#### Topics

##### Number Systems

##### Real Numbers

##### Algebra

##### Pair of Linear Equations in Two Variables

- Linear Equation in Two Variables
- Graphical Method of Solution of a Pair of Linear Equations
- Substitution Method
- Elimination Method
- Cross - Multiplication Method
- Equations Reducible to a Pair of Linear Equations in Two Variables
- Consistency of Pair of Linear Equations
- Inconsistency of Pair of Linear Equations
- Algebraic Conditions for Number of Solutions
- Simple Situational Problems
- Pair of Linear Equations in Two Variables
- Relation Between Co-efficient

##### Arithmetic Progressions

##### Quadratic Equations

- Quadratic Equations
- Solutions of Quadratic Equations by Factorization
- Solutions of Quadratic Equations by Completing the Square
- Nature of Roots of a Quadratic Equation
- Relationship Between Discriminant and Nature of Roots
- Situational Problems Based on Quadratic Equations Related to Day to Day Activities to Be Incorporated
- Application of Quadratic Equation

##### Polynomials

##### Geometry

##### Circles

- Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles
- Tangent to a Circle
- Number of Tangents from a Point on a Circle
- Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles

##### Triangles

- Similar Figures
- Similarity of Triangles
- Basic Proportionality Theorem (Thales Theorem)
- Criteria for Similarity of Triangles
- Areas of Similar Triangles
- Right-angled Triangles and Pythagoras Property
- Similarity of Triangles
- Application of Pythagoras Theorem in Acute Angle and Obtuse Angle
- Triangles Examples and Solutions
- Angle Bisector
- Similarity of Triangles
- Ratio of Sides of Triangle

##### Constructions

- Division of a Line Segment
- Construction of Tangents to a Circle
- Constructions Examples and Solutions

##### Trigonometry

##### Heights and Distances

##### Trigonometric Identities

##### Introduction to Trigonometry

- Trigonometry
- Trigonometry
- Trigonometric Ratios
- Trigonometric Ratios and Its Reciprocal
- Trigonometric Ratios of Some Special Angles
- Trigonometric Ratios of Complementary Angles
- Trigonometric Identities
- Proof of Existence
- Relationships Between the Ratios

##### Statistics and Probability

##### Probability

##### Statistics

##### Coordinate Geometry

##### Lines (In Two-dimensions)

##### Mensuration

##### Areas Related to Circles

- Perimeter and Area of a Circle - A Review
- Areas of Sector and Segment of a Circle
- Areas of Combinations of Plane Figures
- Circumference of a Circle
- Area of Circle

##### Surface Areas and Volumes

- Concept of Surface Area, Volume, and Capacity
- Surface Area of a Combination of Solids
- Volume of a Combination of Solids
- Conversion of Solid from One Shape to Another
- Frustum of a Cone
- Concept of Surface Area, Volume, and Capacity
- Surface Area and Volume of Different Combination of Solid Figures
- Surface Area and Volume of Three Dimensional Figures

##### Internal Assessment

**Theorem**: When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides.

## Theorem

Theorem1: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given: ∆ABC ∼ ∆PQR

Construction: Draw AM ⊥ BC and PN ⊥ QR

To prove: `"ar∆ABC"/"ar∆PQR" = ("AB"/"PQ")^2= ("BC"/"QR")^2= ("AC"/"PR")^2`

Proof: ∆ABC ∼ ∆PQR (Given)

`"AB"/"PQ"= "BC"/"QR" = "AC"/"PR"` (Corresponding parts of Congruent triangles) ...eq1

and ∠B= ∠Q (Corresponding parts of Congruent triangles)

Now, `"ar(∆ABC)"= 1/2 × "BC" × "AM"`

`"ar(∆PQR)"= 1/2 × "QR" × "PN"`

`"ar(∆ABC)"/"ar(∆PQR)"= (1/2 xx "BC" xx "AM")/(1/2 xx "QR" xx "PN")`

`"ar(∆ABC)"/"ar(∆PQR)"`= `"BC"/"QR"×"AM"/"PN"` .......eq2

Again, In ∆ABM and ∆PQN

∠AMB = ∠PNQ = 90°

∠B= ∠Q (Corresponding parts of Congruent triangles)

∆ABM ∼ ∆PQN

So, `"AB"/"PQ"= "AM"/"PN"` (Corresponding parts of Congruent triangles) .....eq3

From eq2 and eq3

`"ar(∆ABC)"/"ar(∆PQR)"= "BC"/"QR"×"AB"/"PQ"` ......eq4

From eq1 and eq4

`"ar(∆ABC)"/"ar(∆PQR)"= "BC"/"QR" × "BC"/"QR"`

`"ar(∆ABC)"/"ar(∆PQR)"= ("BC"/"QR")^2` .........eq5

From eq1 and eq5

`"ar(∆ABC)"/"ar(∆PQR)"= ("AB"/"PQ")^2 = ("BC"/"QR")^2 = ("AC"/"PR")^2`

Hence proved.