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Hunter-Best [27]
3 years ago
7

What is the area of a parallelogram whose vertices are A(−12, 2) , B(6, 2) , C(−2, −3) , and D(−20, −3) ?

Mathematics
2 answers:
Darina [25.2K]3 years ago
7 0

I am finding the answer right now but there is a website that you can use for math that really helps. It's called Desmos.

lukranit [14]3 years ago
5 0

These have two sides with constant y values, i.e. two sides parallel to the x axis.  We'll check they're the same length

AB=6 - -12 = 18

CD = -2 - -20 = 18  good

That's the base.  The height is 2 - -3 = 5 so an area of 18(5)=90

Answer: 90

Even if we didn't know it was a nicely oriented parallelogram we can get the area with the shoelace formula, which says it's half the absolute value of the sum of the cross products of each side.

A(−12, 2) , B(6, 2) , C(−2, −3) , D(−20, −3)

B(6, 2) , C(−2, −3) , D(−20, −3) , A(−12, 2)

A = (1/2) |  (-12)(2) - 2(6) + 6(-3) - 2(-2) + (-2)(-3) - (-3)(-20) + (-20)(2) - (-3)(-12) | = |-180|/2 = 90


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