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what is the area of a parallelogram whose vertices are ( a(-12,2) ), ( …

Question

what is the area of a parallelogram whose vertices are ( a(-12,2) ), ( b(6,2) ), ( c(-2,-3) ), and ( d(-20,-3) )?
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units²

Explanation:

Step1: Calculate the base length

The base \(AB\) can be calculated using the distance formula for two points \((x_1,y_1)\) and \((x_2,y_2)\) which is \(d=\vert x_2 - x_1\vert\) (since \(y\)-coordinates of \(A(-12,2)\) and \(B(6,2)\) are the same).
\(AB=\vert6-(-12)\vert=\vert6 + 12\vert=18\)

Step2: Calculate the height

The height is the vertical distance between the two parallel lines (the \(y\)-coordinates of the two sets of points). The \(y\)-coordinate of \(A\) (and \(B\)) is \(y_1 = 2\), and the \(y\)-coordinate of \(C\) (and \(D\)) is \(y_2=-3\).
The height \(h=\vert2-(-3)\vert=\vert2 + 3\vert=5\)

Step3: Calculate the area of the parallelogram

The area formula of a parallelogram is \(A=\text{base}\times\text{height}\).
\(A = 18\times5=90\)

Answer:

\(90\)