I've been doing a task which says the following:

Area of a circular sector is $3.2\pi cm^2$, arc length is $0.8\pi cm$. What is the central angle?

I've been attacking this from several angles (boo), and finally settled on approaching this from area aspect of it. This is what I did:

$A = \frac{rl}{2}$ -> $r=\frac{2A}{l}$, where r is radius and l arc length.

So then, I did:


From there, I thought arc length formula was appropriate to rearrange into theta.


and then I did it with numbers:


$\frac{\pi}{10}$ radians, hopefully. Which should be $\frac{180}{\pi}*\frac{\pi}{10}=18°$

Problem now is that since I have no solutions provided in this textbook, I can't check the answer by looking at one, so I opted to check it back by plugging in newly found radius and arc length to find if the area matches with given one.

$A=\frac{\frac{8}{1}*\frac{\pi}{10}}{2}=\frac{4\pi}{5}cm^2$ which isn't right.

I am sure that either I'm making a stupid mistake somewhere with units or my approach to this is completely wrong, but I am stuck.


You got the right value for ${\theta}$, but your radius should be squared in the formula for area of sector.

The area of a circle is ${\pi}r^2$.
If ${\theta}$ is in radians, then the area of a sector is ${\pi}r^2$*($\frac{\theta}{2{\pi}}$)

So we have Area of sector = $\frac{r^2\theta}{2}$

So, plugging in r=8 and ${\theta}={\frac{\pi}{10}}$ gets you the required value of 3.2${\pi}cm^2$

  • $\begingroup$ I see. So, I've made a same mistake in the beginning as well where $A = \frac{rl}{2}$ -> $r=\frac{2A}{l}$, where r is radius and l arc length should have actually been $A = \frac{r^2l}{2}$ -> $r^2=\frac{2A}{l}$. $\endgroup$ – MathAgain Mar 30 '15 at 22:21
  • 1
    $\begingroup$ Well no- I think that was correct in your beginning statement. ${l}$ = r${\theta}$, so $\theta = \frac{l}{r}$ Substituting into A=$\frac{r^2{\theta}}{2}$ yields your original formula: A = $\frac{r{l}}{2}$ $\endgroup$ – Mike Andrejkovics Mar 30 '15 at 22:59
  • $\begingroup$ Caught it later on. :) Thanks! $\endgroup$ – MathAgain Mar 30 '15 at 23:13

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