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

- anonymous

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- jiteshmeghwal9

Area of a sector=\(\Large{\frac{\theta}{360} \times \pi r^2}\)

- anonymous

thank you!

- anonymous

would the answer be A?

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## More answers

- anonymous

because it says its wrong

- anonymous

is 60 the theta

- 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

if the radius is 15, then what is the total circumference?

- anonymous

30?

- jim_thompson5910

use the formula
C = 2*pi*r

- jim_thompson5910

you're close

- anonymous

i got 94.24

- anonymous

when i plugged it in

- jim_thompson5910

yeah \[\Large 30\pi \approx 94.2477796\]

- jim_thompson5910

I'm going to leave it as 30pi to keep things as exact as possible

- anonymous

okay so what i do with that

- 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

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

thank you so much!

- jim_thompson5910

|dw:1436594886725:dw|

- jim_thompson5910

no problem

- anonymous

can you help me with another problem? @jim_thompson5910

- jim_thompson5910

sure

- jim_thompson5910

what's your question?

- 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

##### 1 Attachment

- anonymous

wait sorry

- anonymous

A. r=1
B. r= 3
C. r=12
D. r=15
i have to find the length of the radius

- jim_thompson5910

does it give you the entire circumference?

- jim_thompson5910

oh wait nvm, we don't need that info

- anonymous

no it does not

- 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

\[\frac{\theta}{360} \times 2 \pi r= s\]\[\frac{\frac{4}{9}\pi} {360} \times 2 \pi r=\frac{20}{3} \pi\]solve for 'r'.

- anonymous

i have to put theta over 180 pi and then pi*radius

- 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

i know that the pi's cancel out right?

- jim_thompson5910

yep

- anonymous

simplify the fractions then?

- jim_thompson5910

not quite

- jim_thompson5910

we now have 20/3 = (4/9)*r
how would you isolate r ?

- anonymous

divide?

- jim_thompson5910

divide both sides by what

- anonymous

i'm not sure

- jim_thompson5910

|dw:1436596805511:dw|

- 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

i got 15

- jim_thompson5910

|dw:1436596853297:dw|

- jim_thompson5910

yes, r = 15

- anonymous

thank you!

- jim_thompson5910

no problem

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