# How do you find the area and perimeter of a parallelogram with vertices at points (-6,-5), (-2,4), (5,4), and (1, -5)?

Nov 15, 2016

#### Explanation:

Let's move everything to the right 6 and up 5; this makes the vertices become:

$A = \left(0 , 0\right) , B = \left(4 , 9\right) , C = \left(11 , 9\right) , \mathmr{and} D = \left(7 , 0\right)$

Let side AD be the base of the parallelogram; it runs along the x axis for 7 units, therefore, this is the length of the base, b.

Point B is 9 units above side AD, therefore, this is the height.

$\text{Area" = "base" xx "height}$

$\text{Area" = 7 " units" xx 9 " units}$

${\text{Area" = 63 " units}}^{2}$

The length of side AB is:

$A B = \sqrt{{\left(4 - 0\right)}^{2} + {\left(9 - 0\right)}^{2}}$

$A B = \sqrt{16 + 81}$

$A B = \sqrt{97}$

$\text{Perimeter} = 2 A B + 2 A D$

$\text{Perimeter} = 2 \sqrt{97} + 2 \left(7\right)$

$\text{Perimeter" ~~ 33.7 " units}$