How was the area formula for a circle ($A = \pi r^2$) derived before the introduction of calculus? How did mathematicians prior to the coming of calculus derive the area of the circle from scratch, without the use of calculus?
The area, $A$, of a circle is $\pi r^2$. Given radius $r$, diameter $d$ and circumference $c$, by definition, $\pi := \frac cd$.
 A: Monte Carlo method.
Pick a random x/y by throwing lots of dice. Ensure the points fall randomly on a square of width equal to circle's diameter. Repeat many times, or until satisfied, very bored, or until Happy Hour and Martinis are on the Casino (hence 'Monte Carlo' method. It is just about April 1st).
Take the ratio of the points that fall in the circle to those that fall outside. Note how as the number of random points increases, the ratio converges on a certain amount. 
A: There's an interesting method using which you can approximately find the area. Split up the circle into many small sectors, and arrange them as a parallelogram as shown in the image (from wikipedia)

The higher the number of sectors you take, the more it tends to a parallelogram, with one of it’s height the radius $r$, and the other side half the circumference $\pi r$. Thus, its area tends to $\pi r \cdot r = \pi r^2$
A: Calculus depends on the concept of a limit, and does apply that to the problem of determining the area of a curved object. But the concept of a limit and the ability to reason about the area of a curved object using that concept existed before Calculus. 
As far as recorded history is concerned, Archimedes was the first to derive $A = \pi r^2$. Though he didn't call it $\pi$, I think we can still say this is the answer to your question. His proof depends on the concept of a limit. He showed that, given a circle, with radius $r$ and circumference $c$, the area of that circle can't be more than that of a triangle with height $r$ and base $c$, and that it can't be less than the area of that triangle either. He did this by examining the area of polygons with an increasing number of sides, both inside the circle and outside the circle. This involved the concept of a limit and calculating area, which FEELS like calculus, but isn't.
Polygons with increasing number of sides inside:

Polygons with increasing number of sides outside:

A: Since two circles are similar, their areas are proportional to the squares of their radii. Now the area of a circle with radius $1$ is the proportional factor, just call it $\pi$.
Addendum.  If you insist in the usual definition of $\pi$, i. e., the ratio of circumference to diameter, I doubt there will be no connection to my approach without an infinitesimal argument, that is calculus.
One imho nice approach would be to define a discrete $\pi_n$ for every regular $n$-gon, namely $\pi_n:=n\cdot\tan(\pi/n)$, which is the ratio of the $n$-gon’s circumference to its diameter. (Here the radius $r_n$ is defined by 
the distance of the $n$-gon’s center to one of its sides.)
In that case the area $A_n=\pi_n\cdot r_n^2$ and the circumference is $2\pi_n\cdot r_n$.
From here we see that if the proportional factor for the area is $\pi_n$, the factor for the circumference $2\pi_n$.  That’s true for $n$-gons from above.
But how to show that without calculus for circles?
I know that there are some curves which lengths may be calculated purely algebraic, such as Neil’s parabola $t\mapsto(t^2,t^3)$.  But I don’t know an algebraic  way for the circle.  Since a purely algebraic rectification was thought impossible since Aristotle, I presume there is none for the circle.
A: *

*Get two large sheets of papyrus with uniform thickness

*Cut a circle of unit radius from one, and a unit square from the other

*Measure the weight of each shape 

*From their ratio you get the area
of the circle

A: Euclid initiates this is Book XII, Prop. 2., showing indirectly that there is a constant of proportionality between the area of a circle and its diameter.  He does this by showing that the area of two circles goes as the square of their diameters, which forces this constant of proportionality between, in the Question's notation, $A$ and $r^2$.  Lets call that constant $p$: $A = p r^2$.  So Euclid shows that $p$ exists, but does not show that $p$ is related to the ratio of the length of the circumference to that of the diameter.
The preceding proposition in that book, that the areas of similar polygons inscribed in circles are in the same ratio as the squares of the diameters of the circles, gives us a method to approximate $p$ from below, by using sequences of polygons that cover more and more of the circle.  Archimedes uses this method (as well as the result that a circumscribed polygon contains all the area of the circle) to bound $p$ in his Measurement of a Circle.  But first, he uses these facts about inscribed and circumscribed polygons to show $p = \pi$.
His first proposition is the result you ask about: The area of a circle is equal to the area of a right triangle with one leg having the length of the radius and the other leg having the length of the circumference.  That is, using the definition you give that the circumference is $\pi$ times the diameter and also that the diameter is twice the radius, 
$$ A = \frac{1}{2} \cdot r \cdot C = \frac{1}{2} \cdot r \cdot (\pi \cdot 2r) = \pi r^2  \text{,}  $$
so $p = \pi$.
Archimedes proves this result by the method of exhaustion.  It is common to claim that this method is elementary calculus, but that isn't entirely correct.  As it is used here (and in many other places), it is an application of trichotomy to the completeness of the reals and is a method more commonly used in advanced calculus and real analysis.  Archimedes observes that if the area of the triangle is not the area of the circle, it must be either greater or less than the area of the circle.  He shows it cannot be greater, then that it cannot be less.  Therefore, the area of the triangle is the same as the area of the circle.  
(Using completeness of the reals, one would phrase this as: for all $\varepsilon >0$, the area of the triangle differs from the area of the circle by less than $\varepsilon$, so is the same as the area of the circle.  If we imagine a shrinking sequence of $\varepsilon$s, this gives the nested sequence of intervals for the completeness link above.)
A: By trial and error and by numerical approximations. The ancient mathematicians (Babylonians, 1800 BC) tried to square a circle (approximating the area of circle with a square, constructing a square with the same area as a circle, proved impossible in 1882 AD), to calculate $\sqrt{2}$, $\pi$, etc.
The precise calculation of the area of circle requires an infinitesimal analysis (calculus).
A: Another way to derive the formula for the area of a circle is by decomposing it into several concentric rings (annuli) to form a "triangle" as seen in the below diagram.  As the width of the concentric rings approach zero, it forms a triangle of base circumference ($C$ = $\pi$$d$) and height $r$.

Since we know the area of a triangle is $\frac{bh}{2}$, by substitution $$\frac{\pi d\cdot r}{2}$$
Since $d$ = 2$r$, $$A = \frac{\pi \cdot 2r\cdot r}{2} \rightarrow A = \frac{\pi \cdot 2r^2}{2} \rightarrow A = \pi \cdot r^2$$.
A: It is easy to go from the area of a triangle to the area of a regular polygon (one whose sides all
have the same length and whose angles all have the same measure) by breaking the polygon into
triangles and summing the areas of the triangles. But we need to first review the formula for the area of a polygon for reference.
It can be shown that every regular polygon can be inscribed in a circle. If we draw a perpendicular line from the center of the circle to any side of the inscribed polygon, that line is called
an apothem. It is a fact
that the area of a regular polygon is $1/2ap$ where $a$ is the apothem and $p$ is the perimeter of the
polygon. It is also a fact that as the number of sides of the inscribed
regular polygons increases, the lengths of the apothems of the polygons approach the radius of the
 circle. 

A: The area is very much a concept belonging to Calculus. In this regard to find the area of something without using Calculus is almost an oxymoron, but we can have an intuitive (and not too general) notion of area based on the following points:


*

*Any rectangle deserves an area (i.e. it is a measurable set) and the area of a rectangle $R$ with side lengths $a,b,a,b$ is $\mu(R)=ab$;

*Isometric measurable sets have the same area and if a line splits some measurable set $R$ into two components $R_1,R_2$ then $R_1,R_2$ are measurable and $\mu(R_1)+\mu(R_2)=\mu(R)$. This fixes the area of parallelograms, triangles, bounded polygons;

*Not every compact subset of $\mathbb{R}^2$ is a polygon, so we want to extend the notion of measurability in the following way: given $K$, a compact and connected subset of $\mathbb{R}^2$, we define $\mathcal{I}$ as the family of polygons contained in $K$, $\mathcal{E}$ as the family of polygons containing $K$. If
$$ \sup_{P\in\mathcal{I}} \mu(P) = \inf_{P\in\mathcal{E}}\mu(P) = L $$
we say that $K$ is measurable and define its area as $\mu(K)=L$.


Notice that 1.,2.,3. do not allow to measure unbounded sets like $\{(x,y)\in\mathbb{R}^2: x\geq 0, 0\leq y\leq\frac{1}{(x+1)^2}\}$ 
or subsets of $\mathbb{R}^2$ with countable connected components, like $\bigcup_{n\in\mathbb{N}}\{(x,y): (x-n)^2+y^2\leq \frac{1}{(n+1)^3}\}$.
Still we may use this naive notion of area to compute the area of the unit circle. By considering the regular polygons in $\mathcal{I}$ and $\mathcal{E}$ we have
$$ \mu\left(\{(x,y):x^2+y^2\leq 1\right) = \sup_{n\geq 3}n\cdot\sin\frac{\pi}{n} = \inf_{n\geq 3}n\cdot\tan\frac{\pi}{n}=\pi $$
hence the area of the unit circle is just half the length of its boundary and $\pi\in(3,4)$ follows from the fact that

If $A,B$ are two convex, bounded sets in $\mathbb{R}^2$ and $A\subsetneq B$, the length of $\partial A$ is less than the length of $\partial B$. (see also this question)

An Archimedean-like geometric approach for the numerical approximation of $\pi$ is the following: a circle with radius $1$ can be decomposed as the union of an octagon with side length $\sqrt{2-\sqrt{2}}$ and eight circular segments. Such segments can be approximated by parabolic segments, whose area is simply $\frac{2}{3}\text{base}\cdot\text{height}$. The parabolic segments are slightly smaller than the corresponding circle segments, hence the following construction

leads to the lower bound 
$$\pi > \frac{16}{3}\sqrt{2-\sqrt{2}}-\frac{2}{3}\sqrt{2}= 3.13914757\ldots$$
whose accuracy is comparable with the actual Archimedean approximation $\pi\leq\frac{22}{7}$. The parabolic method applied to the regular dodecagon leads to the nice bound
$$ \pi > 4\sqrt{6}-4\sqrt{2}-1 = 3.1411\ldots $$
which also explains the proximity between $\pi$ and $\sqrt{2}+\sqrt{3}$.
A broader notion of area comes from defining the Riemann/Lebesgue integral and defining the area of $K\subset\mathbb{R}^2$ as
$$ \mu(K) = \iint_{\mathbb{R}^2}\mathbb{1}_K(x,y)\,dx\,dy $$
provided that the RHS makes sense. 1,2,3 are fulfilled, and this also allows us to measure the previous troubling sets. Additionally, this gives us new ways for approximating 
$$\pi=4\int_{0}^{1}\sqrt{1-x^2}\,dx = 6\arcsin\frac{1}{2} = 3\sum_{n\geq 0}\frac{\binom{2n}{n}}{16^n(2n+1)}=3+\frac{1}{8}+\frac{9}{640}+\left(0\leq\varepsilon\leq 3\cdot 10^{-3}\right) $$
also through
$$ \frac{1}{5\cdot 4^8}\approx\int_{0}^{1}\frac{x^8(1-x)^8}{1+x^2}\,dx=4\pi-\frac{188684}{15015}.$$

Improving the answer: why is it enough to study the unit circle? Because a circle is highly symmetric, os by applying a dilation with factor $\lambda>0$ the area is multiplied by $\lambda^2$ and the length of the boundary is multiplied by $\lambda$ (the fundamental property of affine maps is that they preserve the ratio of areas). In particular if the area of the unit circle is $\beth$, the area of a circle with radius $R$ is $\beth R^2$.
How is this related with the length of the boundary? Through convexity. If we consider the annulus between two concentric circles with radii $R$ and $R+\varepsilon$, the area of the annulus divided by $\varepsilon$ tends to the length of the boundary of the inner circle as $\varepsilon\to 0^+$. In particular if $\beth R^2$ is the area of the circle with radius $R$, the length of its boundary is given by 
$$ \frac{d}{dR}\beth R^2 = \color{red}{2\beth}R.$$
A: I am going to use a non conventional method here:- 
If you stack circles on top of circles ,  it forms a cylinder of considerable height.
Now the volume of this cylinder is equal to
 $Vol\, (cylinder) =A*H   \quad\\where\, A=\text{Area of the circle (still unknown)}\\ \text{and H = height of the cylinder that you have created}  $
Now the volume of the cylinder is equal to  the increase of liquid level when the e object is immersed in the liquid (usually water).(Application of  Archimedes principle) .This is how you get the volume of the cylinder.
Divide the volume by $H(\text{measurable quantity})$ and you get the Area.
To get the formula of the area of the circle you may have to use numerical methods.
