Question

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

Answer 1

Thanks Bro

Answer 2

I appreciate you helpin'

Answer 3

We know that the triangle formed by the two radii and the chord is
an equilateral triangle, all angles are 60 degrees, which is 1/6 of
360 degrees
:
Find area inside the 60 degree arc
1/6*pi8^2 = 35.51
:
Find the area of the equilateral triangle
1/2*8*SQRT8^2-4^2 = 27.71 sq/in
:
Find the area of the shape enclosed by the 60 degree arc and the chord
35.51 - 27.71 = 7.8 sq/in
:
Find the area of the larger segment
pi*8^2 - 7.8 = 193.26 sq/inches
NO PROBLEM damn caps

Answer 4

Hehehe Lol! I know. It happens to me all the time. It's a real pain

Answer 5

ik rite and gtg ill message u when i get back around 1

Answer 6

Alright. I appreciate the help

Answer 7

i see you later

Answer 8

Yeah I still need a little help