Ingeometry, thearea enclosed by acircle ofradiusr isπr². Here, the Greek letterπ represents theconstant ratio of thecircumference of any circle to itsdiameter, approximately equal to 3.14159.

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

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

Modern mathematics can obtain the area using the methods ofintegral calculus or its more sophisticated offspring,real analysis. However, the area of a disk was studied by theAncient 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 ofEuclidean geometry to show that the area inside a circle is equal to that of aright triangle whose base has the length of the circle's circumference and whose height equals the circle's radius in his bookMeasurement of a Circle. The circumference is 2πr, and the area of a triangle is half the base times the height, yielding the areaπ 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 thelune of Hippocrates,^[2] but did not identify theconstant of proportionality.

A variety of arguments have been advanced historically to establish the equation${\displaystyle A=\pi r^2}$  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 alimit, as well as the origin ofArchimedes' axiom which remains part of the standard analytical treatment of thereal 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.

The area of aregular polygon is half its perimeter times theapothem. 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]

Following Archimedes' argument inThe 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, leaving equality as the only possibility. We useregular polygons in the same way.

Suppose that the areaC enclosed by the circle is greater than the areaT =cr/2 of the triangle. LetE 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 thanE, 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,G_n, is less thanE. Now the area of the inscribed polygon,P_n =C − G_n, must be greater than that of the triangle.

 

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 lengths; then the sum of the sides isns, which is less than the circle circumference. The polygon area consists ofn equal triangles with heighth and bases, thus equalsnhs/2. But sinceh < r andns < c, the polygon area must be less than the triangle area,cr/2, a contradiction. Therefore, our supposition thatC might be greater thanT must be wrong.

Suppose that the area enclosed by the circle is less than the areaT of the triangle. LetD 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,G₄, is greater thanD, slice off the corners with circle tangents to make a circumscribed octagon, and continue slicing until the gap area is less thanD. The area of the polygon,P_n, must be less thanT.

 

This, too, forces a contradiction. For, a perpendicular to the midpoint of each polygon side is a radius, of lengthr. And since the total side length is greater than the circumference, the polygon consists ofn identical triangles with total area greater thanT. Again we have a contradiction, so our supposition thatC might be less thanT must be wrong as well.

Therefore, it must be the case that the area enclosed by the circle is precisely the same as the area of the triangle. This concludes the proof.

Following Satō Moshun (Smith & Mikami 1914, pp. 130–132),Nicholas of Cusa^[4] andLeonardo da Vinci (Beckmann 1976, p. 19), we can use inscribed regular polygons in a different way. Suppose we inscribe ahexagon. 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 aparallelogram, with the hexagon sides making two opposite edges, one of which is the base,s. Two radial edges form slanted sides, and the height,h is equal to itsapothem (as in the Archimedes proof). In fact, we can also assemble all the triangles into one big parallelogram by putting successive pairs next to each other. The same is true if we increase it to eight sides and so on. For a polygon with 2n sides, the parallelogram will have a base of lengthns, and a heighth. 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 widthπr and heightr.

Unit disk area by rearranging n polygons.

polygon			parallelogram
n	side		base	height	area
4	1.4142136		2.8284271	0.7071068	2.0000000
6	1.0000000		3.0000000	0.8660254	2.5980762
8	0.7653669		3.0614675	0.9238795	2.8284271
10	0.6180340		3.0901699	0.9510565	2.9389263
12	0.5176381		3.1058285	0.9659258	3.0000000
14	0.4450419		3.1152931	0.9749279	3.0371862
16	0.3901806		3.1214452	0.9807853	3.0614675
96	0.0654382		3.1410320	0.9994646	3.1393502
∞	1/∞		π	1	π

There are various equivalent definitions of the constant π. The conventional definition in pre-calculus geometry is the ratio of the circumference of a circle to its diameter:

${\displaystyle \pi =\frac{C}{D}.}$ 

However, because the circumference of a circle is not a primitive analytical concept, this definition is not suitable in modern rigorous treatments. A standard modern definition is thatπ is equal to twice the least positive root of thecosine function or, equivalently, the half-period of thesine (or cosine) function. The cosine function can be defined either as apower series, or as the solution of a certaindifferential equation. This avoids any reference to circles in the definition ofπ, so that statements about the relation ofπ to the circumference and area of circles are actually theorems, rather than definitions, that follow from the analytical definitions of concepts like "area" and "circumference".

The analytical definitions are seen to be equivalent, if it is agreed that the circumference of the circle is measured as arectifiable curve by means of theintegral

${\displaystyle C=2{\int }_{-R}^R\frac{Rdx}{\sqrt{R^2-x^2}}=2R{\int }_{-1}^1\frac{dx}{\sqrt{1-x^2}}.}$ 

The integral appearing on the right is anabelian integral whose value is a half-period of thesine function, equal toπ. Thus${\displaystyle C=2\pi R=\pi D}$  is seen to be true as a theorem.

Several of the arguments that follow use only concepts from elementary calculus to reproduce the formula${\displaystyle A=\pi r^2}$ , but in many cases to regard these as actual proofs, they rely implicitly on the fact that one can develop trigonometric functions and the fundamental constantπ in a way that is totally independent of their relation to geometry. We have indicated where appropriate how each of these proofs can be made totally independent of all trigonometry, but in some cases that requires more sophisticated mathematical ideas than those afforded by elementary calculus.

Using calculus, we can sum the area incrementally, partitioning the disk into thin concentric rings like the layers of anonion. This is the method ofshell integration in two dimensions. For an infinitesimally thin ring of the "onion" of radiust, the accumulated area is 2πt dt, the circumferential length of the ring times its infinitesimal width (one can approximate this ring by a rectangle with width=2πt and height=dt). This gives an elementary integral for a disk of radiusr.

 

It is rigorously justified by themultivariate substitution rule in polar coordinates. Namely, the area is given by adouble integral of the constant function 1 over the disk itself. IfD denotes the disk, then the double integral can be computed inpolar coordinates as follows:

 

which is the same result as obtained above.

An equivalent rigorous justification, without relying on the special coordinates of trigonometry, uses thecoarea formula. Define a function${\displaystyle \rho :{\mathbb{R}}^2\to \mathbb{R}}$  by$\rho (x,y)=\sqrt{x^2+y^2}$ . Note ρ is aLipschitz function whosegradient is a unit vector${\displaystyle |\mathrm{\nabla }\rho |=1}$  (almost everywhere). LetD be the disc  in${\displaystyle {\mathbb{R}}^2}$ . We will show that${\displaystyle {\mathcal{L}}^2(D)=\pi }$ , where${\displaystyle {\mathcal{L}}^2}$  is the two-dimensional Lebesgue measure in${\displaystyle {\mathbb{R}}^2}$ . We shall assume that the one-dimensionalHausdorff measure of the circle${\displaystyle \rho =r}$  is${\displaystyle 2\pi r}$ , the circumference of the circle of radiusr. (This can be taken as the definition of circumference.) Then, by the coarea formula,

 

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 disk. Consider unwrapping the concentric circles to straight strips. This will form a right angled triangle with r as its height and 2πr (being the outer slice of onion) as its base.

Finding the area of this triangle will give the area of the disk

 

The opposite and adjacent angles for this triangle are respectively in degrees 9.0430611..., 80.956939... and in radians 0.1578311...OEIS: A233527, 1.4129651...OEIS: A233528.

Explicitly, we imagine dividing up a circle into triangles, each with a height equal to the circle's radius and a base that is infinitesimally small. The area of each of these triangles is equal to${\displaystyle 1/2\cdot r\cdot du}$ . By summing up (integrating) all of the areas of these triangles, we arrive at the formula for the circle's area:

 

It too can be justified by a double integral of the constant function 1 over the disk by reversing theorder of integration and using a change of variables in the above iterated integral:

 

Making the substitution${\displaystyle u=r\theta ,du=rd\theta }$  converts the integral to

${\displaystyle {\int }_0^{2\pi r}\frac{1}{2}\frac{r^2}{r}du={\int }_0^{2\pi r}\frac{1}{2}rdu}$ 

which is the same as the above result.

The triangle proof can be reformulated as an application ofGreen's theorem in flux-divergence form (i.e. a two-dimensional version of thedivergence theorem), in a way that avoids all mention of trigonometry and the constantπ. Consider thevector field${\displaystyle \mathbf{r}=x\mathbf{i}+y\mathbf{j}}$  in the plane. So thedivergence ofr is equal to two, and hence the area of a discD is equal to

${\displaystyle A=\frac{1}{2}{\iint }_D\mathrm{div}\mathbf{r}dA.}$ 

By Green's theorem, this is the same as the outward flux ofr across the circle boundingD:

${\displaystyle A=\frac{1}{2}{\oint }_{\mathrm{\partial }D}\mathbf{r}\cdot \mathbf{n}ds}$ 

wheren is the unit normal andds is the arc length measure. For a circle of radiusR centered at the origin, we have${\displaystyle |\mathbf{r}|=R}$  and${\displaystyle \mathbf{n}=\mathbf{r}/R}$ , so the above equality is

${\displaystyle A=\frac{1}{2}{\oint }_{\mathrm{\partial }D}\mathbf{r}\cdot \frac{\mathbf{r}}{R}ds=\frac{R}{2}{\oint }_{\mathrm{\partial }D}ds.}$ 

The integral ofds over the whole circle${\displaystyle \mathrm{\partial }D}$  is just the arc length, which is its circumference, so this shows that the areaA enclosed by the circle is equal to${\displaystyle R/2}$  times the circumference of the circle.

Another proof that uses triangles considers the area enclosed by a circle to be made up of an infinite number of triangles (i.e. the triangles each have an angle ofd𝜃 at the center of the circle), each with an area of⁠1/2⁠ ·r² ·d𝜃 (derived from the expression for the area of a triangle:⁠1/2⁠ ·a ·b · sin𝜃 =⁠1/2⁠ ·r ·r · sin(d𝜃) =⁠1/2⁠ ·r² ·d𝜃). Note thatsin(d𝜃) ≈d𝜃 due tosmall angle approximation. Through summing the areas of the triangles, the expression for the area of the circle can therefore be found: 

Note that the area of a semicircle of radiusr can be computed by the integral${\int }_{-r}^r\sqrt{r^2-x^2}dx$ .

Bytrigonometric substitution, we substitute${\displaystyle x=r\sin \theta }$ , hence${\displaystyle dx=r\cos \theta d\theta .}$  

The last step follows since the trigonometric identity${\displaystyle \cos (\theta )=\sin (\pi /2-\theta )}$  implies that${\displaystyle {\cos }^2\theta }$  and${\displaystyle {\sin }^2\theta }$  have equal integrals over the interval${\displaystyle [0,\pi /2]}$ , usingintegration by substitution. But on the other hand, since${\displaystyle {\cos }^2\theta +{\sin }^2\theta =1}$ , the sum of the two integrals is the length of that interval, which is${\displaystyle \pi /2}$ . Consequently, the integral of${\displaystyle {\cos }^2\theta }$  is equal to half the length of that interval, which is${\displaystyle \pi /4}$ .

Therefore, the area of a circle of radiusr, which is twice the area of the semi-circle, is equal to${\displaystyle 2\cdot \frac{\pi r^2}{2}=\pi r^2}$ .

This particular proof may appear to beg the question, if the sine and cosine functions involved in the trigonometric substitution are regarded as being defined in relation to circles. However, as noted earlier, it is possible to define sine, cosine, andπ in a way that is totally independent of trigonometry, in which case the proof is valid by thechange of variables formula andFubini's theorem, assuming the basic properties of sine and cosine (which can also be proved without assuming anything about their relation to circles).

The circle is the closed curve of least perimeter that encloses the maximum area. This is known as theisoperimetric inequality, which states that if a rectifiable Jordan curve in the Euclidean plane has perimeterC and encloses an areaA (by theJordan curve theorem) then

${\displaystyle 4\pi A\le C^2.}$ 

Moreover, equality holds in this inequality if and only if the curve is a circle, in which case${\displaystyle A=\pi r^2}$  and${\displaystyle C=2\pi r}$ .

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 ofWillebrord Snell (Cyclometricus, 1621), further developed byChristiaan Huygens (De Circuli Magnitudine Inventa, 1654), described inGerretsen & Verdenduin (1983, pp. 243–250).

Given a circle, letu_n be theperimeter of an inscribed regularn-gon, and letU_n be the perimeter of a circumscribed regularn-gon. Thenu_n andU_n are lower and upper bounds for the circumference of the circle that become sharper and sharper asn increases, and their average (u_n +U_n)/2 is an especially good approximation to the circumference. To computeu_n andU_n for largen, Archimedes derived the following doubling formulae:

${\displaystyle u_{2n}=\sqrt{U_{2n}{u}_n}}$    (geometric mean), and

${\displaystyle U_{2n}=\frac{2U_n{u}_n}{U_n+u_n}}$     (harmonic mean).

Starting from a hexagon, Archimedes doubledn four times to get a 96-gon, which gave him a good approximation to the circumference of the circle.

In modern notation, we can reproduce his computation (and go further) as follows.For a unit circle, an inscribed hexagon hasu₆ = 6, and a circumscribed hexagon hasU₆ = 4√3.Doubling seven times yields

Archimedes doubling seven times;n = 6 × 2^k.

k	n	un	Un	⁠un + Un/4⁠
0	6	6.0000000	6.9282032	3.2320508
1	12	6.2116571	6.4307806	3.1606094
2	24	6.2652572	6.3193199	3.1461443
3	48	6.2787004	6.2921724	3.1427182
4	96	6.2820639	6.2854292	3.1418733
5	192	6.2829049	6.2837461	3.1416628
6	384	6.2831152	6.2833255	3.1416102
7	768	6.2831678	6.2832204	3.1415970

(Here⁠u_n +U_n/2⁠ approximates the circumference of the unit circle, which is 2π, so⁠u_n +U_n/4⁠ approximatesπ.)

The last entry of the table has³⁵⁵⁄₁₁₃ as one of itsbest rational approximations;i.e., there is no better approximation among rational numbers with denominator up to 113.The number³⁵⁵⁄₁₁₃ is also an excellent approximation toπ, attributed to Chinese mathematicianZu Chongzhi, who named itMilü.^[5] This approximation is better than any other rational number with denominator less than 16,604.^[6]

Snell proposed (and Huygens proved) a tighter bound than Archimedes':

 

This forn = 48 gives a better approximation (about 3.14159292) than Archimedes' method forn = 768.

Let one side of an inscribed regularn-gon have lengths_n 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. ByThales' theorem, this is a right triangle with right angle at B. Let the length of A′B bec_n, which we call the complement ofs_n; thusc_n²+s_n² = (2r)². 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 iss_2n, the length of C′A isc_2n, 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 issimilar 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

 

In the first equation C′P is C′O+OP, lengthr + ⁠1/2⁠c_n, and C′C is the diameter, 2r. For a unit circle we have the famous doubling equation ofLudolph van Ceulen,

${\displaystyle c_{2n}=\sqrt{2+c_n}.}$ 

If we now circumscribe a regularn-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 sideS_n; then this isS_n : s_n = 1 : ¹⁄₂c_n. (We have again used that OP is half the length of A′B.) Thus we obtain

${\displaystyle c_n=2\frac{s_n}{S_n}.}$ 

Call the inscribed perimeteru_n =ns_n, and the circumscribed perimeterU_n =nS_n. Then combining equations, we have

${\displaystyle c_{2n}=\frac{s_n}{s_{2n}}=2\frac{s_{2n}}{S_{2n}},}$ 

so that

${\displaystyle u_{2n}^2=u_n{U}_{2n}.}$ 

This gives ageometric mean equation.

We can also deduce

${\displaystyle 2\frac{s_{2n}}{S_{2n}}\frac{s_n}{s_{2n}}=2+2\frac{s_n}{S_n},}$ 

or

${\displaystyle \frac{2}{U_{2n}}=\frac{1}{u_n}+\frac{1}{U_n}.}$ 

This gives aharmonic mean equation.

When more efficient methods of finding areas are not available, we can resort to "throwing darts". ThisMonte 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⁻ⁿ requires about 100ⁿ random samples (Thijssen 2006, p. 273).

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 (Laczkovich 1990) is that we can dissect the disk into a large butfinite number of pieces and then reassemble the pieces into a square of equal area. This is calledTarski'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.

Circles can be defined innon-Euclidean geometry, and in particular in thehyperbolic andelliptic planes.

For example, theunit sphere${\displaystyle S^2(1)}$  is a model for the two-dimensional elliptic plane. It carries anintrinsic metric that arises by measuringgeodesic length. The geodesic circles are the parallels in ageodesic coordinate system.

More precisely, fix a point${\displaystyle \mathbf{z}\in S^2(1)}$  that we place at the zenith. Associated to that zenith is a geodesic polar coordinate system${\displaystyle (\phi ,\theta )}$ ,${\displaystyle 0\le \phi \le \pi }$ , , wherez is the point${\displaystyle \phi =0}$ . In these coordinates, the geodesic distance fromz to any other point${\displaystyle \mathbf{x}\in S^2(1)}$  having coordinates${\displaystyle (\phi ,\theta )}$  is the value of${\displaystyle \phi }$  atx. A spherical circle is the set of points a geodesic distanceR from the zenith pointz. Equivalently, with a fixed embedding into${\displaystyle {\mathbb{R}}^3}$ , the spherical circle of radius${\displaystyle R\le \pi }$  centered atz is the set ofx in${\displaystyle S^2(1)}$  such that${\displaystyle \mathbf{x}\cdot \mathbf{z}=\cos R}$ .

We can also measure the area of the spherical disk enclosed within a spherical circle, using the intrinsic surface area measure on the sphere. The area of the disk of radiusR is then given by

${\displaystyle A={\int }_0^{2\pi }{\int }_0^R\sin (\phi )d\phi d\theta =2\pi (1-\cos R).}$ 

More generally, if a sphere${\displaystyle S^2(\rho )}$  has radius of curvature${\displaystyle \rho }$ , then the area of the disk of radiusR is given by

${\displaystyle A=2\pi {\rho }^2(1-\cos (R/\rho )).}$ 

Observe that, as an application ofL'Hôpital's rule, this tends to the Euclidean area${\displaystyle \pi R^2}$  in the flat limit${\displaystyle \rho \to \mathrm{\infty }}$ .

The hyperbolic case is similar, with the area of a disk of intrinsic radiusR in the (constant curvature${\displaystyle -1}$ ) hyperbolic plane given by

${\displaystyle A=2\pi (1-\cosh R)}$ 

where cosh is thehyperbolic cosine. More generally, for the constant curvature${\displaystyle -k}$  hyperbolic plane, the answer is

${\displaystyle A=2\pi k^{-2}(1-\cosh (kR)).}$ 

These identities are important for comparison inequalities in geometry. For example, the area enclosed by a circle of radiusR in a flat space is always greater than the area of a spherical circle and smaller than a hyperbolic circle, provided all three circles have the same (intrinsic) radius. That is,

 

for all${\displaystyle R>0}$ . Intuitively, this is because the sphere tends to curve back on itself, yielding circles of smaller area than those in the plane, whilst the hyperbolic plane, when immersed into space, develops fringes that produce additional area. It is more generally true that the area of the circle of a fixed radiusR is a strictly decreasing function of the curvature.

In all cases, if${\displaystyle k}$  is the curvature (constant, positive or negative), then theisoperimetric inequality for a domain with areaA and perimeterL is

${\displaystyle L^2\ge 4\pi A-k{A}^2}$ 

where equality is achieved precisely for the circle.^[7]

We can stretch a disk to form anellipse. Because this stretch is alinear transformation of the plane, it has a distortion factor which will change the area but preserveratios of areas. This observation can be used to compute the area of an arbitrary ellipse from the area of a unit circle.

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 isπ/4, which means the ratio of the ellipse to the rectangle is alsoπ/4. Supposea andb are the lengths of the major and minor axes of the ellipse. Since the area of the rectangle isab, the area of the ellipse isπab/4.

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.

• Area-equivalent radius
• Area of a triangle

• Archimedes (1897),"Measurement of a circle", inHeath, T. L. (ed.),The Works of Archimedes, Cambridge University Press
  (Originally published byCambridge University Press, 1897, based on J. L. Heiberg's Greek version.)
• Beckmann, Petr (1976),A History of Pi,St. Martin's Griffin,ISBN 978-0-312-38185-1
• Gerretsen, J.; Verdenduin, P. (1983), "Chapter 8: Polygons and Polyhedra", in H. Behnke; F. Bachmann; K. Fladt; H. Kunle (eds.),Fundamentals of Mathematics, Volume II: Geometry, translated by S. H. Gould,MIT Press, pp. 243–250,ISBN 978-0-262-52094-2
  (OriginallyGrundzüge der Mathematik, Vandenhoeck & Ruprecht, Göttingen, 1971.)
• Laczkovich, Miklós (1990),"Equidecomposability and discrepancy: A solution to Tarski's circle squaring problem",Journal für die reine und angewandte Mathematik,1990 (404):77–117,doi:10.1515/crll.1990.404.77,MR 1037431,S2CID 117762563
• Lang, Serge (1985), "The length of the circle",Math! : Encounters with High School Students,Springer-Verlag,ISBN 978-0-387-96129-3
• Smith, David Eugene; Mikami, Yoshio (1914),A history of Japanese mathematics, Chicago:Open Court Publishing, pp. 130–132,ISBN 978-0-87548-170-8{{citation}}:ISBN / Date incompatibility (help)
• Thijssen, J. M. (2006),Computational Physics, Cambridge University Press, p. 273,ISBN 978-0-521-57588-1

• Science News on Tarski problemArchived 2008-04-13 at theWayback Machine