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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;
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.
The area of a triangle can be demonstrated, for example by means of the congruence of triangles, as half of the area of a parallelogram that has the same base length and height. A graphic derivation of the formula T = h 2 b {\displaystyle T={\frac {h}{2}}b} that avoids the usual procedure of doubling the area of the triangle and then halving it.
The area enclosed between a parabola and a chord (see diagram) is two-thirds of the area of a parallelogram that surrounds it. One side of the parallelogram is the chord, and the opposite side is a tangent to the parabola. [16] [17] The slope of the other parallel sides is irrelevant to the area. Often, as here, they are drawn parallel with the ...
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 ...
Shape Figure ¯ ¯ Area rectangle area: General triangular area + + [1] Isosceles-triangular area: Right-triangular area: Circular area: Quarter-circular area [2]: Semicircular area [3]: Circular sector
The following is a list of second moments of area of some shapes. The second moment of area, also known as area moment of inertia, is a geometrical property of an area which reflects how its points are distributed with respect to an arbitrary axis. The unit of dimension of the second moment of area is length to fourth power, L 4, and should not ...
The apparent triangles formed from the figures are 13 units wide and 5 units tall, so it appears that the area should be S = 13×5 / 2 = 32.5 units. However, the blue triangle has a ratio of 5:2 (=2.5), while the red triangle has the ratio 8:3 (≈2.667), so the apparent combined hypotenuse in each figure is actually bent.