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 jointing 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, divides 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 Sides of Polygon:

There is no theoretical limit to the number of sides of a polygon, but only those with twelve or less are commonly met with. The names of polygons which are mostly in use 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:

Perimeter of a regular polygon = perimeter of square  = 4a

A regular polygon can be divided into congruent triangles having common vertex at the center of the polygon. The number of these congruent triangles is the same as that of its sides.
\therefore 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 Area of one such triangle = \frac{1}{2}ah, h being length of 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, Area of 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 perpendicular is \frac{a}{2}.

Comments

comments