What is the area of a polygone with vertices (-3,4) (2,4) (4,2) (2,-3) and (-3,-3)
1 answer:
By summing the areas of <em>minor</em> rectangles and triangles, the area of the <em>irregular</em> polygon whose vertices are (- 3, 4), (2, 4), (4, 2), (2, - 3) and (- 3, - 3) is equal to 42 square units.
<h3>How to determine the area of a irregular polygon</h3>
In this problem we have to determine the area of a <em>irregular</em> polygon, whose calculation is defined as a sum of the areas of triangles and rectangles. To determine the combination of squares and triangles, we need first to plot the vertices and construct the figure on a <em>Cartesian</em> plane.
The area of the irregular figure is:
A = (5) · (7) + 0.5 · 2² + 0.5 · (2) · (5)
A = 35 + 2 + 5
A = 42
By summing the areas of <em>minor</em> rectangles and triangles, the area of the <em>irregular</em> polygon whose vertices are (- 3, 4), (2, 4), (4, 2), (2, - 3) and (- 3, - 3) is equal to 42 square units.
To know more on polygons: brainly.com/question/10441863
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