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December 6th, 2015, 10:19 AM   #1
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Calculating the radius & angle of sector.

Hi All, wondering if anyone is able to help me solve this problem;

Length of an arc of a circle is 6cm. The corresponding sector area is 150cm squared. Calculate i) the radius of the circle and ii) the angle subtended at the centre of the circle by the arc

I found the radius by setting up a pair of simultaneous equations using arc length =r*theta and area of sector= 1/2*r^2*theta. By using these I calculated the radius to be 50cm which surely cannot be correct?

Any help would be great thanks!!

Last edited by wahhdoe; December 6th, 2015 at 10:37 AM.
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December 6th, 2015, 12:40 PM   #2
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$s = r \cdot \theta \implies \theta = \dfrac{s}{r}$

$A = \dfrac{1}{2}r^2 \theta$

$A = \dfrac{1}{2}r^2 \cdot \dfrac{s}{r}$

$A = \dfrac{1}{2}r \cdot s$

$r = \dfrac{2A}{s} = \dfrac{300}{6} = 50 \text{ cm}$

$\theta = \dfrac{6}{50} = \dfrac{3}{25} \text{ radians } \approx 7^\circ$
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December 6th, 2015, 06:52 PM   #3
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Let me try :
$\displaystyle \frac{6}{150}=\frac{2\cancel{\pi}r}{\cancel{\pi}r^ 2}$
$\displaystyle \frac{1}{25}=\frac{2r}{r^2}$
$\displaystyle \frac{1}{25}=\frac{2}{r}$
$\displaystyle r=2\times25$
r = 50

$\displaystyle \frac{\theta}{360^o}=\frac{6}{2\pi\times50}$
$\displaystyle \frac{\theta}{360^o}=\frac{6}{100\pi}$
$\displaystyle \frac{\theta}{2\cancel{\pi}}=\frac{3}{50 \cancel{\pi}}$
$\displaystyle \theta=\frac{3\times2}{50}$
$\displaystyle \theta=\frac{6}{50}$
$\displaystyle \theta=\frac{3}{25}$
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