Short Proofs

Short Proofs

SHORT PROOFS

Balmoral Software

Circle
Bicylinder
Sphere

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. This sphere is a solid of revolution of a non-equable closed semicircle with perimeter 3π + 6 = 15.42 and area 9π/2 = 14.14.

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