chezmarylou1i
2021-12-12
Answered

Find the area of the parallelogram with vertices A(−3, 0), B(−1, 7), C(9, 6), and D(7, −1).

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limacarp4

Answered 2021-12-13
Author has **39** answers

Given vertices A(−3, 0), B(−1, 7), C(9, 6), D(7, −1)

$\overrightarrow{AB}=(-1-(-3),7-0)=(2,7)$

$\overrightarrow{BC}=(9-(-1),6-7)=(10,-1)$
The area of parallelogram with adjacent sides AB and BC is the length of their cross product $|\overrightarrow{AB}\times \overrightarrow{BC}|$

Now

$|\overrightarrow{AB}\times \overrightarrow{BC}|=\left|\begin{array}{ccc}i& j& k\\ 2& 7& 0\\ 10& -1& 0\end{array}\right|$

$=(0-0)i-(0-0)j+(-2-70)k$

$=-72k$

Therefore, the area is,

$|\overrightarrow{AB}\times \overrightarrow{BC}|=\sqrt{{(-72)}^{2}}=72$

Now

Therefore, the area is,

levurdondishav4

Answered 2021-12-14
Author has **38** answers

Area of parallelogram abcd = Area of triangle abc + Area of triangle acd

Therefore, the area of the parallelogram is

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