#### Topics

##### Number Systems

##### Number Systems

##### Algebra

##### Polynomials

##### Linear Equations in Two Variables

##### Algebraic Expressions

##### Algebraic Identities

##### Coordinate Geometry

##### Geometry

##### Introduction to Euclid’S Geometry

##### Lines and Angles

##### Triangles

##### Quadrilaterals

- Concept of Quadrilaterals - Sides, Adjacent Sides, Opposite Sides, Angle, Adjacent Angles and Opposite Angles
- Angle Sum Property of a Quadrilateral
- Types of Quadrilaterals
- Theorem: A Diagonal of a Parallelogram Divides It into Two Congruent Triangles.
- Another Condition for a Quadrilateral to Be a Parallelogram
- The Mid-point Theorem
- Theorem: A Diagonal of a Parallelogram Divides It into Two Congruent Triangles.
- Property: The Opposite Sides of a Parallelogram Are of Equal Length.
- 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

##### Area

##### Circles

- Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles
- Angle Subtended by a Chord at a Point
- Perpendicular from the Centre to a Chord
- Circles Passing Through One, Two, Three Points
- Equal Chords and Their Distances from the Centre
- Angle Subtended by an Arc of a Circle
- Cyclic Quadrilateral

##### Constructions

##### Mensuration

##### Areas - Heron’S Formula

##### Surface Areas and Volumes

##### Statistics and Probability

##### Statistics

##### Probability

#### theorem

**Theorem:** Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).

Consider a circle of centre O . It has PQ and RS two chords of equal length.

Now , draw a perpendicular from centre to PQ and RS intersecting them at L and M .**To Prove:** OL = OM

Join OQ and OS.**Proof:** We know that PQ =RS (given)

QL = `(PQ )/2` (perpendicular drawn from a centre to a chord bisect the chord)

Similarly, SM =`(RS)/2`

Therefore , QL = SM

Consider ∆OQL and ∆OSM

OQ = OS (radius of the same circle)

QL = SM (proved)

∠OLQ = ∠OMS = 90° (given )

∆OQL ≅ ∆OSM (SSA rule)

OL = OM (By CPCT).

**Theorem:** Chords equidistant from the centre of a circle are equal in length.

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#### Shaalaa.com | Theorem: Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).

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