Find the value of x to the nearest tenth.
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?
not that im aware
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.
im lost on this whole topic tbh
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.
r = 10 20sin(c/2)
C represents the angle The angle in this question is 90 degrees
20sin(45) = 20(sqrt2/2) = 20sqrt(2)/2 =10sqrt(2) Does that look at all familiar to you?
eh kinda
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?
What I might be wrong about is the angle, but I feel confident with that. Let me know if you need any more help.
i just dont understand how i am supposed to find the value for x here :/
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.
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Get the idea?
kinda
yes ok
\[a ^{2} + b ^{2} = c ^{2}\] \[x ^{2} + 8 = 10\] Can you solve this now?
x=6 8^2+B^2=10^2 64+B^2=100 -64 -64 B^2=36 \[\sqrt b^2=\sqrt 36\] b=6
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
thanks
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