Determined the sum of the interior angles of a polygon with vertices located at

Question

Determined the sum of the interior angles of a polygon with vertices located at (-3, 4), (0, 7), (1, 4), (-1, -4), and (-4, 0)
Determined the sum of the exterior angles of a polygon with vertices located at (-11,7), (-8, 9), (-5,7), (-2, 3), (-4, 3), and (-9, 5) Determined the sum of the interior angles of a polygon with vertices located at (-3, 4), (0, 7), (1, 4), (-1, -4), and (-4, 0)
Determined the sum of the exterior angles of a polygon with vertices located at (-11,7), (-8, 9), (-5,7), (-2, 3), (-4, 3), and (-9, 5)
Determined the sum of the exterior angles of a polygon with vertices located at (-11,7), (-8, 9), (-5,7), (-2, 3), (-4, 3), and (-9, 5)

Explanation / Answer

(a) Since there are 5 vertices, the figure is a pentagon.

Sum of interior angles of a polygon is given by the formula (n-2) * 180, where n is the number of sides. Here n = 5.

Therefore sum of the interior angle of the pentagon = (5 - 2) * 180 = 3 * 180 = 540o.

(b) The sum of the exterior angles of any polygon, irrespective of the number of vertices is always equal to 360o.

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