碧莲盛植发价格是多少 Area of a circle

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In geometry, the area enclosed by a circle of radius r is πr2. Here, the Greek letter π represents the constant ratio of the circumference of any circle to its diameter, approximately equal to 3.14159.

One method of deriving this formula, which originated with Archimedes, involves viewing the circle as the limit of a sequence of regular polygons with an increasing number of sides. The area of a regular polygon is half its perimeter multiplied by the distance from its center to its sides, and because the sequence tends to a circle, the corresponding formula–that the area is half the circumference times the radius–namely, A = ⁠1/2⁠ × 2πr × r, holds for a circle.

Terminology[edit]

Although often referred to as the area of a circle in informal contexts, strictly speaking, the term disk refers to the interior region of the circle, while circle is reserved for the boundary only, which is a curve and covers no area itself. Therefore, the area of a disk is the more precise phrase for the area enclosed by a circle.

History[edit]

Modern mathematics can obtain the area using the methods of integral calculus or its more sophisticated offspring, real analysis. However, the area of a disk was studied by the Ancient Greeks. Eudoxus of Cnidus in the fifth century B.C. had found that the area of a disk is proportional to its radius squared.[1] Archimedes used the tools of Euclidean geometry to show that the area inside a circle is equal to that of a right triangle whose base has the length of the circle's circumference and whose height equals the circle's radius in his book Measurement of a Circle. The circumference is 2πr, and the area of a triangle is half the base times the height, yielding the area π r2 for the disk. Prior to Archimedes, Hippocrates of Chios was the first to show that the area of a disk is proportional to the square of its diameter, as part of his quadrature of the lune of Hippocrates,[2] but did not identify the constant of proportionality.

Historical arguments[edit]

A variety of arguments he been advanced historically to establish the equation A = π r 2 {\displaystyle A=\pi r^{2}} to varying degrees of mathematical rigor. The most famous of these is Archimedes' method of exhaustion, one of the earliest uses of the mathematical concept of a limit, as well as the origin of Archimedes' axiom which remains part of the standard analytical treatment of the real number system. The original proof of Archimedes is not rigorous by modern standards, because it assumes that we can compare the length of arc of a circle to the length of a secant and a tangent line, and similar statements about the area, as geometrically evident.

Using polygons[edit]

The area of a regular polygon is half its perimeter times the apothem. As the number of sides of the regular polygon increases, the polygon tends to a circle, and the apothem tends to the radius. This suggests that the area of a disk is half the circumference of its bounding circle times the radius.[3]

Archimedes's proof[edit]

Following Archimedes' argument in The Measurement of a Circle (c. 260 BCE), compare the area enclosed by a circle to a right triangle whose base has the length of the circle's circumference and whose height equals the circle's radius. If the area of the circle is not equal to that of the triangle, then it must be either greater or less. We eliminate each of these by contradiction, leing equality as the only possibility. We use regular polygons in the same way.

Not greater[edit] 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. If the total area of those gaps, G4, is greater than E, split each arc in half. This makes the inscribed square into an inscribed octagon, and produces eight segments with a smaller total gap, G8. Continue splitting until the total gap area, Gn, is less than E. Now the area of the inscribed polygon, Pn = C − Gn, must be greater than that of the triangle.

E = C − T > G n P n = C − G n > C − E P n > T {\displaystyle {\begin{aligned}E&{}=C-T\\&{}>G_{n}\\P_{n}&{}=C-G_{n}\\&{}>C-E\\P_{n}&{}>T\end{aligned}}}

But this forces a contradiction, as follows. Draw a perpendicular from the center to the midpoint of a side of the polygon; its length, h, is less than the circle radius. Also, let each side of the polygon he length s; then the sum of the sides is ns, which is less than the circle circumference. The polygon area consists of n equal triangles with height h and base s, thus equals nhs/2. But since h  G n P n = C + G n