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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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