Circle problem - perimater of a minor sector

[footprints]

                 ..       
            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

 

[HallsofIvy]

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].

 

[saltydog]

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?

 

[HallsofIvy]

Why "[itex]5\theta[/itex]"?

 

[saltydog]

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]