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

Answer:

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°) = [tex]\frac{OP}{ON}[/tex]

OP = ON.cos(25°)  

OP = 6cos(25°)

OP = 5.438 cm

By sine ratio of angle PON,

sin(25) = [tex]\frac{PN}{ON}[/tex]

PN = ON.sin(25)

     = 6sin(25°)

     = 2.536 cm

Since, MN = 2(PN)

MN = 5.071 cm

Area of ΔOMN = [tex]\frac{1}{2}(OP)(MN)[/tex]

                         = [tex]\frac{1}{2}(5.438)(5.071)[/tex]

                         = 13.788 cm²

Area of sector OMN = [tex]\frac{\theta}{360}(\pi r^{2})[/tex]

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

Area of sector OMN = [tex]\frac{50}{360}(\pi )(6^{2})[/tex]

                                  = 15.708 cm²

Area of the shaded region = 15.708 - 13.788

                                             = 1.92

                                             ≈ 1.92 cm²

Answer:

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°) = \frac{OP}{ON}ONOP

OP = ON.cos(25°)  

OP = 6cos(25°)

OP = 5.438 cm

By sine ratio of angle PON,

sin(25) = \frac{PN}{ON}ONPN

PN = ON.sin(25)

     = 6sin(25°)

     = 2.536 cm

Since, MN = 2(PN)

MN = 5.071 cm

Area of ΔOMN = \frac{1}{2}(OP)(MN)21(OP)(MN)

                         = \frac{1}{2}(5.438)(5.071)21(5.438)(5.071)

                         = 13.788 cm²

Area of sector OMN = \frac{\theta}{360}(\pi r^{2})360θ(πr2)

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

Area of sector OMN = \frac{50}{360}(\pi )(6^{2})36050(π)(62)

                                  = 15.708 cm²

Area of the shaded region = 15.708 - 13.788

                                             = 1.92

                                             ≈ 1.92 cm²

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