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Chord Length Formula for the Area of an Annulus

Updated on January 02, 2014
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TR Smith is a product designer and former teacher who uses math in her work every day.

There are three formulas for calculating the area of an annulus using any two of the measurements for outer radius, inner radius, and thickness (width). However, there is another equation for the area of an annulus that requires only the length of a chord. The chord must span across the outer circle and be tangent to the inner circle at the chord's midpoint. All chords that satisfy these conditions have the same length for a given annulus. If you know this chord's length, you can plug it into a simple geometry formula to figure the total area of the annulus.


Chord Length Annulus Equation

Consider a chord whose endpoints lie on the circumference of the outer circle and whose midpoint lies on the circumference of the inner circle. For the purposes of this geometry tutorial, we will refer to such a chord as the inner tangent chord of the annulus. If the length of this chord is X, then the area of the annulus is given by the equation

Area = π(X/2)^2
= (π/4)X^2

For example, if an annulus has inner tangent chord length of 50, then the area of the annulus is (π/4)*50^2 = 625π, or approximately 1963.5.

As it turns out, annuli (annuluses) with different outer and inner radii lengths can have the same inner tangent chord length. Here are several examples of this mathematical phenomenon.

In the figure above the dark line drawn across each annulus is an inner tangent chord. All the lines have the same length even though the annuli are of different sizes. Since the chord length is equal in each annulus, all three have the same area. For example, the following annuli all have a tangent chord length of 120:

  • annulus with R = 61 and r = 11
  • annulus with R = 229 and r = 221
  • annulus with R = 109 and r = 91


Why the Formula Works

Why does knowing the length of the inner tangent chord allow you to compute the area of an annulus? To see why the equation works, we need to related the length of the inner tangent chord of the annulus to the outer radius and inner radius. As you can see from the diagram below, if the chord length is X, the outer radius R, and the inner radius r, then the three lengths are related by the Pythagorean equation

R^2 = r^2 + (X/2)^2

From the diagram above, you can see that the inner radius and half the chord form two legs of a right triangle whose hypotenuse is the outer radius. Thus we get the relation R^2 = r^2 + (X/2)^2, or equivalently

(X/2)^2 = R^2 - r^2

Recall that the area of an annulus can be computed with the formula π(R^2 - r^2), which is the difference in area between two circles. Since R^2 - r^2 = (X/2)^2, this formula can be written in the chord length form

Area = π(X/2)^2

Going back to the examples of annuli with different dimensions and the same chord length, we can see why they all have a chord length of 120:

  • 61^2 - 11^2 = 3721 - 121 = 3600
  • 229^2 - 221^2 = 52441 - 48841 = 3600
  • 109^2 - 91^2 = 11881 - 8281 = 3600

Since sqrt(3600) = 60 and 2*60 = 120, we get the inner tangent chord length for each annulus.


Examples

An annulus has an inner tangent chord length of 120 cm. What is its area in square meters? Since X = 120, we have

Area = (π/4)120^2
= 11309.73 cm^2
= 1.130973 m^2


An annulus has an inner tangent chord length of 15.72. If the outer radius were to increase by 2.1, the inner tangent chord length would increase by 6.2. What are the inner and outer radius of this annulus?

The first sentence of the problem tells us that R^2 - r^2 = (15.72 / 2)^2 = 61.7796. The second sentence tells us (R+2.1)^2 - r^2 = (21.92 / 2)^2 = 120.1216. If we subtract the first equation from the second, we get

(R+2.1)^2 - R^2 = 120.1216 - 61.7796
4.2R + 4.41 = 58.342
4.2R = 53.932
R ≈ 12.840952

Plugging this value of R into the first equation gives us the value of r:

12.840952^2 - r^2 = 61.7796
r^2 = 103.110458
r ≈ 10.154332


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