Find the area of the parallelogram with vertices A(-3, 0), B(-1, 5), C(7, 4), and D(5, -1).

iohanetc 2021-06-11 Answered
Find the area of the parallelogram with vertices A(-3, 0), B(-1, 5), C(7, 4), and D(5, -1).

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un4t5o4v
Answered 2021-06-12 Author has 5066 answers
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content_user
Answered 2021-09-28 Author has 2252 answers

Consider the parallelogram A=(-3;0), B(-1,5), C(7;4) and D=(5;-1)

The objective is to find the area of parallelogram

Find the are as follows:

The area of the parallelogram is the magnitude of the cross product of the adjacent edges.

That is, \(Area=|\overline{AB}\times\overline{AD}|\)

Find the adjacent edges \(\overline{AB}\) and \(\overline{AD}\) as follows

\(\overline{AB}=B-A\)

\(=(-1;5)-(-3;0)\)

\(=(-1+3;5-0)\)

\(=(2;5)\)

\(\overline{AD}=D-A\)

\(=(5;-1)-(-3;0)\)

\(=(8;1)\)

Find \(\overline{AB}\times\overline{AD}\) as,

\(\overline{AB}\times\overline{AD}=\begin{bmatrix}i&j&k\\2&5&0\\8&-1&0\end{bmatrix}\)

\(=5(0)-0(-1)i-(2(0)-0(8))j+(2(-1)-5(8))k\)

\(=0i+0j-42k\)

The area of the parallelogram is,

Area \(=|AB\times AD|\)

\(=|(0i+0j-42k|\)

\(=\sqrt{(0)^2+(0)^2+(-42)^2}\)

\(=\sqrt{(42)^2}\)

= 42 square unit

Hence, the area of parallelogram is 42

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