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Archimedes provides the first attested solution to this problem by focusing specifically on the area bounded by a parabola and a chord. [3] Archimedes gives two proofs of the main theorem: one using abstract mechanics and the other one by pure geometry. In the first proof, Archimedes considers a lever in equilibrium under the action of gravity ...
Archimedes encounters the series in his work Quadrature of the Parabola. He finds the area inside a parabola by the method of exhaustion, and he gets a series of triangles; each stage of the construction adds an area 1 / 4 times the area of the previous stage. His desired result is that the total area is 4 / 3 times the area of ...
Archimedes in his The Quadrature of the Parabola used the sum of a geometric series to compute the area enclosed by a parabola and a straight line. Archimedes' theorem states that the total area under the parabola is 4/3 of the area of the blue triangle. His method was to dissect the area into infinite triangles as shown in the adjacent figure ...
Archimedes used the method of exhaustion to compute the area inside a circle. Archimedes used the method of exhaustion as a way to compute the area inside a circle by filling the circle with a sequence of polygons with an increasing number of sides and a corresponding increase in area.
In Quadrature of the Parabola, Archimedes proved that the area enclosed by a parabola and a straight line is 4 / 3 times the area of a corresponding inscribed triangle as shown in the figure at right. He expressed the solution to the problem as an infinite geometric series with the common ratio 1 / 4 :
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The area of the surface of a sphere is equal to four times the area of the circle formed by a great circle of this sphere. The area of a segment of a parabola determined by a straight line cutting it is 4/3 the area of a triangle inscribed in this segment. For the proofs of these results, Archimedes used the method of exhaustion attributed to ...
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