6.6
Polygons (Rectilinear figures):
Let us look at the figures given below and observe their
properties.
They are on the same plane(surface)
They are closed figures
They are bounded by 3 or more lines.
They have 3 or more intersecting points.
Definition:
‘Polygons’ (figures with many sides) are
figures with three or more line segments which are coplanar (lie on the same
plane), non – collinear and
whose line segments intersect each
other at their end points. The end points where
line segments meet are called ‘vertices’ of the polygon.
The line segments which make the polygon are called ‘sides’ of the polygon.
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The
line segment joining any 2 non consecutive (not next to each other) vertices
of a polygon is called a ‘diagonal’. The adjacent figure
has five sides and hence it is called pentagon. |
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Observation: In a
polygon, number of
sides= number
of angles =
number of vertices. |
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Examples
of polygons are Triangle (3 sides), Quadrilateral (4 sides), Pentagon (5
sides), Hexagon (6 sides), Heptagon (7 sides), Octagon (8 sides) and so on… |
Definition:
1. A ‘regular polygon’ is a polygon which is both
equilateral (all sides equal) and equiangular (all angles same). (Square is a best
example of a
regular polygon as its sides are equal
and angles are all equal to 900)
2. The ‘interior region’
of a polygon is a part of the plane enclosed by the polygon. (the space
which is called area)
Exercise:
1. Draw a regular pentagon – do you notice that all
interior angles are equal to 1080?
2. Draw a regular hexagon – do you notice that all interior
angles are equal to 1200?
An ‘inscribed regular
polygon’ is a regular polygon whose vertices are on a circle.
(Alternatively we can say that an ‘inscribed
regular polygon’ is a regular polygon whose sides are chords (arcs) of a
circle)
As an example let
us look at a regular quadrilateral:
(From each of the vertices of the regular polygon draw
lines to the center of the circle)
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1.
Number of lines connecting to origin of the circle is equal to number of
sides of polygon(In the adjacent figure 4 sides: AO,
BO,CO,DO ) 2.
Number of angles formed at the origin of the circle is equal to number of
sides of polygon( In the adjacent figure 4 angles 3. Angles
at the origin are equal to each other(=90o) |
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6.6
Example 1:
Construct an inscribed regular quadrilateral in a circle of radius 4cm:
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Step 1:
Divide 3600 by the number of sides of the quadrilateral (=4) to
get the angle at origin. Therefore angle at origin = 900 = 360/4. Step 2:
With O as center draw a circle of radius 4cm. Step 3:
Draw 2 lines (OA and Step 4:
With B as center cut an arc of radius =AB to cut the circle at C. Step 5:
With C as center cut an arc of radius = BC to cut the circle at D. Step 6: Join the points A, B, C and D to get the quadrilateral ABCD. |
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General Method for
constructing an inscribed regular polygon in a circle of a given radius:
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Step 1:
Divide 3600 by the number of sides of the polygon to get the angle
at origin. Angle at origin = 360/(No of sides of Polygon) Step 2:
With O as center draw a circle of given radius. Step 3:
Draw 2 lines (OA and Step 4:
With B as center cut an arc of radius=AB
to cut the circle at C. Step 5:
With C as center cut an arc of radius =BC to cut the circle at D. Step 6:
Repeat step 5 till the last arc cuts the circle again at A. Step 7:
Join the points to get the required polygon. |
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Observations:
Angle at origin for different inscribed regular polygons:
|
Polygon
type |
No
of sides |
Angle
at center |
|
Triangle
|
3 |
1200(360÷3) |
|
Quadrilateral |
4 |
900(360÷4) |
|
Pentagon |
5 |
720(360÷5) |
|
Hexagon |
6 |
600(360÷6) |
|
Octagon |
8 |
450(360÷8) |
Think why inscribed polygon of 7 sides is not included in
the above table!
6.6 Summary of learning
|
No |
Points to remember |
|
1 |
Polygons
are figures with three or more line segments which are coplanar, non –
collinear and these line segments intersect each other at their end points |
|
2 |
A
regular polygon is a polygon which is both equilateral and equiangular |
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3 |
An inscribed
regular polygon is a regular polygon whose vertices are on the circle |
Additional
Points:
6.6.1
Theorem 1: In a
polygon of ‘n’ sides, the sum of the interior angles is equal to (2n-4) right
angles
Given: ABCDEFG… is a polygon of n sides (ABC, BCD, CDE…
are interior angles)
To prove: Sum of interior angles = (2n-4) right angles
Construction: Take any point O inside the polygon. From O
draw lines to each of the vertices (A,B,C…)
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Step |
Statement |
Reason |
|
|
1 |
The Polygon is made up of n
triangles |
Polygon has n sides |
|
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2 |
Sum of all angles of n triangles = n*2 right angles |
Sum of angles in a triangle = 2 right
angles |
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3 |
Sum of angles at O = 4 right angles |
Angle at a point = 3600 |
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|
4 |
Sum of all angles
of n triangles = Sum
of interior angles of polygon + Angle at O |
Construction |
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5 |
Sum of interior
angles of polygon = Sum of all angles of n triangles - Angle
at O |
Transposition of Step 4 |
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|
6 |
= 2n
right angles - 4 right angles =
(2n-4)right angles |
Substitution (step2 and 3) |
6.6.1 Corollary:
If the sides of a polygon are produced in order (Clock
wise direction or anti clock wise direction), the sum of exterior angles so
formed is equal
to 4 right angles. In the adjoining
figure a,b,c… are the
exterior angles
|
Step |
Statement |
Reason |
|
|
1 |
Sum of (interior + exterior)
angle at one vertex = 2 right angles |
angles on a straight line = 2 right
angles |
|
|
2 |
Sum of (interior
+ exterior) angle at n vertices = 2n right angles |
Step 1 |
|
|
3 |
Sum of
interior angles + sum of exterior angles
= Sum of (interior angle +
exterior angle) for n sides (vertices) |
Construction(figure) |
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4 |
Sum of exterior angles = Sum of (interior + exterior)
angle for n sides - Sum of interior angles |
Transposition of Step 3 |
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5 |
= 2n right angles – (2n-4) right angles |
(step 2
and 6.6.1 Theorem) |
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6 |
= 4 right angles |
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Observations:
|
No |
Properties |
|
1 |
Each interior angle of a regular polygon = (2n-4)*90/n |
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2 |
Each exterior angle of a regular polygon(x) = 4*90/n = 360/n |
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3 |
If x is
the exterior angle of a regular polygon then the
number of sides(n) = 360/x |
6.6.1
Problem 1: AB, BC and CD are three consecutive sides of
a regular polygon. If 
BAC = 150 Find
(i) Each interior angle of the
polygon
(ii) Each exterior angle of the polygon
(iii) Number of sides of the polygon
Solution:
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Since
AB=BC, ABC is an isosceles triangle, hence Thus Each
interior angle = 1500 Each
exterior angle = 300 Number
of sides = 360/30 = 12 |
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6.6.1
Problem 2: Difference
between the exterior angle of (n-1) sided regular polygon and the interior
angle of (n+1) sided regular polygon is 90.
Find the value of n.
Solution:
1. Exterior angle of (n-1) sided regular polygon = 360/(n-1)
2. Exterior angle of (n+1) sided regular polygon = 360/(n+1)
It is given that the difference between them is 9
360/(n-1)
– 360/(n+1) = 9
i.e. {360(n+1) - 360(n-1)}/{(n+1)(n-1)}
= 9
i.e. 720/n2-1 = 9
i.e. n2-1 = 80
i.e. n2 = 81
i.e. n = 9
The regular polygon has 9 sides.