Topics
Number System
Rational Numbers
- Rational Numbers
- Addition of Rational Number
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- Multiplication of Rational Numbers
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- Inserting Rational Numbers Between Two Given Rational Numbers
- Method of Finding a Large Number of Rational Numbers Between Two Given Rational Numbers
Exponents
- Concept of Exponents
- Law of Exponents (For Integral Powers)
- Negative Integral Exponents
- More About Exponents
Squares and Square Root
- Concept of Square Roots
- Finding Square Root by Division Method
- Finding Square Root Through Prime Factorisation
- To Find the Square Root of a Number Which is Not a Perfect Square (Using Division Method)
- Properties of Square Numbers
Cubes and Cube Roots
Playing with Numbers
- Arranging the Objects in Rows and Columns
- Generalized Form of Numbers
- Some Interesting Properties
- Letters for Digits (Cryptarithms)
- Divisibility by 10
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- Divisibility by 11
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Sets
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- Distributive Laws
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Ratio and Proportion
Percent and Percentage
Profit, Loss and Discount
- Concept of Discount
- Overhead Expenses
- To Find S.P., When C.P. and Gain (Or Loss) Percent Are Given
- To Find C.P., When S.P. and Gain (Or Loss) Percent Are Given
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Interest
- Calculation of Interest
- Concept of Compound Interest
- To Find the Principle (P); the Rate Percent (R) and the Time
- Interest Compounded Half Yearly
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Direct and Inverse Variations
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- Concept for Unitary Method (With Only Direct Variation Implied)
- Concept of Arrow Method
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Algebra
Algebraic Expressions
- Algebraic Expressions
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- Dividing a Polynomial by a Monomial
- Dividing a Polynomial by a Polynomial
- Simplification of Expressions
Identities
- Algebraic Identities
- Product of Sum and Difference of Two Terms
- Expansion Form of Numbers
- Important Formula of Expansion
- Cubes of Binomials
- Application of Formulae
Factorisation
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- Factorisation by Taking Out Common Factors
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- Factorisation by Difference of Two Squares
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- Factorising a Perfect Square Trinomial
- Factorising Completely
Linear Equations in One Variable
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Linear Inequations
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- Operation of Whole Numbers on Number Line
- Concept of Properties
Geometry
Understanding Shapes
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- Concept of Polygons
- Sum of Angles of a Polynomial
- Sum of Exterior Angles of a Polynomial
- Regular Polynomial
- Concept of Quadrilaterals
Special Types of Quadrilaterals
Constructions
- Introduction of Constructions
- Construction of an Angle
- To Construct an Angle Equal to Given Angle
- To Draw the Bisector of a Given Angle
- Construction of an angle bisector of an Angles of 60°,30°,90° and 45°
- Construction of Bisector of a Line
- The Perpendicular Bisector
- Construction of Parallel Lines
- Constructing a Quadrilateral
- Construction of Parallelograms
- Construction of a Rectangle When Its Length and Breadth Are Given.
- Construction of Rhombus
- Construction of Square
- Concept of Reflection Symmetry
Representing 3-D in 2-D
- 2dimensional Perspective of 3dimensional Objects
- Concept of Polyhedron
- Faces, Edges and Vertices
- Euler's Formula
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- Nets for Building 3-d Shapes
Mensuration
Area of a Trapezium and a Polygon
Surface Area, Volume and Capacity
Data Handling (Statistics)
Data Handling
- Concept of Data Handling
- Collecting Data
- Frequency
- Raw Data, Arrayed Data and Frequency Distribution
- Constructing a Frequency Table
- Frequency Distribution Table
- Class Intervals and Class Limits
- Graphical Representation of Data
Probability
- Introduction
- The Perpendicular Bisector of a Line Segment
Notes
The perpendicular bisector of a line segment:
Fold a sheet of paper. Let `bar"AB"` be the fold. Place an ink-dot X, as shown, anywhere.
Find the image X' of X, with AB as the mirror line.
Let `bar"AB"` and `bar"XX’"` intersect at O.
This means that `bar"AB" "divides" bar"XX’"` into two parts of equal length.
`bar"AB" "bisects" bar"XX’" or bar"AB" "is a bisector of" bar"XX’"`.
Hence, `bar"AB" "is the perpendicular bisector of" bar"XX’"`.
Construction using ruler and compasses:
Step 1: Draw a line segment `bar"AB"` of any length.
Step 2: With A as the centre, using compasses, draw a circle. The radius of your circle should be more than half the length of `bar"AB"`.
Step 3: With the same radius and with B as centre, draw another circle using compasses. Let it cut the previous circle at C and D.
Step 4: Join `bar"CD". "It cuts" bar"AB"` at O.
Use your divider to verify that O is the midpoint of `bar"AB"`.
Therefore, `bar"CD" "is the perpendicular bisector of" bar"AB".`
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