MATHEMATICS DIGITAL TEXTBOOK

STANDARD : IX

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BINCY ELSA BABY

B .Ed. MATHEMATICS

Reg. No : 18014350006

CONTENT

REAL NUMBERS ………………………….. 3 - 6

Number line Distance Absolute value

PRISMS ………………………………………. 7 - 12

Lateral surface area of prism Surface area of prism Volume of prism Curved surface area of

cylinder Surface area of cylinder Volume of cylinder

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CHAPTER – 1

REAL NUMBERS

REAL NUMBERS

All rational and irrational numbers needed to represent all lengths, together with its negatives and zero are collectively called Real numbers.

Example: 2 , 5/3 ,√2 , ∏

NUMBER LINE

All real numbers represented as points on a line is called a Number line.

. . . . . . . . . . . . . .

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

Numbers become small when we move to left and numbers become large when we move to right from zero.

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If we take two real numbers as x and y , where x is left from y and y is right from x , then x < y .

The position of x such that x > 2 on the number line is,

. . . . . . . . . . . .

-5 -4 -3 -2 -1 0 1 2 3 4 5 6

The position of x such that -1 < x < 3 on the number line is,

. . . . . . . . . . .

-6 -5 -4 -3 -2 -1 0 1 2 3 4

DISTANCE

Distance between zero and any positive number on a number line is that number itself. If the number is negative, we got the distance by removing the negative sign.

Example: Distance between 0 and 5 is 5

Distance between 0 and -5 is also 5

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Distance between any two numbers on the number line is the difference between the smaller from the larger.

Example: Distance between 2 and 5 is 5-2=3

ABSOLUTE VALUE

Distance between zero and a number on the number line is the value of the number disregarding its sign .This value is called the Absolute value of the number.

Absolute value of x can be written as │x│

Example: │4│=4

│-4│=4

If we consider x as positive number, │x│=x

If we consider x as negative number, │x│=-x

If we consider x as 0, │x│=0

That is,

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x , if x > 0

│x│ = x , if x < 0

x , if x = 0

Distance between the numbers x and y on the number line = │x - y│

Example: Distance between 2 and 5,

│x - y│ = │2 - 5│ = │-3│= 3

Distance between -3 and 6,

│x - y│ = │-3 - 6│ = │-9│ =9

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CHAPTER – 2

PRISMS

Triangular prism Rectangular prism

LATERAL SURFACE AREA OF A PRISM

The lateral surface area of any prism is equal to the product of the base perimeter and height.

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Example: The base of a closed triangular prism has sides 13 , 14 and 15 centimeters long and its height is 20 centimeter. What is its lateral surface area?

Lateral surface area = base perimeter x height

= (13+14+15) x 20

= 840 cm²

SURFACE AREA OF A PRISM

Surface area of any prism = lateral surface area

+ twice base area

Example: A solid triangular prism with equilateral base is 5 centimeters high and its base perimeter is 12 centimeters. Find its surface area?

Surface area = lateral surface area + 2xbase area

Lateral surface area = base perimeter x height

= 12 x 5

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= 60 cm²

Base area = √3a²/4

a = 12/3 = 4

Base area = √3x4x4/4 = 4√3 cm²

Surface area = 60 + 4√3x2

= 60 + 8√3 cm²

VOLUME OF A PRISM

Volume of any prism is the product of its base area and height.

Example: The base of a prism is an equilateral triangle of perimeter 15 cm and its height is 5 cm. What is its volume?

Volume of a prism = base area x height

Base area = √3a²/4

= √3x5x5/4 = 10.8 cm²

Volume = 10.8 x 5 = 54.06 cm³

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CYLINDERS

CURVED SURFACE AREA OF A CYLINDER

Curved surface area of a cylinder is the product of the base perimeter and height.

Example :A cylindrical vessel is 15 cm high and its base radius is 5 cm. Find the curved surface area?

Curved surface area = base perimeter x height

= 2∏r x 5

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= 10∏ x 15

= 150∏ cm²

SURFACE AREA OF A CYLINDER

Surface area of a cylinder = curved surface area

+ twice base area

Example: A cylindrical vessel is 10 cm high and its base radius is 7 cm. Find the surface area of the vessel?

Surface area of a cylinder = curved surface area

+ 2 x base area

Curved surface area = 150∏ cm²

Base area = 49∏ cm²

Surface area = 150∏ + 2 x 49∏

= 248∏ cm³

VOLUME OF A CYLINDER

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Volume of a cylinder is the product of its base area and height.

Example: A cylindrical vessel is 15 cm high and its base radius is 10 cm. Find its volume?

Volume = base area x height

Base area = ∏ x 10²

= 100∏ cm²

Volume = 100∏ x 15

= 1500∏ cm³

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