Find the area of each shaded segment in nearest tenth.

Denote the center of circle as O, the 2 endpoints of shaded segment as AB, then we have:

section OAB area = pi x radius^2 x (120/360) = pi x 6^2 x (1/3) = 37.7 (cm2)

Considering triangle OAB, we have OA = OB = radius, angle AOB = 120 deg. Denote OH as the height of triangle. H lies on AB.

=> Applying cosine and sine theorem in right triangle AHO:

OH = OA x cos 60 = 6 x (1/2) = 3 (cm)

AH = OA x sin 60 = 6 x (sqrt3)/2 = 5.2 (cm)

=> triangle AOB area:

A = triangle AOH area x 2

   = AH x OH x (1/2) x 2 = 3 x 5.2 x (1/2) x 2 = 15.6 (cm2)

=> shaded segment area = section OAB area - triangle AOB

                                          = 37.7 - 15.6 = ~21.1 (cm2)

Hope this helps!

:)

1/2*qr*rs

step-by-step explanation: