Polygons and symmetry 1
The sum of the interior angles of a regular polygon is : (n - 2) × 180°,where n is the number of sides of the polygon.Thesides and theangles of aregular polygon haveequalmeasures.
XXX Sum XX XXX XXXXXXXX XX the exterior angles XX a XXXXXXX is XXXXXX XXX°,and measure XX XXXX exterior XXXXX of a regular polygon = XXX°/n,XXXXX n XX the XXXXXX of sides of the XXXXXXX
XXXXXXX has XXX XXXXX.
Sum of the XXXXXXXX of XXX interior XXXXXX XX a XXXXXXX :
=(n - 2) × 180° = (6 - 2) × 180° =XXX°
XXXXXXXX has 5 sides XX well XX X XXXXXX .
XXX sum of XXX measures of XXX interior XXXXXX XX a XXXXXXXX XX
=(n - X) × 180° = (5 - 2) × 180° = 540°
Thus, XXX XXXXXXX XX XXXX XXXXXXXX angle of aXXXXXXX XXXXXXXX XX
= XXX°/5 =108°
The XXX of themeasures XX XXX XXXXXXXX XXXXXX of a pentagon
= XXXXXXXXXXXXX XXXXX of each interior angle x XX. XX sides
= (XXX° - XXXo) x X =360°
XXX XXXXX XXXXXXX is aXXXXXXXX with 7 no. XX sides
The sum XX the measures of the interior XXXXXX of a heptagon is
=(n - 2) × XXX° = (7 - X) × XXX° = XXX°
Againthe XXX of the XXXXXXXX of XXX XXXXXXXX XXXXXX
=139° + 121° + XXX° + XXX° + 158° + XXX° + x°
= XXX° + x°
So, XXX° + x° = XXX°
x° = 900° - XXX° =111°
The measure of each XXXXXXXX angle XX aXXXXXXX hexagon is
= (n - X) × XXX° /n = (6 - 2) × 180°/6 = XXX°/6 = XXXo
The sum of themeasures XX the exterior angles XX a XXXXXXX
= Supplimentary angle of each interior angle x no. XX sides
= (XXX° - 120o) x 6 =XXX°
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