Chapter 7.3, Problem 11PSB

Solution

a.

To find the name of the polygon which has the sum of the measures of the exterior angles, one per vertex, equal to the sum of the measures of the angles of the polygon.

The polygon which has the sum of the measures of the exterior angles, one per vertex, equal to the sum of the measures of the angles of the polygon is square.

Given: Sum of the measures of the exterior angles, one per vertex, equal to the sum of the measures of the angles of the polygon.

Concept used: Here, we are using the formula:

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

Calculation: The sum of the exterior angles is always ${360}^{\circ }$

The sum of the interior angles is ${180}^{\circ }(n-2)$

Now, ${360}^{\circ }={180}^{\circ }(n-2)$

➔ $\frac{{360}^{\circ }}{{180}^{\circ }}=n-2$

➔ $2=n-2$

➔ $n=4$

Conclusion: Hence, the polygon is square.

b.

To find the name of the polygon which has the sum of the measures of the angles of the polygon equal to twice the sum of the measures of the exterior angles, one per vertex.

The polygon which has the sum of the measures of the angles equal to twice the sum of the measures of the exterior angles, one per vertex is hexagon.

Given: Sum of the measures of the angles of the polygon equal to twice the sum of the measures of the exterior angles, one per vertex.

Concept used: Here, we are using the formula:

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

Calculation: The sum of the exterior angles is always ${360}^{\circ }$

The sum of the interior angles is ${180}^{\circ }(n-2)$

Since, the sum is twice the sum of exterior angles we write:

➔ ${180}^{\circ }(n-2)=2\times {360}^{\circ }$

➔ $n-2=\frac{2\times 360}{180}$

➔ $n-2=4$

➔ $n=6$

Conclusion: Hence, the polygon is hexagon.