Topics
Number Systems
Number Systems
Polynomials
Algebra
Coordinate Geometry
Linear Equations in Two Variables
Geometry
Coordinate Geometry
Introduction to Euclid’S Geometry
Mensuration
Statistics and Probability
Lines and Angles
- Introduction to Lines and Angles
- Basic Terms and Definitions
- Intersecting Lines and Non-intersecting Lines
- Introduction to Parallel Lines
- Pairs of Angles
- Parallel Lines and a Transversal
- Angle Sum Property of a Triangle
Triangles
- Concept of Triangles
- Congruence of Triangles
- Criteria for Congruence of Triangles
- Properties of a Triangle
- Some More Criteria for Congruence of Triangles
- Inequalities in a Triangle
Quadrilaterals
- Concept of Quadrilaterals
- Properties of a Quadrilateral
- Types of Quadrilaterals
- Another Condition for a Quadrilateral to Be a Parallelogram
- Theorem of Midpoints of Two Sides of a Triangle
- Property: The Opposite Sides of a Parallelogram Are of Equal Length.
- Theorem: A Diagonal of a Parallelogram Divides It into Two Congruent Triangles.
- Theorem : If Each Pair of Opposite Sides of a Quadrilateral is Equal, Then It is a Parallelogram.
- Property: The Opposite Angles of a Parallelogram Are of Equal Measure.
- Theorem: If in a Quadrilateral, Each Pair of Opposite Angles is Equal, Then It is a Parallelogram.
- Property: The diagonals of a parallelogram bisect each other. (at the point of their intersection)
- Theorem : If the Diagonals of a Quadrilateral Bisect Each Other, Then It is a Parallelogram
Circles
Areas - Heron’S Formula
Surface Areas and Volumes
Statistics
Algebraic Expressions
Algebraic Identities
Area
Constructions
- Introduction of Constructions
- Basic Constructions
- Some Constructions of Triangles
Probability
Theorem
Theorem : The perpendicular from the centre of a circle to a chord bisects the chord.
Draw a circle. Let O be its centre. Draw a chord AB. Let the crease cut AB at the point M. Then, ∠ OMA = ∠ OMB = 90° or OM is perpendicular to AB. So MA = MB.
Theorem : The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
Let AB be a chord of a circle with centre O and O is joined to the mid -point M of AB.
To prove: OM⊥ AB.
Proof: Join OA and OB
In triangles OAM and OBM,
OA =OB (radii of the same circle)
AM =BM (M is the midpoint of PQ)
OM =OM (common)
Therefore, ∆OAM ≅ ∆OBM (SSS congruence rule)
This gives , ∠OMA = ∠OMB = 90°
(Corresponding parts of congruent triangles are congruent).
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