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| | 38 year old Naval Architect Cameron from Strathroy, has hobbies such as comics, property developers in singapore and films. Has been a travel freak and these days traveled to The Causses and the Cévennes.<br><br>Stop by my blog :: [http://rv4x4.com/?q=singapore-asia-pacific-and-international properties for sale in singapore] |
| The '''area of a disk''', more commonly called the '''area of a circle''', of [[radius]] ''r'' is equal to {{math|{{pi}}''r''<sup>2</sup>}}. Here the symbol {{pi}} ([[Greek alphabet|Greek]] letter [[Pi (letter)|pi]]) denotes the constant ratio of the [[circumference]] of a circle to its [[diameter]] or of the [[area]] of a circle to the square of its [[radius]]. It is easy to deduce the [[area]] of a [[disk (mathematics)|disk]] from basic principles: the area of a [[regular polygon]] is half its [[perimeter]] times its [[apothem]], and a regular polygon becomes a circle as the number of sides increase, so the area of a disk is half its circumference times its radius (i.e. {{frac|1|2}} × 2{{pi}}''r'' × ''r'').
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| ==History==
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| Modern mathematics can obtain the area using the methods of [[integral calculus]] or its more sophisticated offspring, [[real analysis]]. However the area of circles was studied by the [[Ancient Greeks]]. [[Eudoxus of Cnidus]] in the fifth century B.C. had found that the areas of circles are proportional to their radius squared.<ref>{{cite book|last=Stewart|first=James|title=Single variable calculus early transcendentals.|year=2003|publisher=Brook/Cole|location=Toronto ON|isbn=0-534-39330-6|pages=3|url=http://www.stewartcalculus.com/media/8_home.php|edition=5th.|quote=However, by indirect reasoning, Eudoxus (fifth century B.C.) used exhaustion to prove the familiar formula for the area of a circle: <math>A= \pi r^2.</math>}} <!--This quote may be an overstatement. I have not been able to confirm that he discovered the actual formula, but perhaps only the proportionality between A and r-squared.--></ref> The great mathematician [[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 2{{pi}}''r'', and the area of a triangle is half the base times the height, yielding the area {{pi}}''r''<sup>2</sup> 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]],<ref name="heath">{{citation|first=Thomas L.|last=Heath|authorlink=Thomas Little Heath|title=A Manual of Greek Mathematics|publisher=Courier Dover Publications|year=2003|isbn=0-486-43231-9|pages=121–132|url=http://books.google.com/books?id=_HZNr_mGFzQC&pg=PA121}}.</ref> but did not identify the [[constant of proportionality]].
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| ==Using polygons==
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| 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 circle is half its circumference times the radius.<ref>Hill, George. ''[http://books.google.com/books?id=zSMAAAAAYAAJ&lpg=PA124&ots=Z973CjYQgF&dq=%22the%20greater%20the%20number%20of%20sides%20of%20the%20polygon%22%20and%20%22George%20Anthony%20Hill%22&pg=PA124#v=onepage&q&f=false Lessons in Geometry: For the Use of Beginners]'', page 124 (1894).</ref>
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| ==Archimedes's proof==
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| Following {{Harvtxt|Archimedes|c. 260 BCE}}, compare 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, leaving equality as the only possibility. We use [[regular polygon]]s in an essential way.
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| ===Not greater===
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| [[File:Archimedes circle area proof - inscribed polygons.svg|thumb|right|Circle with square and octagon inscribed, showing area gap]]
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| Suppose the circle area, ''C'', may be greater than the triangle area, ''T'' = <sup>1</sup>⁄<sub>2</sub>''cr''. 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, ''G''<sub>4</sub>, 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, ''G''<sub>8</sub>. Continue splitting until the total gap area, ''G<sub>n</sub>'', is less than ''E''. Now the area of the inscribed polygon, ''P<sub>n</sub>'' = ''C'' − ''G<sub>n</sub>'', must be greater than that of the triangle.
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| :<math>\begin{align}
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| E &{}= C - T \\
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| &{}> G_n \\
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| P_n &{}= C - G_n \\
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| &{}> C - E \\
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| P_n &{}> T
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| \end{align}</math>
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| 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 have length ''s''; then the sum of the sides, ''ns'', is less than the circle circumference. The polygon area consists of ''n'' equal triangles with height ''h'' and base ''s'', thus equals <sup>1</sup>⁄<sub>2</sub>''nhs''. But since ''h'' < ''r'' and ''ns'' < ''c'', the polygon area must be less than the triangle area, <sup>1</sup>⁄<sub>2</sub>''cr'', a contradiction. Therefore our supposition that ''C'' might be greater than ''T'' must be wrong.
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| ===Not less===
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| [[File:Archimedes circle area proof - circumscribed polygons.svg|thumb|right|Circle with square and octagon circumscribed, showing area gap]]
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| Suppose the circle area may be less than the triangle area. Let ''D'' denote the deficit amount. Circumscribe a square, so that the midpoint of each edge lies on the circle. If the total area gap between the square and the circle, ''G''<sub>4</sub>, is greater than ''D'', slice off the corners with circle tangents to make a circumscribed octagon, and continue slicing until the gap area is less than ''D''. The area of the polygon, ''P<sub>n</sub>'', must be less than ''T''.
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| :<math>\begin{align}
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| D &{}= T - C \\
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| &{}> G_n \\
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| P_n &{}= C + G_n \\
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| &{}< C + D \\
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| P_n &{}< T
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| \end{align}</math>
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| This, too, forces a contradiction. For, a perpendicular to the midpoint of each polygon side is a radius, of length ''r''. And since the total side length is greater than the circumference, the polygon consists of ''n'' identical triangles with total area greater than ''T''. Again we have a contradiction, so our supposition that ''C'' might be less than ''T'' must be wrong as well.
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| Therefore it must be the case that the area of the circle is precisely the same as the area of the triangle. This concludes the proof.
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| ==Rearrangement proof==
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| [[File:CircleArea.svg|thumb|right|Circle area by rearrangement]]
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| [[File:Pie Are Square.gif|thumb|Animated demonstration of rearrangement]]
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| Following Satō Moshun {{Harv|Smith|Mikami|1914|loc=pp. 130–132}} and [[Leonardo da Vinci]] {{Harv|Beckmann|1976|loc=p. 19}}, we can use inscribed regular polygons in a different way. Suppose we inscribe a [[hexagon]]. Cut the hexagon into six triangles by splitting it from the center. Two opposite triangles both touch two common diameters; slide them along one so the radial edges are adjacent. They now form a [[parallelogram]], with the hexagon sides making two opposite edges, one of which is the base, ''s''. Two radial edges form slanted sides, and the height is ''h'' (as in the Archimedes proof). In fact, we can assemble all the triangles into one big parallelogram by putting successive pairs next to each other. The same is true if we increase to eight sides and so on. For a polygon with 2''n'' sides, the parallelogram will have a base of length ''ns'', and a height ''h''. As the number of sides increases, the length of the parallelogram base approaches half the circle circumference, and its height approaches the circle radius. In the limit, the parallelogram becomes a rectangle with width {{pi}}''r'' and height ''r''.
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| :{| cellpadding="3" cellspacing="0" frame="vsides" style="text-align:center"
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| |+ '''''Unit disk area by rearranging n polygons.'''''</span>
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| |- style="background-color:#eeeeee"
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| ! colspan="3" | polygon !! colspan="6" | parallelogram
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| |- style="background-color:#dddddd"
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| ! ''n'' !! !! side !! !! base !! !! height !! !! area
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| |-
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| | align="right" | 4 || || 1.4142136 || || 2.8284271 || || 0.7071068 || || 2.0000000
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| |-
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| | align="right" | 6 || || 1.0000000 || || 3.0000000 || || 0.8660254 || || 2.5980762
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| |-
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| | align="right" | 8 || || 0.7653669 || || 3.0614675 || || 0.9238795 || || 2.8284271
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| |-
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| | align="right" | 10 || || 0.6180340 || || 3.0901699 || || 0.9510565 || || 2.9389263
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| |-
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| | align="right" | 12 || || 0.5176381 || || 3.1058285 || || 0.9659258 || || 3.0000000
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| |-
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| | align="right" | 14 || || 0.4450419 || || 3.1152931 || || 0.9749279 || || 3.0371862
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| |-
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| | align="right" | 16 || || 0.3901806 || || 3.1214452 || || 0.9807853 || || 3.0614675
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| |-
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| | align="right" | 96 || || 0.0654382 || || 3.1410320 || || 0.9994646 || || 3.1393502
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| |-
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| | ∞ || || 1/∞ || || {{pi}} || || 1 || || {{pi}}
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| |}
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| ==Onion proof==
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| [[File:Circle area rings.svg|thumb|right|Area of the disk via ring integration]]
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| Using calculus, we can sum the area incrementally, partitioning the disk into thin concentric rings like the layers of an [[onion]]. This is the method of [[shell integration]] in two dimensions. For an infinitesimally thin ring of the "onion" of radius ''t'', the accumulated area is 2{{pi}}''t dt'', the circumferential length of the ring times its infinitesimal width (you can approach this ring by a rectangle with width=2{{pi}}''t'' and height=''dt''). This gives an elementary integral for a disk of radius ''r''.
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| :<math>\begin{align}
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| \mathrm{Area}(r) &{}= \int_0^{r} 2 \pi t \, dt \\
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| &{}= \left[ (2\pi) \frac{t^2}{2} \right]_{t=0}^{r}\\
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| &{}= \pi r^2.
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| \end{align} </math>
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| ==Triangle proof==
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| Similar to the onion proof outlined above, we could exploit calculus in a different way in order to arrive at the formula for the area of a circle. In this case, we imagine dividing up a circle into triangles, each with a base of length equal to the circle's radius and a height that is infinitesimally small. The area of each of these triangles is equal to 1/2 * r * dt. By summing up (integrating) all of the areas of these triangles, we arrive at the formula for the circle's area:
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| :<math>\begin{align}
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| \mathrm{Area}(r) &{}= \int_0^{2\pi r} \frac{1}{2} r \, dt \\
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| &{}= \left[ \frac{1}{2} r t \right]_{t=0}^{2 \pi r}\\
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| &{}= \pi r^2.
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| \end{align} </math>
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| ==Fast approximation==
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| The calculations Archimedes used to approximate the area numerically were laborious, and he stopped with a polygon of 96 sides. A faster method uses ideas of [[Willebrord Snell]] (''Cyclometricus''<!--: de circuli dimensione secundum logistarum abacos, & ad mechanicem accuratissima; atque omnium parabilissima: eiusdemque usus in quarumlibet, adscriptarum inventione longe elegantissimus, & quidem ex ratione diametri ad suam peripheriam data--><!--, Lugduni Batavorum: Elzevir-->, 1621), further developed by [[Christiaan Huygens]] (''De Circuli Magnitudine Inventa'', 1654), described in {{Harvtxt|Gerretsen|Verdenduin|1983|loc=pp. 243–250}}.
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| ===Archimedes' doubling method===
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| Given a circle, let ''u<sub>n</sub>'' be the [[perimeter]] of an inscribed regular ''n-''gon, and let ''U<sub>n</sub>'' be the perimeter of a circumscribed regular ''n-''gon. Then ''u<sub>n</sub>'' and ''U<sub>n</sub>'' are lower and upper bounds for the circumference of the circle that become sharper and sharper as ''n'' increases, and their average (''u<sub>n</sub>'' + ''U<sub>n</sub>'')/2 is an especially good approximation to the circumference. To compute ''u<sub>n</sub>'' and ''U<sub>n</sub>'' for large ''n'', Archimedes derived the following doubling formulae:
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| :<math>u_{2n} = \sqrt{U_{2n} u_{n}}</math> ([[geometric mean]])
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| :<math>U_{2n} = \frac{2 U_{n} u_{n}}{ U_{n} + u_{n}}</math> ([[harmonic mean]]).
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| Starting from a hexagon, Archimedes doubled ''n'' four times to get a 96-gon, which gave him a good approximation to the circumference of the circle.
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| In modern notation, we can reproduce his computation (and go farther) as follows.
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| For a unit circle, an inscribed hexagon has ''u''<sub>6</sub> = 6, and a circumscribed hexagon has ''U''<sub>6</sub> = 4√3.
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| Doubling seven times yields
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| :{| frame="vsides" style="text-align:center" cellpadding="3" cellspacing="0"
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| |+ '''''Archimedes doubling seven times; n = 6×2<sup>k</sup>.'''''
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| |- style="background-color:#eeeeee"
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| ! ''k'' !! !! ''n'' !! !! ''u<sub>n</sub>'' !! !! ''U<sub>n</sub>'' !! !! (''u<sub>n</sub>'' + ''U<sub>n</sub>'')/4
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| |-
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| | 0 || || 6 || || 6.0000000 || || 6.9282032 || || 3.2320508
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| |-
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| | 1 || || 12 || || 6.2116571 || || 6.4307806 || || 3.1606094
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| |-
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| | 2 || || 24 || || 6.2652572 || || 6.3193199 || || 3.1461443
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| |-
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| | 3 || || 48 || || 6.2787004 || || 6.2921724 || || 3.1427182
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| |-
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| | 4 || || 96 || || 6.2820639 || || 6.2854292 || || 3.1418733
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| |-
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| | 5 || || 192 || || 6.2829049 || || 6.2837461 || || 3.1416628
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| |-
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| | 6 || || 384 || || 6.2831152 || || 6.2833255 || || 3.1416102
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| |-
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| | 7 || || 768 || || 6.2831678 || || 6.2832204 || || 3.1415970
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| |}
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| (Here (''u<sub>n</sub>'' + ''U<sub>n</sub>'')/2 approximates the circumference of the unit circle, which is 2{{pi}}, so (''u<sub>n</sub>'' + ''U<sub>n</sub>'')/4 approximates {{pi}}.)
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| The last entry of the table has <sup>355</sup>⁄<sub>113</sub> as one of its [[Continued_fraction#Best_rational_approximation|best rational approximations]];
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| i.e., there is no better approximation among rational numbers with denominator up to 113.
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| The number <sup>355</sup>⁄<sub>113</sub> is also an excellent approximation to {{pi}}, better than any other rational number with denominator less than 16604.<ref>[http://shreevatsa.wordpress.com/2011/01/10/not-all-best-rational-approximations-are-the-convergents-of-the-continued-fraction/ Not all best rational approximations are the convergents of the continued fraction!]</ref>
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| ===The Snell–Huygens refinement===
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| Snell proposed (and Huygens proved) a tighter bound than Archimedes':
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| :<math> n \frac{3 \sin \frac{\pi}{n}}{2+\cos\frac{\pi}{n}} < \pi < n [2 \sin \frac{\pi}{3 n} + \tan \frac{\pi}{3 n}]. </math>
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| This for ''n'' = 48 gives a better approximation (about 3.14159292) than Archimedes' method for ''n'' = 768.
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| ===Derivation of Archimedes' doubling formulae===
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| [[File:Huygens + Snell + van Ceulen - regular polygon doubling.svg|thumb|right|Circle with similar triangles: circumscribed side, inscribed side and complement, inscribed split side and complement]]
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| Let one side of an inscribed regular ''n-''gon have length ''s<sub>n</sub>'' and touch the circle at points A and B. Let A′ be the point opposite A on the circle, so that A′A is a diameter, and A′AB is an inscribed triangle on a diameter. By [[Thales' theorem]], this is a right triangle with right angle at B. Let the length of A′B be ''c<sub>n</sub>'', which we call the complement of ''s<sub>n</sub>''; thus ''c<sub>n</sub>''<sup>2</sup>+''s<sub>n</sub>''<sup>2</sup> = (2''r'')<sup>2</sup>. Let C bisect the arc from A to B, and let C′ be the point opposite C on the circle. Thus the length of CA is ''s''<sub>2''n''</sub>, the length of C′A is ''c''<sub>2''n''</sub>, and C′CA is itself a right triangle on diameter C′C. Because C bisects the arc from A to B, C′C perpendicularly bisects the chord from A to B, say at P. Triangle C′AP is thus a right triangle, and is [[similarity (geometry)#Similar triangles|similar]] to C′CA since they share the angle at C′. Thus all three corresponding sides are in the same proportion; in particular, we have C′A : C′C = C′P : C′A and AP : C′A = CA : C′C. The center of the circle, O, bisects A′A, so we also have triangle OAP similar to A′AB, with OP half the length of A′B. In terms of side lengths, this gives us
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| :<math>\begin{align}
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| c_{2n}^2 &{}= \left( r + \frac{1}{2} c_n \right) 2r \\
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| c_{2n} &{}= \frac{s_n}{s_{2n}} .
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| \end{align}</math>
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| In the first equation C′P is C′O+OP, length ''r''+<sup>1</sup>⁄<sub>2</sub>''c<sub>n</sub>'', and C′C is the diameter, 2''r''. For a unit circle we have the famous doubling equation of [[Ludolph van Ceulen]],
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| :<math> c_{2n} = \sqrt{2+c_n} . \,\!</math>
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| If we now circumscribe a regular ''n-''gon, with side A″B″ parallel to AB, then OAB and OA″B″ are similar triangles, with A″B″ : AB = OC : OP. Call the circumscribed side ''S<sub>n</sub>''; then this is ''S<sub>n</sub>'' : ''s<sub>n</sub>'' = 1 : <sup>1</sup>⁄<sub>2</sub>''c<sub>n</sub>''. (We have again used that OP is half the length of A′B.) Thus we obtain
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| :<math> c_n = 2\frac{s_n}{S_n} . \,\!</math>
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| Call the inscribed perimeter ''u<sub>n</sub>'' = ''ns<sub>n</sub>'', and the circumscribed perimenter ''U<sub>n</sub>'' = ''nS<sub>n</sub>''. Then combining equations, we have
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| :<math> c_{2n} = \frac{s_n}{s_{2n}} = 2 \frac{s_{2n}}{S_{2n}} , </math>
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| so that
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| :<math> u_{2n}^2 = u_n U_{2n} . \,\!</math>
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| This gives a [[geometric mean]] equation.
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| We can also deduce
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| :<math> 2 \frac{s_{2n}}{S_{2n}} \frac{s_n}{s_{2n}} = 2 + 2 \frac{s_n}{S_n} , </math>
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| or
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| :<math> \frac{2}{U_{2n}} = \frac{1}{u_n} + \frac{1}{U_n} . </math>
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| This gives a [[harmonic mean]] equation.
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| ==Dart approximation==
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| [[File:Circle area Monte Carlo integration.svg|thumb|right|Unit circle area Monte Carlo integration. Estimate by these 900 samples is 4×<sup>709</sup>⁄<sub>900</sub> = 3.15111...]]
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| When more efficient methods of finding areas are not available, we can resort to “throwing darts”. This [[Monte Carlo method]] uses the fact that if random samples are taken uniformly scattered across the surface of a square in which a disk resides, the proportion of samples that hit the disk approximates the ratio of the area of the disk to the area of the square. This should be considered a method of last resort for computing the area of a disk (or any shape), as it requires an enormous number of samples to get useful accuracy; an estimate good to 10<sup>−''n''</sup> requires about 100<sup>''n''</sup> random samples {{Harv|Thijsse|2006|loc=p. 273}}.
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| ==Finite rearrangement==
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| We have seen that by partitioning the disk into an infinite number of pieces we can reassemble the pieces into a rectangle. A remarkable fact discovered relatively recently {{Harv|Laczkovich|1990}} is that we can dissect the disk into a large but ''finite'' number of pieces and then reassemble the pieces into a square of equal area. This is called [[Tarski's circle-squaring problem]]. The nature of Laczkovich's proof is such that it proves the existence of such a partition (in fact, of many such partitions) but does not exhibit any particular partition.
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| ==Generalizations==
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| We can stretch a disk to form an [[ellipse]]. Because this stretch is a [[linear transformation]] of the plane, it has a distortion factor which will change the area but preserve ''ratios'' of areas. This observation can be used to compute the area of an arbitrary ellipse from the area of a unit circle.
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| Consider the unit circle circumscribed by a square of side length 2. The transformation sends the circle to an ellipse by stretching or shrinking the horizontal and vertical diameters to the major and minor axes of the ellipse. The square gets sent to a rectangle circumscribing the ellipse. The ratio of the area of the circle to the square is {{pi}}/4, which means the ratio of the ellipse to the rectangle is also {{pi}}/4. Suppose ''a'' and ''b'' are the lengths of the major and minor axes of the ellipse. Since the area of the rectangle is ''ab'', the area of the ellipse is {{pi}}''ab''/4.
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| We can also consider analogous measurements in higher dimensions. For example, we may wish to find the volume inside a sphere. When we have a formula for the surface area, we can use the same kind of “onion” approach we used for the disk.
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| ==Triangle method==
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| [[File:TriangleFromCircle.gif|thumb|right|Circle unwrapped to form a triangle]]
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| [[File:Area of circle and triangle.svg|thumb|The circle and the triangle are equal in area.]]
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| This approach is a slight modification of onion proof. Consider unwrapping the concentric circles to straight strips. This will form a right angled triangle with r as its height and 2{{pi}}r (being the outer slice of onion) as its base.
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| Finding the area of this triangle will give the area of circle
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| :<math>\begin{align}
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| Area &{}= \frac{1}{2} * base * height \\
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| &{}= \frac{1}{2} * 2 \pi r * r \\
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| &{}= \pi r^2
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| \end{align}</math>
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| The opposite and adjacent angles for this triangle are respectively in degrees 9.0430611..., 80.956939... and in radians 0.1578311..., 1.4129651....
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| ==Bibliography==
| |
| * {{citation
| |
| | author = Archimedes
| |
| | author-link = Archimedes
| |
| | editor = [[T. L. Heath]] (trans.)
| |
| | title = The Works of Archimedes
| |
| | year = c. 260 BCE<!-- very rough guess based on Wilbur Knorr's claim that 'Measurement' is early work -->
| |
| | publication-date = 2002
| |
| | publisher = [[Dover Publications|Dover]]
| |
| | isbn = 978-0-486-42084-4
| |
| | pages = 91–93
| |
| | chapter = Measurement of a circle
| |
| | url = http://www.archive.org/details/worksofarchimede029517mbp
| |
| }}<br />(Originally published by [[Cambridge University Press]], 1897, based on J. L. Heiberg's Greek version.)
| |
| * {{citation
| |
| | last = Beckmann
| |
| | first = Petr
| |
| | author-link = Petr Beckmann
| |
| | title = A History of Pi
| |
| | year = 1976
| |
| | publisher = [[St. Martin's Press|St. Martin's Griffin]]
| |
| | isbn = 978-0-312-38185-1
| |
| }}
| |
| * {{citation
| |
| | last1=Gerretsen
| |
| | first1=J.
| |
| | last2=Verdenduin
| |
| | first2=P.
| |
| | chapter=Chapter 8: Polygons and Polyhedra
| |
| | pages=243–250
| |
| | title=Fundamentals of Mathematics, Volume II: Geometry
| |
| | editor= H. Behnke, F. Bachmann, K. Fladt, H. Kunle (eds.), S. H. Gould (trans.)
| |
| | publisher=[[MIT Press]]
| |
| | year=1983
| |
| | isbn=978-0-262-52094-2
| |
| }}<br />(Originally ''Grundzüge der Mathematik'', Vandenhoeck & Ruprecht, Göttingen, 1971.)
| |
| * {{citation
| |
| | last =Laczkovich
| |
| | first =Miklós
| |
| | author-link =Miklós Laczkovich
| |
| | title =Equidecomposability and discrepancy: A solution to Tarski's circle squaring problem
| |
| | journal = [[Crelle's Journal|Journal für die reine und angewandte Mathematik]]
| |
| | volume =404
| |
| | pages =77–117
| |
| | year =1990
| |
| | url= http://dz-srv1.sub.uni-goettingen.de/sub/digbib/loader?ht=VIEW&did=D262326
| |
| | mr = 1037431
| |
| }}
| |
| * {{citation
| |
| | last = Lange
| |
| | first = Serge
| |
| | author-link = Serge Lang
| |
| | chapter = The length of the circle
| |
| | title = Math! : Encounters with High School Students
| |
| | year = 1985
| |
| | publisher = [[Springer-Verlag]]
| |
| | isbn = 978-0-387-96129-3
| |
| }}
| |
| * {{citation
| |
| | last1=Smith
| |
| | first1=David Eugene
| |
| | author1-link=David Eugene Smith
| |
| | last2=Mikami
| |
| | first2=Yoshio
| |
| | pages=130–132
| |
| | title=A history of Japanese mathematics
| |
| | place=Chicago
| |
| | publisher=[[Open Court Publishing Company|Open Court Publishing]]
| |
| | year=1914
| |
| | isbn=978-0-87548-170-8
| |
| | url=http://www.archive.org/details/historyofjapanes00smituoft
| |
| }}
| |
| * {{citation
| |
| | title=Computational Physics
| |
| | last1=Thijsse
| |
| | first1=J. M.
| |
| | pages=273
| |
| | publisher=Cambridge University Press
| |
| | year=2006
| |
| | isbn=978-0-521-57588-1
| |
| }}
| |
| | |
| ==References==
| |
| <references />
| |
| | |
| == External links ==
| |
| * [http://www.area-of-a-circle.com Area of a Circle Calculator]
| |
| * [http://www.mathopenref.com/circlearea.html Area enclosed by a circle] (with interactive animation)
| |
| * [http://www.sciencenews.org/articles/20041030/mathtrek.asp Science News on Tarski problem]
| |
| | |
| [[Category:Area]]
| |
| [[Category:Circles]]
| |
| [[Category:Articles containing proofs]]
| |
| | |
| [[de:Kreis#Kreisfläche]]
| |