Question

***PLEASE HELP! WILL MEDAL***Express answer in exact form.

Given a circle with an 8" radius, find the area of the smaller segment whose chord is 8" long

Answer 1

When the radius of the circle is 8" and chord length is also 8" Then this chord forms an equilateral triangle a the center . The angle subtended by the chord/arc at the center is also 60° or π/3 radians
Area of the larger segment = Area of the circle − area of the smaller segment
Area of the smaller segment = Area of the sector − area of the equilateral triangle formed by the chord at the center
Area of a sector = ½ R² θ where θ is the angle in radians subtended by the arc at the center of the circle =½×8²×π/3=32π/3 inch ²
Area of equilateral triangle of side 8 "=½×8×8sin60° =32×√3/2 =16√3 inch² Hence Area of the smaller segment = 32π/3 − 16√3
Area of the circle =πR²=π×8²=64π inch²
Hence area of larger segment =Area of the circle − area of the smaller segment
=64π−{32π/3 − 16√3}= 160π/3 + 16√3 inch²

Answer 2

Does that help?

Answer 3

@Tide_Poole

Answer 4

Woah

Answer 5

Give me a sec to read it all

Answer 6

Or Area = 64pi - (128pi^3-96root3)/6

Answer 7

omg

Answer 8

?

Answer 9

Did you get it right?

Answer 10

Yeah that's what I'm thinking

Answer 11

Ok your welcome :)

Answer 12

didn't really help, but okay

Answer 13

Whatever goodbye

Answer 14

Two radii and the chord all = 8" All three sides of the triangle so formed are equal. Each of the angles of an equilateral triangle = 60 Degree.

Area of a pie piece (called a circle sector) = central angle over entire angle of a circle * r^2
Area of sector = 60/ 360 * pi * r^2
Area of sector = 1/6 * 3.14 * 8^2
Area of sector = 33.51 square inches.

Now find the area of an equilateral Triangle.
The area of a triangle is the base * height /2
The height = sqr(8^2 - 4^2) = sqr(48) = 6.928
The area of the triangle = 8*6.928/2 = 27.712

So the area of the smaller segment = area of the sector - area of the triangle
small segment = 33.1 - 27.712 = 5.798 square inches.

Answer 15

It's fine. I'm just a little slow that's all. I've never been taught any of this before

Answer 16

I really can't help it @AnasAlazzawi