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Example of mathematical radius life
Let the outer diameter of the ring be r and the inner diameter be r.

The area of the excircle is: 3.14× r× r.

The area of the small circle inside is: 3.14× r× r.

Annular area = large circle area-small circle area = 3.14× r-3.14× r = 3.14× (r? -r? )

Extended data

A ring is equivalent to a hollow circle. The radius of hollow circle is small (R) and the radius of whole circle is large (capital R). The radius of the whole circle minus the radius of the hollow circle is the ring width. Examples in life are hollow steel pipes, donuts, rings and so on. Part of the ring is called a fan ring.

Ring circumference: outer circumference+inner circumference (? πx (major diameter+minor diameter))

Huanqu: Outside? Area of circle-area of inner circle (square of πx long radius-square of π x short radius \πx (square of long radius-square of short radius)

Expressed in letters:

Internal s+ external s (πR squared)

Outer s- inner s =π(R square -r square)

There is a second method:

S=π[(R-r)×(R+r)]

R= radius of great circle

R= width of ring = radius of large circle-radius of small circle

There is another way:

It is known that the outer diameter of a circular ring is d, and the thickness of the circular ring (that is, the difference between the inner and outer radii) is d.

d=R-r,

D-d=2R-(R-r)=R+r,

S=π[(R-r)×(R+r)]=π(D-d)×d can be derived from the first and second methods.

Ring area S=π(D-d)×d

This is based on the area of the outer diameter and thickness of the ring (that is, the difference between the outer diameter and the inner diameter). These two data are easy to measure in reality and suitable for calculating physical objects, such as round steel pipes.