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Distance covered is the area under the line. Each time interval is coloured differently. The distance covered in the second and subsequent intervals is the area of its trapezium, which can be subdivided into triangles as shown. As each triangle has the same base and height, they have the same area as the triangle in the first interval.
The area of an isosceles (or any) trapezoid is equal to the average of the lengths of the base and top (the parallel sides) times the height. In the adjacent diagram, if we write AD = a , and BC = b , and the height h is the length of a line segment between AD and BC that is perpendicular to them, then the area K is
It is parallel to the bases. Its length m is equal to the average of the lengths of the bases a and b of the trapezoid, [13] = +. The midsegment of a trapezoid is one of the two bimedians (the other bimedian divides the trapezoid into equal areas). The height (or altitude) is the perpendicular distance between the bases.
The formula for the area of a trapezoid can be simplified using Pitot's theorem to get a formula for the area of a tangential trapezoid. If the bases have lengths a, b, and any one of the other two sides has length c, then the area K is given by the formula [2] (This formula can be used only in cases where the bases are parallel.)
giving the basic form of Brahmagupta's formula. It follows from the latter equation that the area of a cyclic quadrilateral is the maximum possible area for any quadrilateral with the given side lengths. A related formula, which was proved by Coolidge, also gives the area of a general convex
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If a cyclic quadrilateral is also orthodiagonal, the distance from the circumcenter to any side equals half the length of the opposite side. [23] In a cyclic orthodiagonal quadrilateral, the distance between the midpoints of the diagonals equals the distance between the circumcenter and the point where the diagonals intersect. [23]