if you have the circumfrence of a circle how do you find the area?

Answer 1

this is very confusing

The area of a circle with radius r is πr2. For example, the area of the unit circle is π.

[edit] Proof

The following is a proof for the area of a disk and a bounded region by an ellipse using calculus:
An ellipse
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An ellipse
The area of a circle
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The area of a circle

A circle is a form of ellipse, denoted by the equation

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.

Solving the equation of the ellipse for y, the following is derived:

\frac{y^2}{b^2} = 1 - \frac{x^2}{a^2} = \frac{a^2 - x^2}{a^2} \ or \ y = \pm\frac{b}{a}\sqrt{a^2 - x^2}

Because the ellipse is symmetric with respect to both axes, the total area A is four times the area in the first quadrant. The part of the ellipse in the first quadrant is given by the function
y = \frac{b}{a}\sqrt{a^2 - x^2} \ 0 \le x \le a and so:\ \frac{1}{4} A = \int_0^a \frac{b}{a} \sqrt{a^2 - x^2} \,dx

To evaluate this integral we substitute x = asinθ. Then dx = acosθdθ. To change the limits of integration we note that when x = 0, sinθ = 0, so θ = 0; when x = a, sinθ = 1, so \theta = \frac{\pi}{2}.

Also

\sqrt{a^2 - x^2} \ = \ \sqrt{a^2 - a^2 \sin^2 \theta} \ = \ \sqrt{a^2 \cos^2 \theta} \ = \ a |\cos \theta| \ = \ a \cos \theta

since 0 \le \theta \le \frac{\pi}{2}

Therefore

A = 4 \frac{b}{a} \int_0^a \sqrt{a^2 - x^2} \,dx \ = 4 \frac{b}{a} \int_{0}^{\pi/2} a \cos \theta \cdot a \cos \theta \,d\theta

= 4ab \int_{0}^{\pi/2} \cos^2 \theta \,d\theta \ = 4ab \int_{0}^{\pi/2} \frac{1}{2} (1 + \cos 2\theta)\,d\theta

= 2ab\left[\theta + \frac{1}{2} \sin 2\theta\right]_{0}^{\pi/2} \ = 2ab \left(\frac{\pi}{2} + 0 - 0\right)

\ \ \ = \pi ab.

This shows that the area of an ellipse with semiaxes a and b is πab. In particular, taking a = b = r, the formula for the area of a circle with radius r, A = πr2, has been proven.

[edit] Proof using limits

A circle is also an infinite-sided polygon. Therefore, the area bounded by a circle can be derived from the area bounded by a regular polygon as the number of its sides goes to infinity. For a regular polygon, the distance between its geometrical centre and any of its vertices is a constant, which we will refer to as r. The area of a triangle created by the centre and two adjacent vertices is therefore \frac{r^2}{2} \sin \theta, where the central angle is \theta \ = \frac{2\pi}{n}, n being the number of sides of the polygon. The total area of the polygon is equal to the area of n such triangles, or n\frac{r^2}{2} \sin \frac{2\pi}{n}. Taking the limit of that as n goes to infinity, we get

A = \lim_{n \to \infty}n\frac{r^2}{2} \sin \frac{2\pi}{n}

\ \ \ = \lim_{n \to \infty}n\frac{2\pi}{2\pi}\frac{r^2}{2} \sin \frac{2\pi}{n}

\ \ \ = \lim_{n \to \infty}2\pi\frac{r^2}{2}\frac{\sin \frac{2\pi}{n}}{\frac{2\pi}{n}}

\ \ \ = \lim_{n \to \infty}\pi r^2\cdot 1

\ \ \ = \pi r^2

Answer 2

Use pi

Find the equation in your math book for the area and circumfrence of a circle and go from there.

Answer 3

ok, the circumference of a circle is given by the formula C=2(pi)r And the realm is given by A=(pi)(r^2) So plug the circumference into the first equation and sparkling up for r (the radius). Then plug that into the realm equation and also you need to get the realm. For this one, the answer must be about 509.3 sq. mm.

Answer 4

The circumference is equal to pi times d where d is the diameter. Or you could say pi times 2 times the radius. The area is pi times the radius squared (pi*r*r)

so, half the circumference to get pi*r. Divide this by pi and you have the radius. Then multiple up by pi*r

Answer 5

Divide the circumference by pi
Dive the result by 2 (this will give you the radius)
multiply this number by itself and then multiply that result by pi.
P (R2)

Answer 6

Circonference = 2 * PI * radius
This will hell you solve for the radius: i.e. r = Circ / (2 * PI)

Area = PI * r^2

A shortcut: area = Circ^2 / PI