Question

Choose the correct pair(s) of sum of interior angles of a regular polygon and the measure of each interior angle, respectively.

• A. ${720}^o,{120}^o$
• B. ${900}^o,{128.7}^o$
• C. ${360}^o,{90}^o$
• D. ${120}^o,{60}^o$

Answer 1

Answer: (A), (C)

${360}^o,{90}^o$

Sum of interior angles $=(n-2)\times {180}^o$
where $n$ is the total number of sides.

For the first given pair, sum of interior angles $={720}^o$ $\Rightarrow {720}^o=(n-2)\times {180}^o$ $\Rightarrow n-2=\frac{{720}^o}{{180}^o}=4$ $\Rightarrow n=4+2=6$ $\therefore$ The polygon has 6 sides. It also has 6 vertices and hence 6 interior angles. The polygon is regular. Hence, all the interior angles are of same measure. Let, each interior angle is $x.$ $\therefore 6x={720}^o$ $\Rightarrow x=\frac{{720}^o}{6}={120}^o$ Hence, each interior angle is ${120}^o.$ $\therefore$ First given pair is correct.

For the second given pair, the sum of interior angles $={900}^o$ Similarly, we find that $n=7.$ Hence, each interior angle is $\frac{{900}^o}{7}\ne {128.7}^o$ $\therefore$ Second option is wrong.

For the third given pair, the sum of interior angles $={360}^o$ Similarly, we find that $n=4.$ Each interior angle of the regular 4-sided polygon is $\frac{{360}^o}{4}={90}^o$ Hence, it is correct.

We know that a triangle is the smallest possible polygon whose sum of interior angles is ${180}^o$ so, a polygon with its sum of interior angle ${120}^o$ is not possible.