How do you find the area of the rectangle with vertices A(-3,0), B(-2,-1), C(1,2), D(0,3)?

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How do you find the area of the rectangle with vertices A(-3,0), B(-2,-1), C(1,2), D(0,3)?

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Answer 1 · 2016-02-06T13:21:29

Area $= 6$ $=6$ square units

Explanation:

Area $= (\frac{1}{2}) \cdot (x_{a} \cdot y_{b} + x_{b} \cdot y_{c} + x_{c} \cdot y_{d} + x_{d} \cdot y_{a} - x_{b} \cdot y_{a} - x_{c} \cdot y_{b} - x_{d} \cdot y_{c} - x_{a} \cdot y_{d})$ $=(1/2)*(x_a*y_b+x_b*y_c+x_c*y_d+x_d*y_a-x_b*y_a-x_c*y_b-x_d*y_c-x_a*y_d)$

Area $= (\frac{1}{2}) \cdot (- 3 (- 1) + (- 2) (2) + (1) (3) + 0 (0) - [0 (- 2) + (- 1) (1) + 2 (0) + (3 (- 3])$ $=(1/2)*(-3(-1)+(-2)(2)+(1)(3)+0(0)-[0(-2)+(-1)(1)+2(0)+(3(-3])$

Area $= (\frac{1}{2}) \cdot (6 - 4 - (- 10))$ $=(1/2)*(6-4-(-10))$

Area $= 6$ $=6$

Have a nice day !! from the Philippines...

Answer 2 · 2016-02-06T13:41:46

$6$ $6$ sq. units

Explanation:

$"Area"_square = "Length" xx "Width"$

Using $|AB|$ as the "Width"
and $|AD|$ as the "Length"

$|AB| = sqrt((-3-(-2))^2+(0-(-1))^2)=sqrt(2)$
$|AD|=sqrt((-3-0)^2+(0-3)^2)=3sqrt(2)$

$"Area"_square = 3sqrt(2)xxsqrt(2)=6$

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How do you find the area of the rectangle with vertices A(-3,0), B(-2,-1), C(1,2), D(0,3)?

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