How to prove the area formula for a piece of a circle?

Question

I have a really tough question on my homework, any help is appreciated.

What could I do to mathematically prove that the area of any sector of a circle with central angle '@' is A=1/2@r^2 (@ is the measure of the center angle in radians, r is radius, 1/2 is one half). It's a basic formula, but I can't think of any way other than example, and that won't work.

Someone better at math, any suggestions?

Answer 1

If this a homework problem, for SURE the teacher expects you to figure it in this way:

The area A of a full circle of radius r is:

A = π r²

A sector angle of θ is a fraction of the full circle of 2π. This fraction is

θ / 2π

The area S of the sector with central angle θ is that fraction of the full circle, or

S = (θ / 2π) A = (θ / 2π) (π r²) = (θ / 2) r²

Keep in mind that 360 degrees in radians is 2π. That's it.

Addendum: Read the guy's answer below if you're taking a calculus class.

Answer 2

Start with the length of the arc. This would be @*r since ds, an infinitely small piece of arc length equals the radius times the angle the piece sweeps over with respect to the center of the circle . Make that a differential equation. So dA, an infinitely small piece of the sector's area, dA=@*r*dr. Then integrate this to find the area of the arc swept out by the radius. The integral of dA is A and the integral of @*r*dr is 1/2*@*r^2.

Answer 3

Remember that the area of the circle for an angle of 2 pi radians is pi r^2. Then, its a simple proportion for any other angle. for example, for pi/2 (90 degrees), the area is pi/4 r^2, or 1/4 the circle.

Answer 4

The sector and circle have their areas and measures of arcs proportional to each other. Area of sector/Area of circle = m(arc)/m(circle) S/pi*r^2 = theta/ 360 S = Area of sector S = theta/360 * pi*r^2 Let O be the centre and OA and OB be the radii of the circle forming the sector. O-AB. Draw AC perpendicular to OB. theta is m(angle AOB). sin theta = AC/OA = AC/r, since OA = OB = r. AC = r*sin theta. Area of triangle OAB = 1/2 base *height = 1/2 OB*AC = 1/2 r*r* sin theta.= 1/2 r^2 sin theta If theta is 60 degrees, the sin theta is (sq rt 3)/2 A(triangle OAB) = 1/2 r^2 (sq rt 3)/2 Area of segment = Area of sector - Area of triangle = theta/360 * pi*r^2 - 1/2 r^2 sin theta = theta/360 * pi*r^2 - 1/2 r^2 (sq rt 3)/2 if theta is 60 degrees.

Answer 5

(c.a)/(2pi) *Pi*r^2=(c.a)/2 *r^2
What you want to do is multiply the total area of the circle by the fraction of the circle represented by sector--that is where the (c.a)/(2pi) comes from.

Answer 6

The area of a circle is =pi*r^2 (at 360 degrees =2pi (radians))
at 2pi ...........pi*r^2
at @..............x
=>x=(@*pi*r^2)/2pi=>x=1/2@*r^2

Answer 7

you could integrate, but that's nasty, what with finding the bounds over which to integrate. there might be an easier or more elegant way but i can't think of it