#### 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:** Parallelograms on the same base and between the same parallels are equal in area.**Proof :** Two parallelograms ABCD and EFCD, on the same base DC and between the same parallels AF and DC are given in following fig.

We need to prove that ar (ABCD) = ar (EFCD).

In ∆ ADE and ∆ BCF,

∠ DAE = ∠ CBF (Corresponding angles from AD || BC and transversal AF) (1)

∠ AED = ∠ BFC (Corresponding angles from ED || FC and transversal AF) (2)

Therefore, ∠ ADE = ∠ BCF (Angle sum property of a triangle) (3)

Also, AD = BC (Opposite sides of the parallelogram ABCD) (4)

So, ∆ ADE ≅ ∆ BCF [By ASA rule, using (1), (3), and (4)]

Therefore, ar (ADE) = ar (BCF) (Congruent figures have equal areas) (5)

Now, ar (ABCD) = ar (ADE) + ar (EDCB)

= ar (BCF) + ar (EDCB) [From(5)]

= ar (EFCD)

So, parallelograms ABCD and EFCD are equal in area.