Circle questionWatch

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#1
Hi.

I've just started looking at radians and I'm being very slow and dumb at the moment. (Anyone else have those days?)

Anyway, would some kind soul show me their working to the following (very easy) problem:

Two circles of radii 5cm and 12cm are drawn, partly overlapping. Their centres are 13cm apart. Find the area common to the two circles.

Simple. I'm sure. It's just one of those days...

Cheers.
0
13 years ago
#2
Hmmm.. you can find the area of the sectors, and the important part here is what I call the "sector ends", that is to say, the sector minus the triangle. Here's the deal: draw a line straight through the two points of circle intersection, and you should be able to divide the area of the overlap into two parts: the sector end of the first circle, and the sector end of the second circle.

Call the center of the 5cm circle A, the center of the other B, and one of the points of intersection C. ABC forms a 5-12-13 triangle (which is right angled). More importantly, though, the angle 1/2 thetha of sector a (of circle A) is arctan(12/5), and the angle 1/2 thetha of sector b is arctan(5/12).

Once you have the thethas, find the area of the sector ends. A = pi*r^2*thetha / 2pi - 1/2r^2sin thetha = 1/2r^2(thetha - sin thetha)

A total = 25/2(2arctan(12/5) - sin 2arctan(12/5)) +144/2(2arctan(5/12) - sin 2arctan(5/12))
13 years ago
#3
(Original post by Theophilus7)
Two circles of radii 5cm and 12cm are drawn, partly overlapping. Their centres are 13cm apart. Find the area common to the two circles.

Simple. I'm sure. It's just one of those days...

Cheers.
I get Area ≈ 26.26cm²

I Maybe wrong
0
#4
(Original post by Theophilus7)
Hi.

I've just started looking at radians and I'm being very slow and dumb at the moment. (Anyone else have those days?)

Anyway, would some kind soul show me their working to the following (very easy) problem:

Two circles of radii 5cm and 12cm are drawn, partly overlapping. Their centres are 13cm apart. Find the area common to the two circles.

Simple. I'm sure. It's just one of those days...

Cheers.
Thanks folks.

I assume if it wasn't a right-angled triangle the proper procedure would be to use Cosine Rule to get theta (?)
0
13 years ago
#5
(Original post by Theophilus7)
Thanks folks.

I assume if it wasn't a right-angled triangle the proper procedure would be to use Cosine Rule to get theta (?)

ya, right....
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