Thread: Arc length

hey...

if a sector of a circle has a perimeter of 20cm,

how do you prove the "area = 25.(r-5)^2"

thankyou

Remember that

S = arc length
r = radius
theta = angle of arc/sector

And the area of a sector is

which simplifies to

we know from the given information that S = 20cm so we can write theta in terms of r.




Lets substitute theta into the previous equation and we get:



which simplifies to:



I beleive I have made a mistake somewhere . . . can anyone find it?

MathGuru, perimeter of sector is
P = r+r+s
P = 2r +r*theta
So,
2r +r*theta = 20
r*theta = 20 -2r
theta = (20 -2r)/r

Using your area A = [(r^2)theta]/2,
A = [(r^2)/2](theta)
A = [(r^2)/2][(20 -2r)/r]
A = r(10-r) ----***

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I'd do it another way.

P = 2r +s
2r +s = 20
s = 20 -2r
s = 2(10-r)

A = (1/2)(r)(s)
A = (1/2)(r)[2(10-r)]
A = r(10-r) ----same as in your way.

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That A is not the same as the A = 25(r-5)^2, which is given in the question.

I say then that A = 25(r-5)^2 is wrong.
And so, the question is wrong.

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Another reason why A = 25(r-5)^2 is wrong.

When r = 5:

A = 25(5 -5)^2 = 0.
There is no area, no sector, when r = 5?
Incredible.

P = 20 = 2r +s
20 = 2*5 +s
s = 10
So, if r = 5, and s = 10, why is there no sector?

soz i typed it incorrectly :S its A = 25 - (r - 5)^2

there lol soz bout that :S

Umm, so A = 25 -(r-5)^2 ?

That figures.

I found A = r(10-r).
I will convert that into A = 25 -(r-5)^2.

A = r(10 -r)
A = 10r -r^2
r^2 -10r +A = 0
Complete the square of the r-terms,
[r^2 -10r] + A = 0
[r^2 -10r +(-10/2)^2 -(-10/2)^2] +A = 0
[r^2 -10r +(-5)^2 -(-5)^2] +A = 0
[(r -5)^2 -(25)] +A = 0
(r-5)^2 -25 +A = 0
A = 25 -(r-5)^2 -----***

Proven.