Find the area of the parallelogram with vertices a (-3, 0), b(-1, 7), c(9, 6), and d(7, -1)?

Solution:

The vertices of a parallelogram are a (-3, 0), b(-1, 7), c(9, 6), and d(7, -1)

Area of triangle abc = \( \frac{1}{2}\begin{vmatrix} -3 & 0 & 1\\ -1 & 7 & 1\\ 9 & 6 & 1 \end{vmatrix} \)

= 1/2 [-3 (7 - 6) - 1 (- 6 - 63)]

= 1/2 [-3 (1) - 1 (-66)]

= 1/2 [-3 + 66]

= 1/2 [63]

= 63/2 sq units

Area of triangle acd = \( \frac{1}{2}\begin{vmatrix} -3 & 0 & 1\\ 9 & 6 & 1\\ 7 & -1 & 1 \end{vmatrix} \)

= 1/2 [-3 (6 + 1) - 1 (- 9 - 42)]

= 1/2 [-3 (7) - 1 (-51)]

= 1/2 [-21 + 51]

= 1/2 [30]

= 12 sq units

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

= 63/2 + 12

= (63 + 24)/2

= 87/2 sq units.

Therefore, the area of the parallelogram is 87/2 sq. units.

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Find the area of the parallelogram with vertices a (-3, 0), b(-1, 7), c(9, 6), and d(7, -1)?

Summary:

The area of the parallelogram with vertices a (-3, 0), b(-1, 7), c(9, 6), and d(7, -1) is 87/2 sq. units.