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The ratio of the areas of each pair of adjacent triangles is the same as that between the lengths of the parallel sides. [18] Let the trapezoid have vertices A, B, C, and D in sequence and have parallel sides AB and DC. Let E be the intersection of the diagonals, and let F be on side DA and G be on side BC such that FEG is parallel to AB and CD.
If the midpoints of two triangle sides are connected then the resulting line segment is parallel to the third triangle side (Midpoint theorem of triangles). If the midpoints of the two non-parallel sides of a trapezoid are connected, then the resulting line segment is parallel to the other two sides of the trapezoid.
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.
Any non-self-crossing quadrilateral with exactly one axis of symmetry must be either an isosceles trapezoid or a kite. [5] However, if crossings are allowed, the set of symmetric quadrilaterals must be expanded to include also the crossed isosceles trapezoids, crossed quadrilaterals in which the crossed sides are of equal length and the other sides are parallel, and the antiparallelograms ...
where K is the area of the quadrilateral and s is its semiperimeter. For a tangential quadrilateral with given sides, the inradius is maximum when the quadrilateral is also cyclic (and hence a bicentric quadrilateral). In terms of the tangent lengths, the incircle has radius [8]: Lemma2 [14]
This is not true for other lines through the centroid; the greatest departure from the equal-area division occurs when a line through the centroid is parallel to a side of the triangle, creating a smaller triangle and a trapezoid; in this case the trapezoid's area is that of the original triangle. [18]
The ratio of the area of the incircle to the area of an equilateral triangle, , is larger than that of any non-equilateral triangle. [ 37 ] The ratio of the area to the square of the perimeter of an equilateral triangle, 1 12 3 , {\displaystyle {\frac {1}{12{\sqrt {3}}}},} is larger than that for any other triangle.
All problems that can be solved using mass point geometry can also be solved using either similar triangles, vectors, or area ratios, [2] but many students prefer to use mass points. Though modern mass point geometry was developed in the 1960s by New York high school students, [ 3 ] the concept has been found to have been used as early as 1827 ...