Find the area of a pentagon having the vertic | vedclass

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(A) To find the area of a polygon with vertices $(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$, we use the shoelace formula:Area $= \frac{1}{2} |(x_1y_2

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$40$

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(A) To find the area of a polygon with vertices $(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$, we use the shoelace formula:Area $= \frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_5 + x_5y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_5 + y_5x_1)|$Given vertices: $(1, 5), (-2, 4), (-3, -1), (2, -3), (5, 1)$.Sum $1 = (1 	imes 4) + (-2 	imes -1) + (-3 	imes -3) + (2 	imes 1) + (5 	imes 5) = 4 + 2 + 9 + 2 + 25 = 42$Sum $2 = (5 	imes -2) + (4 	imes -3) + (-1 	imes 2) + (-3 	imes 5) + (1 	imes 1) = -10 - 12 - 2 - 15 + 1 = -38$Area $= \frac{1}{2} |42 - (-38)| = \frac{1}{2} |42 + 38| = \frac{1}{2} |80| = 40$ square units.

Find the area of a pentagon having the vertices $(1, 5), (-2, 4), (-3, -1), (2, -3),$ and $(5, 1).$

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