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.)

  ((160/3)π + 16√3) in²

Step-by-step explanation:

Together with the radii of 8 inches, the 8-inch chord creates an equilateral triangle with a central angle of 60°, or π/3 radians.

In problem 1, you found the area of the smaller segment to be ...

  A = (1/2)r²·(θ -sin(θ))

For θ = π/3 and r = 8 in, this is (1/2)(8 in)²(π/3 -(√3)/2) = ((32/3)π -16√3) in².

The remaining segment of the circle is the area of the circle less this amount, so is ...

  π·(8 in)² -(32/3π -16√3) in² = (160/3π +16√3) in² . . . . larger segment area

guy above is correct, i used it on my school, and i was right.

A=(160over3 pi + 16 sqr root 3)inches squared

the transverse axis of a hyperbola is the line that contains the two vertices and the two focuses.

in this example (hyperbola of equation x^2/2 - y^2/4 = 1), the transverse axis is the  x-axis  . 

step-by-step explanation: