MATHEMATICS

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IX

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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: sajujayan1980@gmail.com Phone: 9497011959

Dear Children,

One more step in the journey towards knowledge.

A book to learn math and learn it right

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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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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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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

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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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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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