Area of a sector question (1 Viewer)

k150764 · May 30, 2009

Hi i'm really stuck on this..and if anyone can help it would be great

Q. The area of the sector of a circle is 3pie units^2 and the length of the arc bounded by this sector is pie/6 units.
Find the radius of the circle and the angle subtended at the centre.

CountVonChocula · May 30, 2009

check your textbook!

k150764 · May 30, 2009

CountVonChocula said:
check your textbook!

The textbook shows no soulutions...they only have answers which is kinda crap
arghhhhhhhh

Drongoski · May 30, 2009

I'll try:

$\100dpi A\ =\ \frac{1}{2}\theta r^{2}\ =\ 3\pi \\ \\ arc\ =\ r\theta \ =\ \frac{\pi }{6}\\ \\ \therefore \ \ \frac{\frac{1}{2}\theta r^{2}}{r\theta }\ \ =\ \ \frac{3\pi }{(\frac{\pi }{6})}\\ \\ \therefore \ \ \frac{1}{2}r\ \ =\ \ 18\ \ \Rightarrow \ \ r \ =\ 36\\ \\ \therefore \ \ \theta \ =\ \frac{\pi }{6\times 36}\ \ =\ \ \frac{\pi }{216}\ \ radians$

k150764 · May 30, 2009

Drongoski said:
I'll try:

$\100dpi A\ =\ \frac{1}{2}\theta r^{2}\ =\ 3\pi \\ \\ arc\ =\ r\theta \ =\ \frac{\pi }{6}\\ \\ \therefore \ \ \frac{\frac{1}{2}\theta r^{2}}{r\theta }\ \ =\ \ \frac{3\pi }{(\frac{\pi }{6})}\\ \\ \therefore \ \ \frac{1}{2}r\ \ =\ \ 18\ \ \Rightarrow \ \ r \ =\ 36\\ \\ \therefore \ \ \theta \ =\ \frac{\pi }{6\times 36}\ \ =\ \ \frac{\pi }{216}\ \ radians$

Thank you soo much!
you got it right
^^

Area of a sector question (1 Viewer)

k150764

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CountVonChocula

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k150764

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Drongoski

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k150764

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