SOLUTION: A sector of a circle is enclosed by two 12.0-inch radii and a 9.0-inch arc. Its perimeter is therefore 33.0 inches. There is another circular sector — part of a circle of a dif

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Question 1164064: A sector of a circle is enclosed by two 12.0-inch radii and a 9.0-inch arc. Its perimeter is therefore 33.0 inches.
There is another circular sector — part of a circle of a different size — that has the same 33-inch perimeter and that encloses the same area. Find its central angle, radius, and arc length, rounding to the nearest tenth.
Answer by math_helper(2461) About Me  (Show Source):
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Let s=arc length, r=radius, and alpha = central angle (in radians)



We set up the problem with these two constraints:

The area of the given sector is (9*12)/2 = 54 in%5E2

The perimeter is 33in





So, 

    2r + s  =  33    (1)

     rs/2   =  54    (2)   

and arc-length relates to r and alpha by:

       alpha*r = s     (3)





Equations (1) and (2) can be reduced (by substitution) to one equation in variable r:

    2r + (108/r) = 33

    2r - 33 + 108/r = 0



Muliply by r:

    2r^2 - 33r + 108 = 0



The LHS factors to:

    (2r-9)(r-12) = 0    



Solutions:  r=9/2   and r=12  (discard r=12, as that is the already-known circle)



Plug in r=(9/2)in (=4.5in) into (1) to get s=24in, and then use these values for r and s in (3) to get alpha+=+48%2F9 rad  (=5.3 rad, to one tenth)





NOTICE that the radius of the first circle multiplied by 2 is the arc length found in the 2nd circle, and the arc length from the first circle divided by 2 is the radius of the 2nd circle.