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Question

We know that the sum of the interior angles of a triangle is 180°. Show that the sums of the interior angles of polygons with 3, 4, 5, 6, ... sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.                                            [NCERT EXEMPLAR]


Solution

We know that,
the sum of the interior angles of a polygon with 3 sides, a1 = 180°,
the sum of the interior angles of a polygon with 4 sides, a2 = 360°,
the sum of the interior angles of a polygon with 5 sides, a3 = 540°,
.
.
.

As, a2-a1=360°-180°=180° and a3-a2=540°-360°=180°i.e. a2-a1=a3-a2So, a1, a2, a3, ... are in A.P.Also, a=180° and d=180°Since, the sum of the interior angles of a 3 sided polygon=aSo, the sum of the interior angles of a 21 sided polygon=a19Now,a19=a+19-1d=180°+18×180°=180°+3240°=3420°

So, the sum of the interior angles for a 21 sided polygon is 3420°.

Mathematics
RD Sharma XI (2019)
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