Circle problem - perimater of a minor sector

In summary, the conversation is about finding the values of P and Q in the equation P{\pi} + Q = 5\theta, where theta is the radian measure of the angle inside the slice of pie. The perimeter of the minor sector AOB is also mentioned, and a possible solution is suggested by setting Q = 10 and P = \frac{5\theta-Q}{\pi}. However, it is noted that there may be other solutions depending on the given conditions. Another suggestion is to use the whole perimeter (arc length + 2*radius) in the equation, resulting in P = \frac{5\theta+10-Q}{\pi}.
  • #1
footprints
100
0
Code:
                 ..       
            A .'    '.B   
             . \    / .   
            .   \  /   .  
            .    \/    .  
             .   o    .   
              .      .    
               ' . .'
The radius of the circle is 5. The perimeter of the minor sector AOB is [tex]P{\pi} + Q[/tex]. Find P and Q
 
Last edited:
Physics news on Phys.org
  • #2
The circumference of the circle is, of course, [itex]10\pi[/itex]. I don't see any way of determining P and Q without knowing what part of the entire circle AOB is. Are you given the central angle? Surely there must be some conditions on P and Q? If P and Q could be any real numbers, then even if we know exactly what the perimeter is, we could choose one of P and Q to be anything we want and then solve for the other.

I take it by "perimeter" you mean the whole perimeter, both the curved part and the two radii. A plausible answer would be that Q= 10 (the length of the two radii) and P would be [itex]\frac{5\theta}{\pi}[/itex].
 
  • #3
How about this:

[tex] P \pi +Q=5\theta [/tex]

with theta being the radian measure of the angle (inside the slice of pie):

Thus for "any" Q:

[tex] P=\frac {5\theta-Q}{\pi} [/tex]

Is is really for any Q?
 
Last edited:
  • #4
Why "[itex]5\theta[/itex]"?
 
  • #5
HallsofIvy said:
Why "[itex]5\theta[/itex]"?

Ohhh, they mean the whole perimeter around the slice of pie and not just the arc length. In that case may I suggest:


[tex] P \pi +Q=(5\theta+10) [/tex]
(the arc length +2*radius)

with theta being the radian measure of the angle (inside the slice of pie):

Thus for "any" Q:

[tex] P=\frac {5\theta+10-Q}{\pi} [/tex]
 
Last edited:

FAQ: Circle problem - perimater of a minor sector

What is the formula for finding the perimeter of a minor sector?

The formula for finding the perimeter of a minor sector is (θ/360) x 2πr, where θ is the central angle in degrees and r is the radius of the circle.

How do you find the measure of the central angle in a minor sector?

To find the measure of the central angle in a minor sector, you can use the formula θ = (arc length/radius) x 180/π. Alternatively, you can use the proportion θ/360 = arc length/2πr.

Can the perimeter of a minor sector be greater than the circumference of the circle?

Yes, the perimeter of a minor sector can be greater than the circumference of the circle. This occurs when the central angle of the sector is greater than 180 degrees.

How do you calculate the arc length of a minor sector?

The formula for calculating the arc length of a minor sector is (θ/360) x 2πr, where θ is the central angle in degrees and r is the radius of the circle. You can also use the proportion arc length/2πr = θ/360.

Can the perimeter of a minor sector be negative?

No, the perimeter of a minor sector cannot be negative. The perimeter is a measurement of distance and therefore must be a positive value.

Back
Top