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The area A of the parabolic segment enclosed by the parabola and the chord is therefore =. This formula can be compared with the area of a triangle: 1 / 2 bh. In general, the enclosed area can be calculated as follows. First, locate the point on the parabola where its slope equals that of the chord.
A parabolic segment is the region bounded by a parabola and line. To find the area of a parabolic segment, Archimedes considers a certain inscribed triangle. The base of this triangle is the given chord of the parabola, and the third vertex is the point on the parabola such that the tangent to the parabola at that point is parallel to the chord.
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 ...
The area enclosed by the lemniscate is a 2 = 2c 2. The lemniscate is the circle inversion of a hyperbola and vice versa. The two tangents at the midpoint O are perpendicular, and each of them forms an angle of π / 4 with the line connecting F 1 and F 2. The planar cross-section of a standard torus tangent to its inner equator is a ...
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 ...
Construction of the limaçon r = 2 + cos(π – θ) with polar coordinates' origin at (x, y) = ( 1 / 2 , 0). In geometry, a limaçon or limacon / ˈ l ɪ m ə s ɒ n /, also known as a limaçon of Pascal or Pascal's Snail, is defined as a roulette curve formed by the path of a point fixed to a circle when that circle rolls around the outside of a circle of equal radius.
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 :
The area bounded by the intersection of a line and a parabola is 4/3 that of the triangle having the same base and height (the quadrature of the parabola); The area of an ellipse is proportional to a rectangle having sides equal to its major and minor axes;