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# Find the area of the parallelogram with vertices a(-3, 0), b(-1, 5), c(7, 4), and d(5, -1)?

**Solution:**

The vertices of a parallelogram are a(-3, 0), b(-1, 5), c(7, 4), and d(5, -1)

Area of triangle abc = \( \frac{1}{2}\begin{vmatrix} -3 & 0 & 1\\ -1 & 5 & 1\\ 7 & 4 & 1 \end{vmatrix} \)

= 1/2 [-3 (5 - 4) - 1 (- 4 - 35)]

= 1/2 [-3 (1) - 1 (-39)]

= 1/2 [-3 + 39]

= 1/2 [36]

= 18 sq units

Area of triangle acd = \( \frac{1}{2}\begin{vmatrix} -3 & 0 & 1\\ 7 & 4 & 1\\ 5 & -1 & 1 \end{vmatrix} \)

= 1/2 [-3 (4 + 1) - 1 (- 7 - 20)]

= 1/2 [-3 (5) - 1 (-27)]

= 1/2 [-15 + 27]

= 1/2 [12]

= 6 sq units

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

= 18 + 6

= 24 sq units.

Therefore, the area of the parallelogram is 24 sq. units.

## Find the area of the parallelogram with vertices a(-3, 0), b(-1, 5), c(7, 4), and d(5, -1)?

**Summary: **

The area of the parallelogram with vertices a(−3, 0), b(−1, 5), c(7, 4), and d(5, −1) is 24 sq. units.

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