#### 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

#### notes

Two triangles ABC and PBC on the same base BC and between the same parallels BC and AP. in following fig.

Draw CD || BA and CR || BP such that D and R lie on line AP in following fig.

From this, you obtain two parallelograms PBCR and ABCD on the same base BC and between the same parallels BC and AR.

Therefore, ar (ABCD) = ar (PBCR)

Now ∆ ABC ≅ ∆ CDA and ∆ PBC ≅ ∆ CRP

So, ar (ABC) = `1/2` ar (ABCD) and ar (PBC) = `1/2` ar (PBCR)

Therefore, ar (ABC) = ar (PBC)

#### theorem

**Theorem :** Two triangles on the same base (or equal bases) and between the same parallels are equal in area.**Proof:** Now, suppose ABCD is a parallelogram whose one of the diagonals is AC (see Fig.).

Let AN ⊥ DC. Note that

∆ ADC ≅ ∆ CBA

So, ar (ADC) = ar (CBA)

Therefore, ar (ADC) = `1/2` ar (ABCD)

=`1/2` (DC × AN)

So, area of ∆ ADC = `1/2` × base DC × corresponding altitude AN

In other words, area of a triangle is half the product of its base (or any side) and the corresponding altitude (or height).

The two triangles with same base (or equal bases) and equal areas will have equal corresponding altitudes.