SOLUTION: In a question I do not know what is the first step in solving the question. Given that the perimeter of the sector AOB is 18cm and that the area is 8cm2, calculate the values of r

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Question 203503: In a question I do not know what is the first step in solving the question.
Given that the perimeter of the sector AOB is 18cm and that the area is 8cm2, calculate the values of r and (theta).

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Area of any circle: A+=+pi%2Ar%5E2
Since we know the area of this circle we can use this equation to find the radius (r):
8+=+pi%2Ar%5E2
Divide by pi
8%2Fpi+=+r%5E2
Find the square root of each side:
sqrt%288%2Fpi%29+=+r
(We can discard the possible negative value for r since it represents the length of the radius.)

The perimeter of a sector is r + r + the length of the arc, which we'll call "a":
P = 2r + a
We know what P and r are so we can use this equation to find a:
18+=+2%28sqrt%288%2Fpi%29%29+%2B+a
Subtract 2%28sqrt%288%2Fpi%29%29 from each side:
18+-+2%28sqrt%288%2Fpi%29%29+=+a

And finally we can use the length of the arc, a, to find the central angle. To do this we will need the circumference of the circle. The circumference of any circle is: C+=+2%2Api%2Ar For this circle:
C+=+2%2Api%2A%28sqrt%288%2Fpi%29%29
Now we can set up a proportion between the number of radians in a whole circle, the circumference of the whole circle, the number of radians of the central angle (which we'll call "x" (Sorry, Algebra.com's software doesn't do "theta".)) and the length of the arc:

Multiplying both sides by %2818+-+2%28sqrt%288%2Fpi%29%29%29 we get:

%2818+-+2%28sqrt%288%2Fpi%29%29%29%2F%28sqrt%288%2Fpi%29%29+=+x
18%2F%28sqrt%288%2Fpi%29%29+-+2+=+x
Now we use our calculators for pi and the square root:
18%2Fsqrt%282.5464790894703255%29+-+2+=+x
18%2F1.5957691216057307+-+2+=+x
11.2798272358395023+-+2+=+x
9.2798272358395023+=+x
So the central angle is approximately 9.28 radians. (If you want the answer in degrees, multiply this by 180%2Fpi).