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 $\displaystyle S = r\theta$
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?
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?
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.