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.)
and is there any way to put the answer in this format?
https://suwannee.owschools.com/media/g_geo_2013/8/q255_x.gif

- anonymous

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- schrodinger

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- triciaal

|dw:1437507590402:dw|

- triciaal

|dw:1437507884957:dw|

- anonymous

Okay

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## More answers

- anonymous

Sorry, but the internet here is running extremely slow

- anonymous

It's taking me like five minutes to send a reply

- anonymous

Alright I thin I understand how you got that

- anonymous

So what about the second one?

- triciaal

problems here too maybe it was this website

- triciaal

what second one?

- triciaal

remember I did not see problem 1. I don't know what you have for the value of the area of the smaller segment shaded in my sketch

- anonymous

Oh. It's A= (64/6)pi = 16 times sqrt3) inches squared

- anonymous

@triciaal

- triciaal

do you have a question for me?

- anonymous

|dw:1437569371363:dw|

- anonymous

I need to find the area of the part that's shaded in

- anonymous

@Paintking95

- anonymous

so you must first find the area of
a) the circular pie shaped wedge
b) the triangle
then you subtract the two areas

- anonymous

that how you find the first one im still trying to figure out the second one

- anonymous

Okay

- anonymous

Yeah I had to get help with the last one yesterday too

- anonymous

0.0145

- anonymous

I got that from this site if it helps http://www.mathopenref.com/segmentarea.html

- anonymous

Oh. I'm gonna go ahead and write down the site then

- anonymous

But. I'm supposed to have the answer in this format here:

- anonymous

https://suwannee.owschools.com/media/g_geo_2013/8/q254_x.gif

- anonymous

That's what I don't know how to do

- anonymous

are you still here

- anonymous

i got you hold up ill see what i can do

- anonymous

Alright, i appreciate it

- anonymous

A[16pi-8]inches ^2 i think

- anonymous

So how would you put it in that format

- anonymous

A={(32over2)pi - ? } inches squared

- anonymous

wait a sec

- anonymous

that what im still trying to figure out

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