prove that the sum of the interior angles of a pentagon is 540°
dhruvbadaya1:
In△ABE∠EAB+∠AEB+∠ABE=180°In△BEC∠BEC+∠BCE+∠EBC=180°In△CDE∠CDE+∠CED+∠ECD=180°Addingthese3equations,weget∠EAB+∠AEB+∠ABE+∠BEC+∠BCE+∠EBC+∠CDE+∠CED+∠ECD=540°∠EAB+∠ABE+∠EBC+∠BCE+∠ECD+∠CDE+∠AEB+∠BEC+∠CED∠A+∠B+∠C+∠D+∠E=540°So,sumoffiveanglesofpentagonis540°.
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since the sum of all the interior angle of a polygon =(n-2)×180°,
therefore
sum of interior angles of a pentagon =(5-2)×180°,
=3×180°,
=540°
therefore
sum of interior angles of a pentagon =(5-2)×180°,
=3×180°,
=540°
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