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We know that the sum of the interior angles of a triangle is`180^0`. Show that the sums of eth interior angles of polygons with `3,4,5,6,`sides for an arithmetic progression. Find the sum of the interior angles ofor a 21 sided polygon. |
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Answer» We know that sum of interior angles of a polygon of side `n=(2n-4)xx90^(@)=(n-2)xx180^(2)` Sum of interior angles a polygon with sides 3 is 180. Sum of interior angles of polygon with side `4=(4-2)xx180^(@)=360^(2)` Similarly , sum of interior angles of polygon with side 5,6,7,. are `540^(@) ,720 ^(@),900^(@)`....... the series will be `180^(@),360^(@),540^(@),720^(@),900^(@),......` Here `a=180^(@)` ` and d=360^(2)-180^(@)=180^(@)` since common difference is same between two consecutive terms of the series . so ir from an AP . we have to find the of interior angles of a 21 sides polygen . it means , we have to find the 19 th term of the above series . `a_(19)=a+(19-1)d` `=180+18xx180=3420` |
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