# Find the area of a parallelogram with vertices A(1,5,0), B(6,10,−3), C(−4,5,−2), and D(1,10,−5)? Show steps.

Jan 6, 2016

${S}_{\text{parallelogram}} = 15 \cdot \sqrt{6} \cong 36.742$

#### Explanation:

Note that it doesn't matter whether the parallelogram is formed by the vertices A, B, C, D in this order or by the vertices A, B, D. C in this order, any diagonal of the parallelogram divides it in two triangles with equal areas. Then S_("parallelogram" =2*S_(triangle). In this explanation the $\triangle A B C$ is selected (but any other possible triangle formed with the points A, B. C or D, would do).
Repeating the coordinates of the points of $\triangle A B C$:
A(1,5,0)
B(6,10,-3)
C(-4,5,-2)

Obtaining sides of the triangle ABC:
$A B = \sqrt{{\left(6 - 1\right)}^{2} + {\left(10 - 5\right)}^{2} + {\left(- 3 - 0\right)}^{2}} = \sqrt{25 + 25 + 9} = \sqrt{59}$
$A C = \sqrt{{\left(- 4 - 1\right)}^{2} + {\left(5 - 5\right)}^{2} + {\left(- 2 - 0\right)}^{2}} = \sqrt{25 + 0 + 4} = \sqrt{29}$
$B C = \sqrt{{\left(- 4 - 6\right)}^{2} + {\left(5 - 10\right)}^{2} + {\left(- 2 + 3\right)}^{2}} = \sqrt{100 + 25 + 1} = \sqrt{126}$

Using Heron's Formula
Triangle ABC ($a = A B , b = A C \mathmr{and} c = B C$)
$s = \frac{a + b + c}{2} = \frac{\sqrt{59} + \sqrt{29} + \sqrt{126}}{2} \cong 12.14564$
$\left(s - a\right) = \frac{- \sqrt{59} + \sqrt{29} + \sqrt{126}}{2} \cong 4.46450$
$\left(s - b\right) = \frac{\sqrt{59} - \sqrt{29} + \sqrt{126}}{2} \cong 6.76048$
$\left(s - c\right) = \frac{\sqrt{59} + \sqrt{29} - \sqrt{126}}{2} \cong 0.92067$
${S}_{\triangle A B C} = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)} = \sqrt{12.141564 \times 4.46450 \times 6.76048 \times 0.92067} = 18.371$

${S}_{\text{parallelogram}} = 2 \cdot {S}_{\triangle A B C} = 2 \cdot 18.371 = 36.712$

Other way to find the area of the kind of triangle involved in this question is described in:
when the triangle is embedded in three-dimensional space