The diagram shows a vector of a circle centre O and radius 6 cm. Mn is a chord of the circle. Angle MON is 50degrees. Calculate the area of the shaded segment. Give your answer to three significant figures

Area of the shaded region = 1.92 cm²

Step-by-step explanation:

Area of the shaded region = Area of sector OMN - Area of isosceles triangle OMN

Drop a perpendicular OP from vertex O to opposite side MN.

Perpendicular OP will bisect angle MON.

m∠MOP = 25°

By cosine ratio in triangle OPN,

cos(25°) =

OP = ON.cos(25°)  

OP = 6cos(25°)

OP = 5.438 cm

By sine ratio of angle PON,

sin(25) =

PN = ON.sin(25)

     = 6sin(25°)

     = 2.536 cm

Since, MN = 2(PN)

MN = 5.071 cm

Area of ΔOMN =

                         =

                         = 13.788 cm²

Area of sector OMN =

Here 'θ' is the angle subtended by the arc MN at the center.

Area of sector OMN =

                                  = 15.708 cm²

Area of the shaded region = 15.708 - 13.788

                                             = 1.92

                                             ≈ 1.92 cm²

c

step-by-step explanation:

it would be see because your formula is f(x)=6x, so 6 times 1 is 6 and you go on and the best answer is c.

$25

step-by-step explanation:

25 - 4.50 = 20.5

20.5 / 2 = 10.25

10.25 + 10 = 20.25

20.25 - 12.50 = 7.75

therefore the answer is $25 as shown as the first number.