hey...

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

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

thankyou

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- Jul 4th 2005, 07:35 AMslurch901radians....
hey...

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

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

thankyou - Jul 4th 2005, 09:31 AMMathGuruArea of a sector of a circle
Remember that $\displaystyle S = r\theta$

S = arc length

r = radius

theta = angle of arc/sector

And the area of a sector is $\displaystyle \pi*r^2\frac{\theta}{2\pi}$

which simplifies to $\displaystyle r^2\frac{\theta}{2}$

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

$\displaystyle 20cm = r\theta $

$\displaystyle \theta = \frac{20cm}{r} $

Lets substitute theta into the previous equation and we get:

$\displaystyle r^2\frac{\frac{20cm}{r}}{2}$

which simplifies to:

$\displaystyle r*10cm$

I beleive I have made a mistake somewhere . . . can anyone find it? - Jul 4th 2005, 10:39 PMticbol
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.

-----------------

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.

--------------

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? - Jul 5th 2005, 01:07 PMslurch901
soz i typed it incorrectly :S its A = 25 - (r - 5)^2

there lol soz bout that :S - Jul 6th 2005, 03:51 AMticbol
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.