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