ems asked in Science & MathematicsMathematics · 1 decade ago

sector area of a circle?

How do I find the radius of a sector of a circle if:

the area of the sector is 9.817 and the degree measure of the arc is 22.5.

How do I find the Radius??

6 Answers

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  • Anonymous
    1 decade ago
    Favorite Answer

    Area of sector = Angle/360 x (pi x radius^2)

    (Area of sector x 360)/ (Angle x pi) = radius^2

    Take the root of this.

    So it would be:

    rt [(9.817 x 360)/(22.5 x 3.14)]

    You work it out. I don't have a calculator.

  • 1 decade ago

    the formula is

    Area of sector = 1/2 x radius^2 x angle in radians

    so

    let a = radius

    Area = 9.817

    pi = 3.14159265358979

    Angle = 22.5 deg = 22.5 x pi/180 = 0.39269908169872 rad

    so 1/2 x a^2 x 0.39269908169872 = 9.817

    a^2 = 9.817 x 2 / 0.39269908169872

    = 49.99757044266072

    a = 7.0708960141315

  • 5 years ago

    The circle has center (0,0) with a radius of 2. The sector is defined by the angle between the hypotenuse of a triangle whose vertices are at O(0,0), B(^/2,^/2) and C(^/2,0) which is a right triagle with /_BOC = 45 degrees. and the hypotenuse of a triangle whose vertices are O(0,0) A(1,^/3) and D(0,1) which is a right triangle whose /_AOD = 60 degrees. The angle of the sector is /_AOD - /_BOC = 60 - 45 = 15 degrees. The area of the circle is (pi)r**2= (pi)(2)**2 = 4(pi) sq units. The circle entails 360 degrees and the sector is 15 degrees. The area of the sector AOB is (15/360)*(pi)=(pi)/24 sq units.

  • 1 decade ago

    Area of circle = PI * radius^2

    Area of sector = Area of circle * angle/360

    Use your algebra skills to solve these two equations for radius. Then plug and chug.

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  • Anonymous
    1 decade ago

    Area circle= pi*r*r Sector area= circle area*22.5/360

    r=sq root (Sector area *360/pi *arc degrees)

    r= sq root( 9.817*360/pi*22.5) = 7.070896

  • 1 decade ago

    The area of the entire circle is 360/22.5 *9.817 = pir^2.

    So r = sqrt(360*9.817/(22.5 pi)) = 7.07 approximately.

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