Short Proofs

SHORT PROOFS

Balmoral Software


Equable Circle

Solutions: 1

For a circle of radius r, we have

A = π r2 = 2π r = P
or
r = 2
The associated area and perimeter are 4π.


Equable Bicylinder

Solutions: 1

A bicylinder is the orthogonal intersection of two cylinders having the same radius. If r is the radius, then

V = (16/3)r3 = 16r2 = S,
or
r = 3
The associated volume and surface area are 144.


Equable Sphere

Solutions: 1

For a sphere of radius r, we have

V = (4/3)π r3 = 4π r2 = S

r = 3

The associated volume and surface area are 36π.


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