The sum of interior angles of a polygon with “n” sides =180°(n-2) Number of diagonals of a “n-sided” polygon = [n(n-3)]/2. The measure of interior angles of a regular n-sided polygon = [(n-2)180°]/n. The measure of exterior angles of a regular n-sided polygon = 360°/n.
From the above figure, the area of the quadrilateral ABCD = area of ΔBCD + area of ΔABD. Thus, the area of the quadrilateral ABCD = (1/2) × d × h1 h 1 + (1/2) × d × h2 h 2 = (1/2) × d × (h1+h2 h 1 + h 2 ).
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Answer:
The sum of interior angles of a polygon with “n” sides =180°(n-2) Number of diagonals of a “n-sided” polygon = [n(n-3)]/2. The measure of interior angles of a regular n-sided polygon = [(n-2)180°]/n. The measure of exterior angles of a regular n-sided polygon = 360°/n.
Step-by-step explanation:
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From the above figure, the area of the quadrilateral ABCD = area of ΔBCD + area of ΔABD. Thus, the area of the quadrilateral ABCD = (1/2) × d × h1 h 1 + (1/2) × d × h2 h 2 = (1/2) × d × (h1+h2 h 1 + h 2 ).
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