Question

The sum of all the angles of a polygon except one angle is $2220{}^\circ $ , then the number of sides of the polygon is:
(a) 12
(b) 13
(c) 14
(d) 15

Answer 1

Hint: As it is given that the sum of all the angles of a polygon except one angle is $2220{}^\circ $ , so we can say that the sum of all the angles must be greater than $2220{}^\circ $ . We also know that the sum of all the interior angles of an n-sided polygon is given by $\left( n-2 \right)180{}^\circ $ , so form the inequality and solve it to get the answer.

Complete step-by-step answer:
As it is given that the sum of all the angles of a polygon except one angle is $2220{}^\circ $ , so we can say that the sum of all the angles must be greater than $2220{}^\circ $ . We also know that the sum of all the interior angles of an n-sided polygon is given by $\left( n-2 \right)180{}^\circ $ , where n is the number of sides, so if we represent this in form of inequality, we get
$\left( n-2 \right)180 > 2220$
Now we will divide both sides of the inequality by 180. On doing so, we get
$n-2 > \dfrac{2220}{180}$
$\Rightarrow n > \dfrac{222}{18}+2$
Now, if we take the LCM of the RHS as 18, we get
$n>\dfrac{222+36}{18}$
$\Rightarrow n > \dfrac{258}{18}$
$\Rightarrow n>14.33$
As n is the number of sides, so it must be an integer. Also, n is an integer greater than 14.33, so n can be 15, 16 and all other integers greater than 15.
Hence, the answer is option (d), because the only integer greater than 14.33 among the options is 15.

Note: Remember that the sum of all the exterior angles of a polygon is always equal to $360{}^\circ $ , no matter whether it is a regular polygon or not. Also, whenever dealing with an inequality be very careful while you multiply, square or perform other operations, as there are cases where the sign of inequality changes. For example: x > y implies $-y > -x$ , i.e. , when both sides of an inequality are multiplied by a negative number, the sign of inequality changes.