Question

Find the sum of the measures of the interior angles of each of the following convex polygons: a pentagon, an octagon, a dodecagon, a 40-gon, a 52-gon, a 100-gon.

Answer 1

Answer:

$540^{\circ},\ 1080^{\circ},\ 1800^{\circ},\ 6840^{\circ},\ 9000^{\circ},\ 17640^{\circ}$

Step-by-step explanation:

The sum of the measures of the interior angles of each convex n-sided polygon is always equal to

$(n-2)\cdot 180^{\circ}.$

1. A pentagon is 5-sided polygon, then the sum of the measures of the interior angles of pentagon is

$(5-2)\cdot 180^{\circ}=540^{\circ}.$

2. An octagon is 8-sided polygon, then the sum of the measures of the interior angles of octagon is

$(8-2)\cdot 180^{\circ}=1080^{\circ}.$

3. A dodecagon is 12-sided polygon, then the sum of the measures of the interior angles of dodecagon is

$(12-2)\cdot 180^{\circ}=1800^{\circ}.$

4. For 40-sided polygon the sum of the measures of the interior angles is

$(40-2)\cdot 180^{\circ}=6840^{\circ}.$

5. For 52-sided polygon the sum of the measures of the interior angles is

$(52-2)\cdot 180^{\circ}=9000^{\circ}.$

6. For 100-sided polygon the sum of the measures of the interior angles is

$(100-2)\cdot 180^{\circ}=17640^{\circ}.$