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MATHEMATICS
Standard
IX
1
THE NATIONAL ANTHEMJana-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!
PLEDGE
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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: [email protected] Phone: 9497011959
Dear Children,
One more step in the journey towards knowledge.
A book to learn math and learn it right
3
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
Textbook Development Committee
Mathematics IX
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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
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1
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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
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.
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.
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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.
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.
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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
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 +
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2
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:
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.
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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?
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.
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 11
3
Look at this figure:
C
A B P
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
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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
triangles; that is (n-2) * 1800.
B
A B
Q
What is the relation between the external angle ABQ we
Now got and the external angle CBP got earlier?
C
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.
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And like this, we can draw external angles at each vertex of any polygon. For example, look at the external angles of a quadrilateral:
Here also, an external angle and the angle of the quadrilateral at this vertex are supplementary, right?
UNCHANGING SUM
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4
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
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?
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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
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.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?
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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.
Now if the angles of a quadrilateral are all equal, is it necessary that its sides are also equal?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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