# Properties of a Regular Polygon

These are the properties of a regular polygon:

• The sides and interior angles of a regular polygon are all equal.
• The bisectors of the interior angles of a regular polygon meet at its center.
• The perpendiculars drawn from the center of a regular polygon to its sides are all equal.
• The lines joining the center of a regular polygon to its vertices are all equal.
• The center of a regular polygon is the center of both the inscribed and circumscribed circles.
• Straight lines drawn from the center to the vertices of a regular polygon divide it into as many equal isosceles triangles as there are sides in it.
• The angle of a regular polygon of $n$ sides $= \left( {\frac{{2n - 4}}{n}} \right) \times {90^ \circ }$.

Detail of the Sides of a Polygon

There is no theoretical limit to the number of sides of a polygon, but only those with twelve or less are commonly seen. The names of polygons which are used most often are as follows:

 Number of sides Polygon name $5$ Pentagon $6$ Hexagon $7$ Heptagon $8$ Octagon $9$ Nonagon $10$ Decagon $\cdots$ $\cdots$ $\cdots$ $\cdots$ $n$ $n -$gon

Example:

The perimeter and area of a regular polygon are respectively equal to those of a square of sides $a$. Find the length of the perpendicular from the center of a regular polygon to any of its sides.

Solution:

The perimeter of a regular polygon = perimeter of square $= 4a$

A regular polygon can be divided into congruent triangles having a common vertex at the center of the polygon. The number of these congruent triangles is the same as that of its sides.

$\therefore$ the area of one such triangle $= \frac{1}{2} \times$ sides of polygon $\times$ length of perpendicular from the center to any side of polygon.

$\therefore$ the area of one such triangle$= \frac{1}{2}ah$, $h$ being the length of the perpendicular

$\therefore$ Area of polygon = Sum of areas of all such triangles

$\therefore$ Area of polygon $= \frac{1}{2} \times$ Perimeter of polygon $\times {\text{ }}h$

$\therefore$ Area of polygon $= \frac{1}{2} \times 4a \times h = 2ah$ --- (1)

Now, the area of the polygon = Area of square$= {a^2}$ (given) --- (2)

$\therefore$ from (1) and (2), we have
$\therefore$ $2ah = {a^2} \Rightarrow h = \frac{a}{2}$

Hence, the length of the perpendicular is $\frac{a}{2}$.