The area of the shaded region can be gotten by multiplying the area of the circle by the subtended angle of the sector.
It looks like 3/4 of the area of a circle and it is called a reflex angle
Area of a sector of a circle.
Multiply ( pi R2 ) by [ (angle included in the sector) / 360 ].
A central angle can subtend (form) an arc of a circle. That has an area of 2 x pi x r x (angle A) / 360.
Area of sector/Area of circle = Angle of sector/360o Area of sector = (Area of circle*Angle of sector)/360o
6.46
Area of whole circle = pi*r2 = 64*pi Area of Sector = Area of Whole Circle * Angle of Sector/Angle of Whole Circle = Area of Whole Circle * 120/360 = Area of Whole Circle / 3 = 64*pi/3 = 67.0 to the nearest tenth.
If the sector of a circle has a central angle of 50 and an area of 605 cm2, the radius is: 37.24 cm
The area of the shaded region can be gotten by multiplying the area of the circle by the subtended angle of the sector.
6.46
It looks like 3/4 of the area of a circle and it is called a reflex angle
area of sector = (angle at centre*area of circle)/360
This question is too vague to have an answer, but here is one.For the shaded area (pie wedge) of a circle, find the area of the circle and multiply by the ratio of the wedge angle to the entire circle (angle/360).For the shaded region of a triangle, find the area of the smaller triangle, if necessary using trig functions to define a known angle or length of a side.For other polygons, you may be able to divide the area into triangles separately, then sum their areas.
Area of a sector of a circle.
The area of a sector in a circle if the radius is 4 cm and the arc has degree 120 is: 16.76 cm2
This question is too vague to have an answer, but here is one.For the shaded area (pie wedge) of a circle, find the area of the circle and multiply by the ratio of the wedge angle to the entire circle (angle/360).For the shaded region of a triangle, find the area of the smaller triangle, if necessary using trig functions to define a known angle or length of a side.For other polygons, you may be able to divide the area into triangles separately, then sum their areas.