 -64 B^2=36 \[\sqrt b^2=\sqrt 36\] b=6

OpenStudy (disco619):

Oh yea, i forgot to put the ^2 with 8 and 10... \[a ^{2} + b ^{2} = c ^{2}\] \[x ^{2} + 8^{2} = 10^{2}\] \[x ^{2} + 64 = 100\] \[x ^{2} = 100 - 64\] \[x ^{2} = 36\] \[\sqrt{x} ^{2} = \sqrt{36}\] \[x = 6\] LOL Srry, i just have a habit of doing everything myself in the end to be sure :P

OpenStudy (highschoolmom2010):

thanks

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