anonymous
  • anonymous
HELP PLEASE!!!! Find the area of a sector with an arc length of 60 in. and a radius of 15 in. A. 117.75 in.² B. 282.60 in.² C. 450 in.² D. 6,750 in.²
Geometry
jamiebookeater
  • jamiebookeater
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jiteshmeghwal9
  • jiteshmeghwal9
Area of a sector=\(\Large{\frac{\theta}{360} \times \pi r^2}\)
anonymous
  • anonymous
thank you!
anonymous
  • anonymous
would the answer be A?

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anonymous
  • anonymous
because it says its wrong
anonymous
  • anonymous
is 60 the theta
jim_thompson5910
  • jim_thompson5910
60 is not the theta 60 is the distance around the curve. It's the length of the arc this shaded distance |dw:1436594438367:dw|
jim_thompson5910
  • jim_thompson5910
if the radius is 15, then what is the total circumference?
anonymous
  • anonymous
30?
jim_thompson5910
  • jim_thompson5910
use the formula C = 2*pi*r
jim_thompson5910
  • jim_thompson5910
you're close
anonymous
  • anonymous
i got 94.24
anonymous
  • anonymous
when i plugged it in
jim_thompson5910
  • jim_thompson5910
yeah \[\Large 30\pi \approx 94.2477796\]
jim_thompson5910
  • jim_thompson5910
I'm going to leave it as 30pi to keep things as exact as possible
anonymous
  • anonymous
okay so what i do with that
jim_thompson5910
  • jim_thompson5910
30pi is the entire circumference the arc length is 60 we divide the two to get the fraction of the circle in which this arc is applied to \[\Large \frac{60}{30\pi} = \frac{2}{\pi}\]
jim_thompson5910
  • jim_thompson5910
now what we do is multiply that by the area pi*r^2 = pi*15^2 = 225pi \[\large 225\pi*\frac{2}{\pi} = 225\cancel{\pi}*\frac{2}{\cancel{\pi}} = 225*2 = 450\]
anonymous
  • anonymous
thank you so much!
jim_thompson5910
  • jim_thompson5910
|dw:1436594886725:dw|
jim_thompson5910
  • jim_thompson5910
no problem
anonymous
  • anonymous
can you help me with another problem? @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
sure
jim_thompson5910
  • jim_thompson5910
what's your question?
anonymous
  • anonymous
Find the length of the radius of . A. 2004-05-03-07-00_files/i0360003.jpg B. 2004-05-03-07-00_files/i0360004.jpg C. 2004-05-03-07-00_files/i0360005.jpg D. 2004-05-03-07-00_files/i0360006.jpg
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anonymous
  • anonymous
wait sorry
anonymous
  • anonymous
A. r=1 B. r= 3 C. r=12 D. r=15 i have to find the length of the radius
jim_thompson5910
  • jim_thompson5910
does it give you the entire circumference?
jim_thompson5910
  • jim_thompson5910
oh wait nvm, we don't need that info
anonymous
  • anonymous
no it does not
jim_thompson5910
  • jim_thompson5910
notice how the central angle is 4pi/9 what happens when you divide that central angle by 2pi? what do you get?
jiteshmeghwal9
  • jiteshmeghwal9
\[\frac{\theta}{360} \times 2 \pi r= s\]\[\frac{\frac{4}{9}\pi} {360} \times 2 \pi r=\frac{20}{3} \pi\]solve for 'r'.
anonymous
  • anonymous
i have to put theta over 180 pi and then pi*radius
jim_thompson5910
  • jim_thompson5910
I guess a shortcut is to use the formula \[\Large s = \theta*r\] this formula only works if theta is in radian mode. In this case, theta = 4pi/9 and s = 20pi/3 \[\Large s = \theta*r\] \[\Large \frac{20\pi}{3} = \frac{4\pi}{9}*r\] do you see how to solve for r from here?
anonymous
  • anonymous
i know that the pi's cancel out right?
jim_thompson5910
  • jim_thompson5910
yep
anonymous
  • anonymous
simplify the fractions then?
jim_thompson5910
  • jim_thompson5910
not quite
jim_thompson5910
  • jim_thompson5910
we now have 20/3 = (4/9)*r how would you isolate r ?
anonymous
  • anonymous
divide?
jim_thompson5910
  • jim_thompson5910
divide both sides by what
anonymous
  • anonymous
i'm not sure
jim_thompson5910
  • jim_thompson5910
|dw:1436596805511:dw|
jim_thompson5910
  • jim_thompson5910
or you can multiply both sides by the reciprocal of 4/9 the reciprocal of 4/9 is 9/4 |dw:1436596829027:dw|
anonymous
  • anonymous
i got 15
jim_thompson5910
  • jim_thompson5910
|dw:1436596853297:dw|
jim_thompson5910
  • jim_thompson5910
yes, r = 15
anonymous
  • anonymous
thank you!
jim_thompson5910
  • jim_thompson5910
no problem

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