Question

Can the sum of the measures of all the interior angles of a polygon be ${1170}^{\circ }$?

• A. Yes
• B. No

Answer 1

The correct option is B No

Sum of interior angles of polygon = $(n-2)\times {180}^{\circ }$

$\therefore$ sum of the measures of the all the interior angles of a polygon must be the multiple of ${180}^{\circ }$

Now,
${180}^{\circ }\times 1={180}^{\circ }$
${180}^{\circ }\times 2={360}^{\circ }$
${180}^{\circ }\times 3={540}^{\circ }$
${180}^{\circ }\times 4={720}^{\circ }$
${180}^{\circ }\times 5={900}^{\circ }$
${180}^{\circ }\times 6={1080}^{\circ }$
${180}^{\circ }\times 7={1260}^{\circ }$

From, above data we can see ${1170}^{\circ }$ is not the multiple of ${180}^{\circ }$,
hence, the sum of the measures of all the interior angles of a polygon can never be ${1170}^{\circ }$.