How do you find the area of the parallelogram with vertices (4,5), (9, 9), (13, 10), and (18, 14)?

Feb 23, 2015

Calculate the areas of the four trapezoids formed by the line segments, the X-axis, and the line segments joining the given vertices to the X-axis.

Subtract the lower two trapezoid areas from the upper two. General form of Trapezoid Area = $\Delta x + a v {e}_{y}$

$A 1 = \left(9 - 4\right) \times \left(\frac{9 + 5}{2}\right)$
$= 5 \times 7 = 35$

$A 2 = \left(18 - 9\right) \times \left(\frac{9 + 14}{2}\right)$
$= 9 \times 11 \frac{1}{2} = 103 \frac{1}{2}$

$B 1 = \left(13 - 4\right) \times \left(\frac{5 + 10}{2}\right)$
$= 9 \times 7 \frac{1}{2} = 67 \frac{1}{2}$

$B 2 = \left(18 - 13\right) \times \left(\frac{14 + 10}{2}\right)$
$= 5 \times 12 = 60$

Area of parallelogram
$= \left(A 1 + A 2\right) - \left(B 1 + B 2\right)$
$= \left(35 + 103 \frac{1}{2}\right) - \left(67 \frac{1}{2} + 60\right)$
$= 138 \frac{1}{2} - 127 \frac{1}{2}$
$= 11$

(always assuming my basic arithmetic is correct).