Answer
area of parallelogram ABCD is 16
Work Step by Step
area of parallelogram = $base\times height$.
base = the distance of AB or CD
AB = $D=\sqrt (x_{1}-x_{2})^2 + (y_{1}-y_{2})^2$
$D=\sqrt (1-5)^2 + (2-2)^2$
$D=\sqrt (-4)^2 + (0)^2$
$D=\sqrt (16 + 0)$
$D=\sqrt (16)$
$D=4$
AB = 4
$height^2$= the distance between AC and A to (3,2) squared OR the distance between BD and D to (5,6) squared which is the same distance
AC = $D=\sqrt (x_{1}-x_{2})^2 + (y_{1}-y_{2})^2$
$D=\sqrt (1-3)^2 + (2-6)^2$
$D=\sqrt (-2)^2 + (-4)^2$
$D=\sqrt (4 + 16)$
$D=\sqrt 20$
A to (3,2) = (1,2) , (3,2)
$D=\sqrt (3-1)^2 + (2-2)^2$
$D=\sqrt (2)^2 + (0)^2$
$D=\sqrt (4 + 0)$
$D=\sqrt 4$
$D=2$
$\sqrt 20= height^2 + 2^2$
$20=height^2 + 4$
$16=height^2$
4= height
area of parallelogram = $base\times height$.
area of parallelogram = $4\times 4$.
area of parallelogram = 16
