Question

A circle has a circumference of 14 ft. Its area is halved. Describe the effect on the circumference of the circle.

1)The circumference is divided by 2.
2)The circumference doubles.
3)The circumference is divided by √3
4)The circumference is divided by √2

Answer 1

@Jhannybean

Answer 2

21.98? but I still dont understand.

Answer 3

One minute.

Answer 4

rearrange both eqns so r = r
area = pi r^2
circum = 2 pi r

so rearranging:
assume radius of 1:
C^2 = 4 * A * pi

Answer 5

i'm not understanding this . :|

Answer 6

so what happens to circumference if we assume first area = 10 m^2
half area = 5m^2

circumf 1 = 2 * (sqrt (10 pi)) approx 11.2

circumf 2 = 2 * (sqrt (5 pi)) approx 7.9

Answer 7

r = r?

Answer 8

sorry:
r = r
rearrange both the circum equation and the area eqn in terms of r
C/(2*pi) = sqrt(A/pi)

Answer 9

thanks:)

Answer 10

radius in terms of circumference would be the same radius from the equation in terms of area

Answer 11

so did you see the relationship...?

Answer 12

Nope.

Answer 13

I think i got it now.

Answer 14

area 2 / area 1 = 2 / 1
circumference 2 / circumference 2 = 1.41414 = sqrt 2

... sorry, had the area ratio value backwards before

Answer 15

Your Area and Circumference have a direct relationship, so if one increases, the other does to, vise versa.

You're given C = 14. and the area is reduced by half of the original area of the circle.

\[\large C=2 \pi r \quad \quad \quad A=\pi r^2\]\[\large \text{Circle 1:}\]\[\large 14= 2 \pi r\]\[\large R_{1}=\frac{C}{2\pi}\]\[\large R=\frac{14}{2 \pi}\]\[\large\text{Area}=\pi r^2=\frac{14^2}{4\pi^2}* \pi= \frac{14^2}{4\cancel{\pi^2}\pi}*\cancel{\pi}=\frac{14^2}{4\pi}\]\[\large \text{now make half the area}=\frac12*\frac{14^2}{4\pi}= \frac{14^2}{8 \pi}\]\[\large\text{Circle 2}:\]\[\large \text{get new reduced r}:\]\[\large \text{Area}= \pi r^2 = \frac{14^2}{8 \pi} \implies r^2=\frac{14^2}{8 \pi} * \frac1\pi=\frac{14^2}{8\pi^2} \implies r_{2}=\frac{14}{\pi \sqrt{8}}\]\[\large \text{New Circumference}= 2 \pi r =\cancel{2 \pi}*\frac{14}{\cancel{ 2 \pi} \sqrt{2}}=\frac{14}{\sqrt{2}}\]

Now we can see how the circumference changes when we reduce the area by half.

Answer 16

For circle 1
find radius
find area
take half the area

Circle 2 : use the halved area to find NEW radius,
use NEW radius to find NEW circumference.

Answer 17

Another way is to notice that area is related to an exponent of 2.

So if the perimeter/circumference increases by a factor of 2, then the area increases by a factor of 4.

If p/c increases by a factor of 4, then the area increases by a factor of 16.

And so we can say that if area increases/decreases by a factor of \(n\), then the p/c has decreased by a factor of \(\sqrt{n}\). Here, \(n = 2\) so the circumference is being divided by \(\sqrt2\).

Answer 18

That is true for any regular figure.

Answer 19

And ontop of that,you can misread the question as well, it says the area is halved.

This could mean that since the circumference is directly correlated with the area, a "halved" area could mean that the circumference is divided by 2. lol.

14/2 = 7 :|

Answer 20

It's a matter of saying "the area is reduced by half" to saying "the area is halved"

Answer 21

@Jhannybean I said that the area has decreased by a factor 2.

Answer 22

No i wasn't referring to you, i was referring to the question, lol.

Answer 23

The question isn't wrong at all.

Answer 24

hehehe

Answer 25

lol

Answer 26

\[\begin{align}\left.\begin{array}{cc}C=2\pi r&\implies&r=\frac C{2\pi}\\A=\pi r^2&\implies&r=\sqrt\frac A\pi\end{array}\right\}\implies C&=2\pi\sqrt\frac A\pi\\C'&=2\pi\sqrt\frac {A'}\pi\\A'=A/2\qquad&=2\pi\sqrt\frac {A/2}\pi\\&=\frac2{\sqrt2}\pi\sqrt\frac {A}\pi\\&=\frac C{\sqrt2}\end{align}\]

Answer 27

Rhaukus, derivatives? your way is more complicated D: or so it seems....

Answer 28

Derivatives? I don't see a derivative anywhere.

Answer 29

I've used the 'primes' to denote the NEW Circumference and New Area

Answer 30

\(C' \) is just a notation for talking about the 2nd circumference.

Answer 31

Exactly.

Answer 32

What the derivative of the area with respect to radius @ParthKohli ?

Answer 33

Its circumference lol

Answer 34

This feels like you two are trading secrets. And i'm not in on it. :(

Answer 35

Did i get it? ^_^ lol.