The document discusses various topics related to polygons: - It defines different types of polygons from triangles to dodecagons based on their number of sides. - It explains how to calculate the sum of interior angles of any polygon by dividing it into smaller polygons and triangles. - It establishes that the sum of interior angles of an n-sided polygon is (n-2)*180 degrees. - It describes exterior angles of polygons and establishes that the sum of exterior angles of any polygon is always 360 degrees.

1. MATHEMATIC
S
Standard
IX
1

2. THE NATIONAL ANTHEM
Jana-gana-mana adhinayaka, jaya he
Bharatha-bhagya- vidhata
Punjab- Sindh- Gujarat- Maratha
Dravida-Utkala-Banga
Vindhya-Himachala-Yamuna-Gana
Uchchala-Jaladhi-taranga
Tava subha name jage,
Tava subha asisa mage,
Gahe tava jaya gatha.
Jana-gana-mangala-dayaka jaya he
Bharatha-bhagya-vidhata.
Jaya he,jaya he, jaya he,
Jaya jaya jaya, jaya he!
2

3. PLEDGE
India is my country. All Indians are my brothers and sisters. I love my
country, and I am proud of its rich and varied heritage. I shall always to
be worthy of it. I shall give respect to my parents, teachers and all elders
and treat everyone with courtesy. I pledge my devotion to my country and
my people. In their well-being and prosperity alone lies my happiness.
Prepared By
Saju Kumari
e-mail:
sajujayan1980@gmail.com
Phone: 9497011959
3

4. Dear Children,
One more step in the journey towards knowledge.
A book to learn math and learn it right
New thoughts, new deeds
New meanings of old ideas
Recognizing the joy of learning
And the power of action
Moving yet forward……
With regards
Prof. K.A. Hashim
Director
SCERT
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5. Textbook Development Committee
Mathematics IX
Saju Kumari
B.N.V. College, Thiruvananthapuram
Members
*Ajeesh *Nazrian Banu
Student of B.N.V TVPM Student of B.N.V TVPM
*Renjith Nair *Vibitha
Student of B.N.V TVPM Student of B.N.V TVPM
*Padmaja *Sruthi
Student of B.N.V TVPM Student of B.N.V TVPM
*Renju *Viji
Student of B.N.V TVPM Student of B.N.V TVPM
5

6. 1
6
Unit Contents Page
No
1 Growing Shapes 07-08
2 Sum of the Angles 09-11
3 Exterior Angles 12-14
4 Unchanging Sum 15-16
5 Regular Polygons 17

7. GROWING SHAPES .
Polygons are one of the most all encompassing shapes in
geometry. It includes Squares, Rectangles, and Trapeziods to Dodecagons
and beyond. A triangle has three sides and three angles; a quadrilateral
has four.
A Pentagon is a figure of five sides and five angles, six sides and
six angles form a Hexagon.
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1

8. A Figure of Seven sides and angles is a called a Heptagon, and of Eight
sides and angles an Octagon.
There is a common name for all these together polygons.
We have seen that a quadrilateral can be split into two triangles by
drawing a diagonal.
Similarly in a Pentagon, if be choosing a vertex, skip the one near it
and joint with the next vertex, we can divide it into a quadrilateral and
triangle.
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9. How about a hexagon?
Thus we can divide any polygon into another polygon with one side less
and a triangle by drawing a line starting at any vertex, skipping one vertex
and joining with the next.
Sum of the Angles
The Sum of the angles of a triangle is 180 0 and that the
sum of the angles of a quadrilateral is 3600.
When we draw a diagonal of a quadrilateral it is split into two
triangles. The diagonal also divides into two, the angle at each of the
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2

10. vertex it passes through, and one part in one triangle and the other part in
the other triangle. Thus the angles of the quadrilateral become the angles
of these two triangles and so their sum is 2 x 1800 = 3600.
Can’t we compute the sum of the angles of a pentagon in the same
fashion? We can split it into a quadrilateral and a triangle, as we saw
earlier:
Sum of the angles of the pentagon
= Sum of the angles of the quadrilateral +
Sum of the angles of the triangle
= 3600+1800
= 5400
Now can you find like this, the sum of the angles of a hexagon, by
splitting it into a pentagon and a triangle?
Generally speaking, we can divide any polygon into a previous
polygon (meaning one side less) and triangle, and so its sum of angles
will be 180 more than sum of the angles of the previous one.
Thus we have a scheme for computing sum of the angles of a
polygon:
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11. Triangle 1800
Quadrilateral 1800 + 1800 = 2*1800
Pentagon 1800+ 2*1800 = 3*1800
Hexagon 1800+ 3*180 = 4*1800
Heptagon 1800+4*1800 = 5*1800
Octagon 1800+5*1800 = 6*1800
Nonagon 1800+ 6*1800 = 7*1800
Decagon 1800+ 7*1800 = 8*1800
…………. ………………. ……………
…………………………..…………….
How about a Twenty - sided Polygon?
Starting with the sum 180 for a triangle,
For each additional side, the sum increases by 1800.
20 sided means, 17 sides more than a triangle;
and this means the sum increases by 17*1800.
The sum of the angles of a 20 sided polygon
= 1800+ (17*1800)
= 18*1800
= 32400
We can write this as a general principle,
using a bit of algebra. To get n- sided polygon,
We start with a triangle and add n-3 sides more.
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Triangular Division
Just as we divide a
quadrilateral into two
triangles, we can
divide a pentagon into
three triangles.
What about a
hexagon?
In general, what is the
relation between the
number of sides of the
polygon and the
number of triangles we
get like this?

12. Adding each side increases the sum of angles by 180.
So the sum of the angles of n sided polygon is
(n-3)*180 more than the sum of the angles
of a triangle.
i.e., The sum of the angles is
1800 + ((n-3) *1800) = (1+n-3)) *1800
= (n-2) *1800
The sum of the angles of an n sided polygon is
(n-2) *1800
EXTERIOR ANGLES
Look at this figure:
C
A B P
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3

13. In ABC, the side AB is extended and this produces a new angle
outside the triangle. This angle CBP is called an external angle of the
triangle.
What is the relation between this angle and the (internal) angle
CBA of the triangle?
C
A B
C
CBA and CBP from a linear pair right?
So, CBP = 1800 - CBA
Now instead of AB, if we extend CB, then also
we get an exterior angle at B.
C
B
A B
Q
What is the relation between the external angle ABQ we
Now got and the external angle CBP got earlier?
C
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Another Way
We can divide
any polygon into
triangles, by joining
any one vertex to all
other vertices, except
the adjacent ones on
either side.
A
quadrilateral can be
thus divided into two
triangles, a pentagon
into four and so on.
In general, each
additional vertex
gives another
triangle.
An n-sided
polygon can be
divided thus into how
many triangles?
n-2 triangles, right?
The sum of the
angles of the polygon
is the sum of the
angles of all these

14. P
A Q
They are the opposite angles made by the
Lines .AP and CQ intersecting at B,
Aren’t they? So,
ABQ = CBP.
Thus when we speak only about the
measureres of external angles, it doesn’t matter of
these two external angles we choose.
As with B,we can draw external angles at each of the other two vertices also.
And like this, we can draw external angles at each vertex of any polygon. For
example, look at the external angles of a quadrilateral:
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triangles; that is
(n-2) * 1800.

15. Here also, an external angle and the angle of the quadrilateral at this vertex are
supplementary, right?
UNCHANGING SUM
We have learnt a trick to compute the sum of the angles of a polygon. Is there a
way to compute the sum of the external angles?
Let’s start with a triangle:
C
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Another division
We can divide into
triangles in another
fashion, by drawing
lines from a point inside
to the vertices.
An n- sided polygon
gives n triangles like
this, right? The sum of
the angles o these
triangles is n* 1800.
Among all these
angles, those other than
the angles at O add up
to the sum of the angles
of the polygon; and the
angles at O add up to
3600 , as we have
already seen. Thus the
sum of the angles of the
polygon is
(n*1800) – (2*1800) =
(n-2) *1800
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16. A B
The sum of the external angle at A and the angle of the
triangle itself at A, is 1800, right?
The same thing happens at the vertices B and C also.
So, the sum of these three pairs of external and internal
angles is 3* 1800 = 5400. That is, the sum of the three
external angles and the three angles of the triangle is
5400.
In this, the sum of the angles of the triangle is 1800.
From this, we see that the sum of the external angles
only is 5400 _ 1800 = 3600.
What about a quadrilateral, instead of a triangle?
Here, at each of the four vertices, there is a linear pair formed by an external
angle and an angle of the quadrilateral. The sum of the angles in each such pair
is 1800. So, the four external angles and the four angles of the quadrilateral
together make 4* 1800 = 7200.
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17. In this, the sum of the four angles of the quadrilateral is 3600.
So, the sum of the four external angles is
7200 – 3600 = 3600
Similarily, if we take one external angle from each vertex of a pentagon and
add up, what do we get? Try it!
Let’s think about an n-sided polygon in general. There are n vertices in all; and
at each vertex, a linear pair formed by an external angle and an angle of the
polygon itself. So, the sum of all these angles n*1800. In this, the sum of the
angles of the polygon is (n-2)*1800, as seen earlier.
So,
the sum of the external angles = n*1800 – ((n-2) *1800 )
= 2*1800
= 3600
Thus in any polygon (whatever be the number of sides), the sum of the external
angles, one at each vertex, is 3600. How do we state this in short?
The sum of the external angles of any polygon is
3600.
REGULAR
POLYGONS.
If all angles of a triangle are equal, how much is each angle?
Since the angles are equal, the sides of this triangle must also be equal.
On the other hand, what if the sides of a triangle are all equal? Then its angles
are also equal.
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Now if the angles of a quadrilateral are all equal, is it necessary that its sides
are also equal?
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18. In any rectangle, the angles are all equal; but the sides may not be equal. If the
sides are also equal, it becomes a square.
Polygons like this, with equal angles and lengths of sides also equal are called
Regular Polygons.
The figures below show a regular pentagon and a regular hexagon:
How much is each angle of a regular pentagon?
The sum of the angles is 3* 1800 = 5400;
And since it is regular, this is the sum of
Five equal angles.So, each angle
is 1/5*5400 = 1080.
Similarly, we can easily see that each angle of a regular hexagon is 1/6*4*1800 = 1200.
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