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If the trapezoid is divided into four triangles by its diagonals AC and BD (as shown on the right), intersecting at O, then the area of AOD is equal to that of BOC, and the product of the areas of AOD and BOC is equal to that of AOB and COD. The ratio of the areas of each pair of adjacent triangles is the same as that between the lengths of the ...
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.)
The only equable rectangles with integer sides are the 4 × 4 square and the 3 × 6 rectangle. [4] An integer rectangle is a special type of polyomino, and more generally there exist polyominoes with equal area and perimeter for any even integer area greater than or equal to 16. For smaller areas, the perimeter of a polyomino must exceed its area.
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
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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Placing the point P on any of the four vertices of the rectangle yields the square of the diagonal of the rectangle being equal to the sum of the squares of the width and length of the rectangle, which is the Pythagorean theorem.
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