Express answer in exact form.

a regular hexagon with sides of 3" is inscribed in a circle. find the area of a segment formed by a side of the hexagon and the circle.

(hint: remember corollary 1--the area of an equilateral triangle is 1/4 s2 √3.)

Area segment = (3/2)π - (9/4)√3 inches²

Step-by-step explanation:

∵ The hexagon is regular

∴ Then it consists of 6 equilateral triangles with side length 3"

∴ the radius of the circle = the side of the triangle = 3"

∵ The measure of each angle of the triangle is π/3

∴ Area each triangle = (1/4)(3)²√3 = (9/4)√3

∵ Area segment = area sector - area triangle

∵ Area sector = (1/2)r²Ф

∵ r = 3" and Ф = π/3 ⇒ (60° = π/3 rad)

∴ Area sector = (1/2)(3)²(π/3) = (3/2)π

∴ The area of the segment = (3/2)π - (9/4)√3 inches²

other person is correct please give them brainliest

all it means is function 3

idk

step-by-step explanation: