Area of a Sector (1 Viewer)

[Avenger6]

Hi, I need a bit of help with the following question:

I tried to solve simultaneously but ended up with r^2-r which I couldn't really use. Any help is greatly appreciated.

Thanks.

 

[foram]

area of sector is 1/2 r^2 @ = 3pi/10 (Equation 1)
arc length is r@ = pi/5 (Equation 2)

(1/2 r^2 @) / r@ = (3pi/10) / (pi/5) (Dividing 1 and 2)
1/2r = 3/2
r = 3

(substituting r=3 into 1/2 r^2 @)
9/2 @ = 3pi/10
@ = 3pi/10 x 2/9
@ = pi/15

 

[lyounamu]

Hi, I need a bit of help with the following question:

I tried to solve simultaneously but ended up with r^2-r which I couldn't really use. Any help is greatly appreciated.

Thanks.

A =
1/2 . r^2 . @ = 3pi/10 ... (1)

l =
r@ = pi/5 ... (2)

(1) divided by (2): 1/2r = (3pi/10)/(pi/5)
1/2r = 3/2

Therefore, r = 3.
Substitue 3 for r into the equation (1):
1/2 . 3^2 . @ = 3pi/10
9/2@ = 3pi/10
@ = 3pi/45
= pi/15

EDIT: Why do I always end up posting up second??? Well done, Foram!

 

[foram]

I got there before you lyounamu!

 

[Avenger6]

Thanks guys, i was subracting instead of dividing when using them simulataneously. Thanks again for the help.