Find the area of the parallelogram with vertices A(−3, 0),

chezmarylou1i

chezmarylou1i

Answered question

2021-12-12

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

Answer & Explanation

limacarp4

limacarp4

Beginner2021-12-13Added 39 answers

Given vertices A(−3, 0), B(−1, 7), C(9, 6), D(7, −1)
AB=(1(3),70)=(2,7)
BC=(9(1),67)=(10,1) The area of parallelogram with adjacent sides AB and BC is the length of their cross product |AB×BC|
Now
|AB×BC|=|ijk2701010|
=(00)i(00)j+(270)k
=72k
Therefore, the area is,
|AB×BC|=(72)2=72
levurdondishav4

levurdondishav4

Beginner2021-12-14Added 38 answers

=12[3(76)1(663)]
=12[3(1)1(66)]
=12[3+66]
=12[63]
=632 sq units
=12[3(6+1)1(942)]
=12[3(7)1(51)]
=12[21+51]
=12[30]
=12 sq units
Area of parallelogram abcd = Area of triangle abc + Area of triangle acd
=632+12
=63+242
=872 sq units.
Therefore, the area of the parallelogram is 872 sq. units.

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