How many sides does a polygon have the of the interior angles is- 900-o

Question

Toppr Admin · Accepted Answer

Since sum of the interior angles is 9000- therefore- -2n-x2013-4-900-90002n-x2013-4-90090-x21D2-2n-x2013-4-10-x21D2-2n-10-4-x21D2-2n-14-x21D2-n-142-7Hence- the polygon has 7 sides-

How many sides does a polygon have if the sum of the interior angles is: $900^{o}$

Since sum of the interior angles is $900^{0}$ , therefore, $(2 n - 4) 90^{0} = 900^{0}$

$2 n - 4 = \frac{900}{90} \Rightarrow 2 n - 4 = 10 \Rightarrow 2 n = 10 + 4 \Rightarrow 2 n = 14 \Rightarrow n = \frac{14}{2} = 7$

Hence, the polygon has $7$ sides.

How many sides does a polygon have if the sum of the interior angles is: 900o

Since sum of the interior angles is 9000, therefore, (2n–4)900=90002n–4=90090⇒2n–4=10⇒2n=10+4⇒2n=14⇒n=142=7Hence, the polygon has 7 sides.

How many sides does a polygon have if the sum of the interior angles is: $900^{o}$

Since sum of the interior angles is $900^{0}$ , therefore, $(2 n - 4) 90^{0} = 900^{0}$

$2 n - 4 = \frac{900}{90} \Rightarrow 2 n - 4 = 10 \Rightarrow 2 n = 10 + 4 \Rightarrow 2 n = 14 \Rightarrow n = \frac{14}{2} = 7$

Hence, the polygon has $7$ sides.