Circle problem using radians

[jonnburton]

I wondered if anyone could help me with this problem which I have been trying to solve.

The perimeter ofa sector of a circle of radius r and angle $\displaystyle \theta$ is the same as the perimeter of a square of side r.

a. Show that $\displaystyle \theta$ = 2 radians
b. Show that the area of the sector is the same as the area of the square.

a.

perimieter of square = 4r.
Perimeter of sector of circle = $\displaystyle 2r + \frac {1}{2}r^2\theta$

so $\displaystyle 4r = 2r + \frac {1}{2}r^2\theta$

$\displaystyle \frac {1}{2}r^2\theta = 2r$

$\displaystyle r^2\theta = 4r$

$\displaystyle \theta = \frac {4}{r}$

I am not sure how to show that $\displaystyle \theta$ = 2 radians because I do not seem to be able to get rid of r from the equations above.

 

[JeffM]

I wondered if anyone could help me with this problem which I have been trying to solve.

The perimeter ofa sector of a circle of radius r and angle $\displaystyle \theta$ is the same as the perimeter of a square of side r.

a. Show that $\displaystyle \theta$ = 2 radians
b. Show that the area of the sector is the same as the area of the square.

a.

perimieter of square = 4r.
Perimeter of sector of circle = $\displaystyle 2r + \frac {1}{2}r^2\theta$

But where did you get this formula? You have mixed up the formulas for area and perimeter of a sector.

$\displaystyle Area\ of\ sector = \dfrac{r^2 \theta}{2},\ where\ \theta\ is\ measured\ in\ radians.$

$\displaystyle Length\ of\ arc\ of\ sector = r \theta ,\ where\ \theta\ is\ measured\ in\ radians.$

$\displaystyle Length\ of\ perimeter\ of\ sector = 2r + r \theta = r(2 + \theta ),\ where\ \theta\ is\ measured\ in\ radians.$

so $\displaystyle 4r = 2r + \frac {1}{2}r^2\theta$ So this equation is false.

$\displaystyle \frac {1}{2}r^2\theta = 2r$

$\displaystyle r^2\theta = 4r$

$\displaystyle \theta = \frac {4}{r}$

I am not sure how to show that $\displaystyle \theta$ = 2 radians because I do not seem to be able to get rid of r from the equations above.

Try redoing this problem using the correct formulas.

 

[jonnburton]

Thanks Jeff. It's very simple when the correct formula is used! It only took a couple of minutes to figure out the question.