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a sector of a circle has perimeter of 32cm and an area of 63cmsquare. find the radius length and the magnitude of the angle subtended .

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2 Answers

First draw a picture to help understand the problem. For this problem that should be a circle with an angle drawn in it. Label the picture, r for the radius of the circle and θ for the angle. Now look back to the problem and see what it is talking about.

The line "a sector of a circle has perimeter of 32cm" is talking about a the perimeter of the sector highlight that in your picture. You should have traced from the center of the circle along the radius (a distance r), then around the arc spanned by θ (a distance equal to the proportion of the circle times circumference which is θ/(2π)*(2πr) which reduces to θr), then back down a radius to the center again (a distance r). Thus the perimeter of the sector is r+θr+r. Therefore, by factoring out the r we get the formula P=(2+θ)r. Plug in the perimeter of 32 for P and we get 32=(2+θ)r.

The line "and an area of 63cmsquare" is talking about the area of the sector we just outlined. Lets shade it lightly. The area of this sector is the proportion of the circle times the area of the circle. In other words it's θ/(2π) times the area of the circle, which is πr^2. Thus we get the formula A=θ/(2π)*πr^2. But we can cancel out the π to get A=θ/(2)*r^2. Plug in the area of 63 for A and we get 63=θ/(2)*r^2.

Now we have two equations with two unknowns.

32=(2+θ)r

63=θ/(2)*r^2

Solve one of the equations for one of the unknowns. I picked 63=θ/(2)*r^2 and solved for θ. Which gave me:

θ = 2*63/r^2 ,or θ = 126/r^2

Plug that into the other equation (32=(2+θ)r). we get

32=(2+126/r2)r

Distribute the r

32=2r+126/r^2*r

reduce the r over r^2

32=2r+126/r

Agh!, we have a fraction. No problem just multiply both sides of the equation by the least common denominator, in our case that is r.

32r=(2r+126/r)r

distribute the r:

32r=2r^2+126

This is a quadratic, we solve that by getting it equal to 0

0=2r^2-32r+126

Factor, Lets factor by grouping. We need to find a pair of numbers that multiply to get 2*126 and add to get -32. We can find those numbers by listing all the pairs of factor of 252 and find the pair that adds to -32. We find that that pair is 14 and 18, ie -14*-18=252 and -14+(-18)=-32. Now replace -32 in the problem.

0=2r^2 +(-14+(-18))r+126

distribute the r

0=2r^2-14r+(-18)r+126

group the first two terms and the last two terms

0=(2r^2-14r)+((-18)r+126)

factor out the common factors from each parenthesis

0=2r(r-7)-18(r-7)

now factor out the common factor of (r-7)

0=(r-7)(2r-18)

Use the zero factor property to set each factor equal to 0

r-7=0 and 2r-18=0

solve

r-7=0

r-7+7=0+7       

r=7                 

and

2r-18=0

2r-18+18=0+18

2r=18

2r/2=18/2

r=9

Ok, so now we have r, it could be 7 or 9, lets plug it in to find θ. Recall we know θ=126/r^2

lets assume r=7 first, we will do r=9 in a bit.

θ=126/7^2

θ=126/49

θ=18/7

check to make sure θ<2π, it is. So this is valid answer.

so we have one pair r=7 and θ=18/7

now lets assume r=9 then

θ=126/9^2

θ=126/81

θ=14/9

check to make sure θ<2π, it is. So this is valid answer.

so we have another answer r=9 and θ=14/9

Don't forget your units. The length is in cm and the angle is in radians. Thus

the radius is 7cm and the angle is 18/7 radians

or the radius is 9cm and the angle is 14/9 radians

 

Start first by remembering that the total area of circle =Πr2   

 

The area of sector is a fraction of the total area of the circle , and is calculated by multiplying the circle's total area (Πr2) to the ratio of the angle (θ) to 360 degrees (or 2Π in radians)  

 

Area of sector = Πr2 (θ/360) = Πr(θ/2Π) = r2θ/2

 

The area of a sector can also be found by multiplying the total area of the circle with a ratio of the arc length (L) to the circumference (2Πr).

Area of sector = Πr2(L/2Πr) ... cancel like terms, and simplify --> rL/2

 

 

Now, if we set these two equations for area of sector equal to one another, we can cross-multiply and solve for L. 

rL/2 = r2θ/2  --> 2rL = 2r2θ 

L = θr

 

 

Lastly...

The circumference of the total circle = 2Πr, and the perimeter of the sector is a fraction of that total AND the perimeter of sector = arc length (L) + 2r

 

Remembering that L = 2Πr(θ/2Π) = θr   we can substitute L in equation for perimeter of sector

Perimeter of sector =  L + 2r = θr + 2r = r(θ +2)