Question

Find the sum of the measures of the interior angles of a polygon with 24 sides. Sum of the measures of the interior angles of any $n$-gon = $(n-2)180$ For any regular $n$-gon; Measure of an interior angle = $\frac{(n-2)180}{n}$ Measure of an exterior angle = $\frac{360}{n}$ A = $\frac{1}{2}ap$ A 3960° B 4680° C 4140° D 4320°

          Find the sum of the measures of the interior angles of a polygon with 24 sides.
Sum of the measures of the interior angles of any $n$-gon = $(n-2)180$
For any regular $n$-gon;
Measure of an interior angle = $\frac{(n-2)180}{n}$
Measure of an exterior angle = $\frac{360}{n}$
A = $\frac{1}{2}ap$
A 3960°
B 4680°
C 4140°
D 4320°

Find the sum of the measures of the interior angles of a polygon with 24 sides.
Sum of the measures of the interior angles of any n-gon = (n-2)180
For any regular n-gon;
Measure of an interior angle = ((n-2)180)/(n)
Measure of an exterior angle = (360)/(n)
A = (1)/(2)ap
A 3960°
B 4680°
C 4140°
D 4320°

Added by Karen A.

Geometry A Common Core Curriculum

Ron Larson, Laurie Boswell 1st Edition

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The sum of the measures of the interior angles of any n-gon is (n-2)180. Find the sum of the measures of the interior angles of a polygon with 24 sides. For any regular n-gon: Measure of an (n-2)180 interior angle. 39609 46809 Measure of an exterior angle 360 41409 ap 4320