Area of a sector from its permiter

[the tree]

There is a circle with the radius r around the centre O, there is a minor sector with a perimiter of 100cm and an area of Acm

²

, show that the area of the sector is A=50r-r

²

I figured that as [math]P=2r+\theta^{c} r=100[/math] I should find some way of fitting that in the same equation as [math]A=\frac{1}{2}r^{2}\theta^{c}[/math], but I couldn't do that without getting a useless mess.

Any help?

 

edit: could some mod take out the dual post please?

[5614]

Ah... got it!

 

[math]P = 2r + r \theta[/math]

Therefore:

[math]\theta = \frac{P - 2r}{r}[/math]

from

[math]A = \frac{1}{2} r^2 \theta[/math]

we get

[math]\theta = \frac{2A}{r^2}[/math]

combining the two theta equals equations we get

[math]\frac{P - 2r}{r} = \frac{2A}{r^2}[/math]

simplifying gives:

[math]Pr - 2r^2 = 2A[/math]

 

[math]A = \frac{Pr}{2} - r^2[/math]

we know that P = 100 therefore sticking that in (P/2 = 100/2 = 50)

[math]A = 50r - r^2[/math]

QED

 

Basically we had the P=100 and the 0.5 in the A= formula so I was looking for something there, although couldn't initially do much with that. In fact when I was calculating as I did above it was only at the very last step did I remember I could substitute 100 for P.

 

I tried getting two equations for r= but kept getting square roots everywhere do I moved onto theta, as the only other variable in both the P= and the A= formula.

[the tree]

Wow thanks, I would never have thought to get two equations for theta. That's awesome.

[5614]

I just knew I had to combine them somehow so looked at how I could do that.

 

The only way of really doing that would be if each equation equalled the same thing. So if both equations equalled 10 then the two equations would equal each other, at which point you can combine them. In this case combining them using r turned out stupid square roots everywhere so I tried using theta, which worked.