An Explanatory Approach to. Archimedes’s Quadrature of the Parabola. by. A. Kursat ERBAS. Have you ever been in a situation where you are trying to show the. Archimedes’ Quadrature of the Parabola is probably one of the earliest of Archimedes’ extant writings. In his writings, we find three quadratures of the parabola. Archimedes, Quadrature of the Parabola Prop. 18; translated by Henry Mendell ( Cal. State U., L.A.). Return to Vignettes of Ancient Mathematics ยท Return to.

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I say that a tangent to the section at B will be parallel to base AG and that a line drawn from B to parabla base AG parallel to the diameter will bisect AG. Go to theorem In a segment quadrathre enclosed by a straight-line and section of a right-angled cone, the line drawn from the middle of the base will be a third quadratjre in length that drawn from the middle of the half.

The first quadrature,takes a triangle circumscribing the segment and divides up the segment and the triangle into trapezoids and triangle constructed by lines parallel to archmiedes base of the segment and lines parallel to the diameter, which is also parallel to one side of the circumscribing triangle. I say that Z is larger than L, but less than M. This page was last edited on 25 Decemberat Go to theorem Every segment enclosed by a straight line and a section of a right-angled is a third again of a triangle having the same base as it and an equal height.

Recalling that the light blue area in Figure-2 is. And I think this will allow to pupils to get a broad vision of mathematics as well as a well built mathematical thinking.

### The Quadrature of the Parabola – Wikipedia

Return to Vignettes of Ancient Mathematics. The reductio is based on a summation of a series, a 1Each successive purple square has one fourth the area of the previous square, with the total purple area being the sum. I call base the straight line of segments enclosed by a straight-line and a curved line, and height the largest perpendicular drawn from the curve line to the base of the segment, and vertex the point from which the largest perpendicular is drawn.

It is adequate given that those presented by us have been raised to a conviction similar to these. Look up quadrature in Wiktionary, the free dictionary. Go to theorem If a straight line is drawn from the middle of the base in a segment which is enclosed by a straight line and a section of a right-angled cone, the point will be a vertex of the segment at which the line drawn parallel to the diameter cuts the section of the cone.

Because, you just have to use your ingenuity.

Archimedes may have dissected the area into infinitely archimsdes triangles whose areas form a geometric progression. The statement about the height follows from the geometric properties of a parabola, and is easy to prove using modern analytic geometry.

First, let, in fact, BG be at right angles to the diameter, and let BD be drawn from point B parallel to the diameter, and let GD from G be a tangent to the section of the cone at G. But first there are also proved drawn conic elements quadratuge are useful for the demonstrations. These approximate the segment from within.

The significance of the Archimedes’ solution to this problem is hidden in the fact that none of differention, integration, or coordinate geometry were known in his time. The Quadrature of the Parabola Greek: This picture shows a unit square which has been dissected into an infinity of smaller squares. Quadrature of the Parabola.

The converse is easy to prove: Go to theorem If a triangle is inscribed in a segment which is enclosed by a straight line and a section of a right-angled and has the same base as the segment and the same height, the inscribed triangle will be more than half the segment.

Corollary With this proved, it is clear that it is possible to inscribe a polygon in the parwbola so that the left over segments are less than any proposed area.

The second, archiimedes famous proof uses pure geometry, specifically the method of exhaustion. Theorem 0 B Case where BD is parallel to the diameter. And these have been proved in the Conic Elements. For they use this lemma itself to demonstrate that circles have to one another quadraturr ratio of the diameters, and that spheres have triple ratio to one another of the diameters, and further that every pyramid is a third part of the prism having the same base as the pyramid and equal height.

Among these trapezoids and triangles is a set on the curve, which are used to fuel the reductio. Quadrature by the mechanical means props.

Go to theorem Again, let segment BQG be enclosed by a straight-line and section of a right-angled cone, but let BG not be at right angles to the diameter.

### Quadrature of the parabola, Introduction

The two here take very different approaches, and yet more different from that in the Method. It will be proved similarly to that earlier that Z is larger than L and smaller than M.

Quadrature by geometrical means props. Archimedes evaluates the sum using an entirely geometric method, [2] illustrated in the adjacent picture. Hence, quaddrature are two vertices of the segment, which is impossible as noted above, we may want to prove this from the properties of cones. In fact, triangle BGD will be right-angled.

Hence, B and Z are the same point. Go to theorem If a segment is enclosed by a straight line and a section quadrture a right-angled, and areas are positioned successively, however many, in a ratio of four-times, and the largest of the areas is equal to the triangle having the base having the same base as the triangle and height the same, then the areas altogether will be smaller than the segment.

## Quadrature of the Parabola

Here T represents the area archimexes the large blue triangle, the second term represents the total quadratre of the two green triangles, the third term represents the total area of the four yellow triangles, and so forth. Theorem 0 C Case where BD is parallel to the diameter. For this reason, these were condemned by most people as not being discovered by them. Go to a sample proof. Go to theorem If magnitudes are placed successively in a ratio of four-times, all the magnitudes and yet the third part of the least composed into the same magnitude will be a third-again the largest.

History has lots of examples of this kind of situation.

In his writings, we find three quadratures of the parabola or segment enclosed by a straight-line and a section of a right-angled conetwo here and one in the Method 1probably one of his last works among extant texts. With this respect, I think we must teach mathematics with a little bit history of mathematics. I hope you have not.