anonymous
  • anonymous
***WILL MEDAL AND FAN*** Express answer in exact form. Find the area of the larger segment whose chord is 8" long in a circle with an 8" radius. (Hint: A chord divides a circle into two segments. In problem 1, you found the area of the smaller segment.)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
I practically already know (How) to get the answer, I just need to make it look like this::
anonymous
  • anonymous
|dw:1437655413905:dw|
anonymous
  • anonymous
And it says to drag numbers out of the box and place the apropiately in the boxes

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More answers

anonymous
  • anonymous
Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers.
anonymous
  • anonymous
That's what it tells me to do
phi
  • phi
Did you find the area of the "small" segment?
anonymous
  • anonymous
yES
anonymous
  • anonymous
dO YOU WANT IT
anonymous
  • anonymous
Woops. Caps lock again
phi
  • phi
yes
anonymous
  • anonymous
Okay. Gimmee a sec
anonymous
  • anonymous
|dw:1437655965362:dw|
anonymous
  • anonymous
There you go.
phi
  • phi
ok. the area on the other side of the chord |dw:1437656246763:dw|
phi
  • phi
the area of the entire circle \( \pi \ r^2\) = \( 64 \pi\)
phi
  • phi
thus you want to do area of circle - area of segment \[ 64 \pi - \left( \frac{64 \pi}{6} - 16 \sqrt{3}\right) \]
anonymous
  • anonymous
Okay
phi
  • phi
I would simplify that
phi
  • phi
\[ 64 \pi - \left( \frac{64 \pi}{6} - 16 \sqrt{3}\right) \\ 64 \pi -\frac{64 \pi}{6} + 16 \sqrt{3}\]
phi
  • phi
multiply the first term by 6/6 \[ \frac{6}{6} \cdot 64 \pi - \frac{64 \pi}{6} + 16 \sqrt{3}\\ \frac{6\cdot 64 \pi - 64 \pi}{6} + 16 \sqrt{3} \] can you finish?
phi
  • phi
to make it easier, I would "factor out" 64 pi so it looks like this \[ \frac{(6 -1) 64 \pi}{6} + 16 \sqrt{3} \]
anonymous
  • anonymous
So we use distribution?
phi
  • phi
distribution would "undo" the factoring out. Instead, simplify (6-1)
anonymous
  • anonymous
So 5
phi
  • phi
yes. also, we can divide top and bottom by 2
anonymous
  • anonymous
That would make the 5 uneven
phi
  • phi
you have \[ \frac{5 \cdot 64 \pi}{6} + 16 \sqrt{3} \] you can divide the 6 by 2 and the 64 by 2
anonymous
  • anonymous
Oh that makes sense
anonymous
  • anonymous
So 32 and 3?
phi
  • phi
yes, so we now have \[ \frac{5 \cdot 32 \pi}{3} + 16 \sqrt{3} \] 32 is 2*2*2*2*2 so we can't divide any more. multiply 5*32 to get 160 and the final version is \[ \frac{160}{3} \pi+ 16 \sqrt{3} \]
phi
  • phi
now you have to fill in your boxes with those numbers
anonymous
  • anonymous
Okay. Thanks that really helps
anonymous
  • anonymous
And btw,
anonymous
  • anonymous
Is there any way to possibly save conversations for future reference?
phi
  • phi
You could save the link ... I don't think OS will delete this post. Or maybe take screen shots?
anonymous
  • anonymous
Yeah, but this really isn't my personal computer though
anonymous
  • anonymous
Maybe I'll just save the link to my Google Docs account then
phi
  • phi
yes, save the link.
anonymous
  • anonymous
I will
anonymous
  • anonymous
Thanks for the medal

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