Answered

At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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

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 class=

Sagot :

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

View image eudora

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

We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.