Area of a sector (1 Viewer)

Shazer2 · Feb 25, 2013

$The\quad area\quad of\quad a\quad sector\quad is\quad \frac { 3\pi }{ 10 } { cm }^{ 2 }\quad and\quad the\quad arc\quad length\quad cut\quad off\quad by\quad the\quad sector\quad is\quad \frac { \pi }{ 5 } cm.\quad Find\quad the\quad angle\quad subtended\quad at\quad the\quad centre\quad of\quad the\quad circle\quad and\quad find\quad the\quad radius\quad of\quad the\quad circle.$

So far I have tried subbing in the given information into the area formula and the length of an arc, but I am not getting anywhere.

Thanks

Drewk · Feb 25, 2013

Make sure ur calc is in RAD mode
L=r x theta
theta = ((pi)/5)/r
Sub into area of a sector formula
and rearrange to find r
then sub r in any of the formule to give theta

TheMathStudent · Feb 27, 2013

Area of sector: $\pi r^2\frac{\theta}{2\pi}=\frac{r^2\theta}{2}=\frac{3\pi}{10}$

Length of Arc $r\theta=\frac{\pi}{5}\Rightarrow r=\frac{\pi}{5\theta}$ sub this into above equation and you get:

$\frac{\pi^2}{25\theta^2}\frac{\theta}{2}=\frac{3 \pi}{10}$

Solving the rest should be simple, and do the same for radius

Area of a sector (1 Viewer)

Shazer2

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