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The interior angles of a polygon …

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The interior angles of a polygon are in AP. The smallest angle is 120 and c. d. is 5. Find the no. of sides of polygon
  • 1 answers

Sia ? 4 years, 10 months ago

a = 1200, d = 50
{tex}s _ { n } = \frac { n } { 2 } [ 2 a + ( n - 1 ) d ]{/tex}
{tex}s _ { n } = \frac { n } { 2 } [ 2 \times 120 + ( n - 1 ) 5 ]{/tex}
{tex}= \frac { n } { 2 } [ 240 + 5 n - 5 ]{/tex}
{tex}= \frac { n } { 2 } [ 235 + 5 n ]{/tex}
Also
sum of interior angles of a polygon with n sides = (2n - 4) {tex}\times{/tex} 90
ATQ {tex}\frac { n } { 2 } [ 235 + 5 n ] = ( 2 n - 4 ) \times 90{/tex}
n = 9, 16
but n = 16 not possible
n = 9
{tex}s _ { n } = \frac { n } { 2 } [ 2 a + ( n - 1 ) d ]{/tex}
{tex}s _ { n } = \frac { n } { 2 } [ 2 \times 120 + ( n - 1 ) 5 ]{/tex}
{tex}= \frac { n } { 2 } [ 240 + 5 n - 5 ]{/tex}
{tex}= \frac { n } { 2 } [ 235 + 5 n ]{/tex}
Also
Sum of interior angle of polygon with n sides {tex}= ( 2 \mathrm { n } - 4 ) \times 90{/tex}

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