As proved by my good friend Archimedes,
in his _Measurement of a Circle_,
the area enclosed by a circle is equal
to that of a triangle whose base has the length
of the circle's circumference &
whose height equals the circle's radius,
which comes to π multiplied by the radius squared:
Area = pi r^2.
Equivalently, denoting diameter by _d_
Area =pi d^2/4 approx 0.7854d^2,
that is, approximately
79% of the circumscribing
square whose side is of length _d_
The circle is the plane curve enclosing
the maximum area for a given arc length.
This relates the circle to a problem
in the calculus of variations,
namely the isoperimetric inequality [of course]
Sep 1, 2018
Sep 1, 2018 at 6:54 PM UTC
As proved by my good friend Archimedes,
in his _Measurement of a Circle_,
the area enclosed by a circle is equal
to that of a triangle whose base has the length
of the circle's circumference &
whose height equals the circle's radius,
which comes to π multiplied by the radius squared:
Area = pi r^2.
Equivalently, denoting diameter by _d_
Area =pi d^2/4 approx 0.7854d^2,
that is, approximately
79% of the circumscribing
square whose side is of length _d_
The circle is the plane curve enclosing
the maximum area for a given arc length.
This relates the circle to a problem
in the calculus of variations,
namely the isoperimetric inequality [of course]
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