Question

Is it possible to have a polygon; whose sum of interior angles is :

(i) ${870}^o$

(ii) ${2340}^o$

(iii) 7 right - angles

(iv) ${4500}^o$ ?

Answer 1

(i) Let no. of sides = n

sum of angles = ${870}^o$

$\therefore (n-2)\times {180}^o={870}^{o}n-2=\frac{870}{180}n-2=\frac{29}{6}n=\frac{29}{6}+2n=\frac{41}{6}$

Which is not a whole number.

Hence it is not possible to have a polygon, the sum of whose interior angles is ${870}^o$

(ii) Let no. of sides = n

Sum of angles = ${2340}^o$

$\therefore (n-2)\times {180}^o={2340}^{o}n-2=\frac{2340}{180}n-2=13n=13+2=15$

Which is a whole number.

Hence it is possible to have a polygon, the sum of whose interior angles is ${2340}^o$.

(iii) Let no. of sides = n

Sum of angles = 7 right angles

$=7\times 90={630}^o\therefore (n-2)\times {180}^o={630}^{o}n-2=\frac{630}{180}n-2=\frac{7}{2}n=\frac{7}{2}+2n=\frac{11}{2}$

Which is not a whole number. Hence it is not possible to have a polygon, the sum of whose interior angles is 7 right - angles.

(iv) Let no. of sides $=n\therefore (n-2)\times {180}^o={4500}^{o}n-2=\frac{4500}{180}n-2=25n=25+2n=27$

Which is a whole number.

Hence it is possible to have a polygon, the sum of whose interior angles is ${4500}^o$.