4/9/2023 0 Comments Area of a circle formula![]() ![]() But I don’t know an algebraic way for the circle. I know that there are some curves which lengths may be calculated purely algebraic, such as Neil’s parabola $t\mapsto(t^2,t^3)$. That’s true for $n$-gons from above.īut how to show that without calculus for circles? In that case the area $A_n=\pi_n\cdot r_n^2$ and the circumference is $2\pi_n\cdot r_n$.įrom here we see that if the proportional factor for the area is $\pi_n$, the factor for the circumference $2\pi_n$. The distance of the $n$-gon’s center to one of its sides.) 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. e., the ratio of circumference to diameter, I doubt there will be no connection to my approach without an infinitesimal argument, that is calculus. If you insist in the usual definition of $\pi$, i. Now the area of a circle with radius $1$ is the proportional factor, just call it $\pi$.Īddendum. Since two circles are similar, their areas are proportional to the squares of their radii. That is, using the definition you give that the circumference is $\pi$ times the diameter and also that the diameter is twice the radius, 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. But first, he uses these facts about inscribed and circumscribed polygons to show $p = \pi$. 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. 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. 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. Lets call that constant $p$: $A = p r^2$. 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 $r^2$. 2., showing indirectly that there is a constant of proportionality between the area of a circle and its diameter. ![]() Polygons with increasing number of sides outside:Įuclid initiates this is Book XII, Prop. Polygons with increasing number of sides inside: This involved the concept of a limit and calculating area, which FEELS like calculus, but isn't. He did this by examining the area of polygons with an increasing number of sides, both inside the circle and outside the circle. 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. His proof depends on the concept of a limit. Though he didn't call it $\pi$, I think we can still say this is the answer to your question. ![]() But the concept of a limit and the ability to reason about the area of a curved object using that concept existed before Calculus.Īs far as recorded history is concerned, Archimedes was the first to derive $A = \pi r^2$. Calculus depends on the concept of a limit, and does apply that to the problem of determining the area of a curved object. ![]()
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