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The area of the parallelogram is the area of the blue region, which is the interior of the parallelogram. The base × height area formula can also be derived using the figure to the right. The area K of the parallelogram to the right (the blue area) is the total area of the rectangle less the area of the two orange triangles. The area of the ...
In a parallelogram, where both pairs of opposite sides and angles are equal, this formula reduces to = . Alternatively, we can write the area in terms of the sides and the intersection angle θ of the diagonals, as long as θ is not 90°: [18]
An arbitrary quadrilateral and its diagonals. Bases of similar triangles are parallel to the blue diagonal. Ditto for the red diagonal. The base pairs form a parallelogram with half the area of the quadrilateral, A q, as the sum of the areas of the four large triangles, A l is 2 A q (each of the two pairs reconstructs the quadrilateral) while that of the small triangles, A s is a quarter of A ...
In speaking about these processes, the measure (length or area) of a figure's base is often referred to as its "base." By this usage, the area of a parallelogram or the volume of a prism or cylinder can be calculated by multiplying its "base" by its height; likewise, the areas of triangles and the volumes of cones and pyramids are fractions of ...
For the general quadrilateral (with four sides not necessarily equal) Euler's quadrilateral theorem states + + + = + +, where is the length of the line segment joining the midpoints of the diagonals. It can be seen from the diagram that x = 0 {\displaystyle x=0} for a parallelogram, and so the general formula simplifies to the parallelogram law.
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 extended parallelogram sides DE and FG intersect at H. The line segment AH now "becomes" the side of the third parallelogram BCML attached to the triangle side BC, i.e., one constructs line segments BL and CM over BC, such that BL and CM are a parallel and equal in length to AH.
The angle depends on the segment length a between the right angle and the vertex of the angle of parallelism. Given a point not on a line, drop a perpendicular to the line from the point. Let a be the length of this perpendicular segment, and () be the least angle such that the line drawn through the point does not intersect the given line ...