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Circle with square and octagon inscribed, showing area gap. Suppose that the area C enclosed by the circle is greater than the area T = cr/2 of the triangle. Let E denote the excess amount. Inscribe a square in the circle, so that its four corners lie on the circle. Between the square and the circle are four segments.
The solution of the problem of squaring the circle by compass and straightedge requires the construction of the number , the length of the side of a square whose area equals that of a unit circle. If π {\displaystyle {\sqrt {\pi }}} were a constructible number , it would follow from standard compass and straightedge constructions that π ...
Squircle centred on the origin (a = b = 0) with minor radius r = 1: x 4 + y 4 = 1. A squircle is a shape intermediate between a square and a circle. There are at least two definitions of "squircle" in use, one based on the superellipse, the other arising from work in optics. The word "squircle" is a portmanteau of the words
Any square can be inscribed in a circle whose center is the center of the square. If the common length of its four sides is equal to a {\displaystyle a} then the length of the diagonal is equal to a 2 {\displaystyle a{\sqrt {2}}} according to the Pythagorean theorem , and Ptolemy's relation obviously holds.
Circle packing in a square is a packing problem in recreational mathematics, where the aim is to pack n unit circles into the smallest possible square. Equivalently, the problem is to arrange n points in a unit square aiming to get the greatest minimal separation, d n , between points. [ 1 ]
Area enclosed by a circle = π × area of the shaded square Main article: Area of a circle As proved by 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 and whose height equals the circle's radius, [ 11 ] which comes to π ...
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The area within a circle is equal to the radius multiplied by half the circumference, or A = r x C /2 = r x r x π.. Liu Hui argued: "Multiply one side of a hexagon by the radius (of its circumcircle), then multiply this by three, to yield the area of a dodecagon; if we cut a hexagon into a dodecagon, multiply its side by its radius, then again multiply by six, we get the area of a 24-gon; the ...