Do the two smaller arcs of an arbelos add up to the largest arc of the arbelos?

Question

For example, if I calculated the two smaller arcs to be 6 cm in length total, would that be the same as the arc above it?

Answer 1

I had to Google arbelos to learn that new word. Thanks for that.

 

Now let's make one, beginning with a line segment AB of length 2(R+r).  Then construct a semicircle using that as diameter.  (All semicircles will be above AB.)

 

Using 2R as a diameter having its left endpoint at A, and naming the right endpoint of this (lesser) diameter E, construct a semicircle.  We'll call its arclength S.

 

Using 2r as a diameter having its left endpont at E (and right endpoint coinciding with point B), construct another semicircle.  We'll call its arclength s.

 

What is the length of the largest semicircle?  one half of the circumference of a circle whose radius is R+r:

½(2pi)(R+r) = pi(R+r)...let' identify this arclength by letter T.

 

What is the length of S?  S = ½(2pi)(R) = pi(R).

 

What is the length of s? s = ½(2pi)(r) = pi(r).

 

We have shown that T = S + s.  Your answer is YES.

 

 

 

Answer 2

The arbelos contains three semi-circles: 1, 2, and 3

They have diameters: d₁, d₂, and d₁ + d₂

The arc of each (half of a circumference) is:

a₁ = 0.5d₁π

a₂ = 0.5d₂π

a₃ = 0.5(d₁ + d₂)π, or 0.5d₁π + 0.5d₂π

This is equal to a₁ + a₂

 

In your figure is the combined measures of the interior arcs is 6 cm, the measure of the large arc is 6 cm.