Lesson Area of n-sided polygon circumscribed about a circle

Algebra ->  Surface-area -> Lesson Area of n-sided polygon circumscribed about a circle      Log On


   

This Lesson (Area of n-sided polygon circumscribed about a circle) was created by by ikleyn(52784) About Me : View Source, Show
About ikleyn:

Area of n-sided polygon circumscribed about a circle


In this lesson you will learn how to calculate the area of an  n-sided polygon circumscribed about a circle.

Theorem 1

The area of an  n-sided polygon circumscribed about a circle equals half the product of the perimeter of the polygon and the radius of the circle.

Proof

Let us consider the circle in a plane with the radius  R  and with the center                  
at the point  O  (Figure 1a).  Let  A1A2A3...A(n-1)An  be an  n-sided polygon
in the plane circumscribed about the circle,  where the points  A1,  A2,
A3, . . . ,  A(n-1),  An  are the vertices of the polygon.

Figure 1a  shows the  5-sided polygon circumscribed about the circle.
Note that the polygon in the  Theorem  is not necessary a regular polygon.

For the proof, let us connect the center of the circle with the vertices by the
straight segments  OA1,  OA2,  OA3, . . . ,  OA(n-1),  and  OAn  (blue lines in
the  Figure 1b).  Then the interior of the polygon is divided in  n  triangles
DELTAOA1A2,  DELTAOA2A3, . . . ,  DELTAOA(n-1)An,  DELTAOAnA1.



  Figure 1a.  To the  Theorem 1        



Figure 1b. To the proof of the Theorem 1

Therefore,  the area of the polygon  A1A2A3...A(n-1)An  is the sum of the areas of the triangles  DELTAOA1A2, DELTAOA2A3, . . . ,  DELTAOA(n-1)An,  DELTAOAnA1:
S = S%5BOA1A2%5D + S%5BOA2A3%5D + . . . + S%5BOA%28n-1%29An%5D + S%5BOAnA1%5D.

The area of each of the triangles  DELTAOA1A2, DELTAOA2A3, . . . ,  DELTAOA(n-1)An  is half the product of the measure of the corresponding side of the polygon  (the base of the triangle)  and the radius  R,  because the radius drawn to the tangent point is perpendicular to the tangent segment  (see the lesson  A tangent line to a circle is perpendicular to the radius drawn to the tangent point  under the topic  Circles and their properties  of the section  Geometry  in this site). So, you have

S%5BOA1A2%5D = |A1A2|*R%2F2 + |A2A3|*R%2F2 + . . . + |A(n-1)An|*R%2F2 + |AnA1|*R%2F2= ( |A1A2| + |A2A3| + . . . + |A(n-1)An| + |AnA1| ) * R%2F2 = 1%2F2.P%2AR,

where  P  is the perimeter of the polygon.  It is exactly what the  Theorem  states.


For triangles,  this Theorem was proved in the lesson  Proof of the formula for the area of a triangle via the radius of the inscribed circle  under the current topic in this site.
For quadrilaterals,  it was proved in the lesson  Area of a quadrilateral circumscribed about a circle.


Example 1

Find the area of a polygon circumscribed about a circle,  if the radius of the circle is of  5 cm and the perimeter of the polygon is of  50 cm.

Solution

Apply the  Theorem 1.  According to this  Theorem,  the area of the polygon is half the product of the perimeter and the radius of the circle.
So,  the area of the polygon is   1%2F2.50*5 = 125 cm%5E2.

Answer.  The area of the polygon is  125 cm%5E2.


My other lessons on the area of polygons in this site are
    - Area of a regular n-sided polygon via the radius of the circumscribed circle  and
    - Area of a regular n-sided polygon via the radius of the inscribed circle
under the topic  Area and surface area  of the section  Geometry,  and
    - Solved problems on area of regular polygons
under the topic  Geometry  of the section  Word problems.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link  GEOMETRY - YOUR ONLINE TEXTBOOK.

This lesson has been accessed 6388 times.