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

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Answer 1 · 2016-11-15T16:01:04

Please see the explanation.

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

$A = (0, 0), B = (4, 9), C = (11,9), and D =(7,0)$

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.

$"Area" = "base" xx "height"$

$"Area" = 7 " units" xx 9 " units"$

$"Area" = 63 " units"^2$

The length of side AB is:

$AB = sqrt((4 - 0)^2 + (9 - 0)^2)$

$AB = sqrt(16 + 81)$

$AB = sqrt(97)$

$"Perimeter" = 2AB + 2AD$

$"Perimeter" = 2sqrt(97) + 2(7)$

$"Perimeter" ~~ 33.7 " units"$