Asked by Pilar
Find the area of rectangle ABCD with vertices A(-3, 0), B(3, 2), C(4, -1), and D(-2, -3).
Answers
Answered by
oobleck
AB = √((3+3)^2 + 2^2) = √40
BC = √((4-3)^2 + (-1-2)^2) = √10
So the area = length * width = √40 * √10 = √400 = 20
I'm assuming that point D is correctly placed, and that AB⊥BC, since they said it was a rectangle.
BC = √((4-3)^2 + (-1-2)^2) = √10
So the area = length * width = √40 * √10 = √400 = 20
I'm assuming that point D is correctly placed, and that AB⊥BC, since they said it was a rectangle.
Answered by
tdog
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Answered by
Someone
I have no idea
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