#### Topics

##### Number Systems

##### Real Numbers

##### Algebra

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

- Linear Equations 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
- Relationship Between Discriminant and Nature of Roots
- Situational Problems Based on Quadratic Equations Related to Day to Day Activities to Be Incorporated
- Quadratic Equations Examples and Solutions

##### 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 Or Thales Theorem
- Criteria for Similarity of Triangles
- Areas of Similar Triangles
- Right-angled Triangles and Pythagoras Property
- Similarity Triangle Theorem
- Application of Pythagoras Theorem in Acute Angle and Obtuse Angle
- Triangles Examples and Solutions
- Angle Bisector
- Similarity
- Ratio of Sides of Triangle

##### Constructions

##### Trigonometry

##### Heights and Distances

##### Trigonometric Identities

##### Introduction to Trigonometry

##### Statistics and Probability

##### Probability

##### Statistics

##### Coordinate Geometry

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

##### Mensuration

##### Areas Related to Circles

##### Surface Areas and Volumes

#### notes

In the previous section, we have discussed how to find the surface area of solids made up of a combination of two basic solids. Here, we shall see how to calculate their volumes.

Volume of combination can be found out by adding volumes of different solids or by subtracting volumes of different solids.

Example 1- Shanta runs an industry in a shed which is in the shape of a cuboid surmounted by a half cylinder. If the base of the shed is of dimension 7 m × 15 m, and the height of the cuboidal portion is 8 m, find the volume of air that the shed can hold. Further, suppose the machinery in the shed occupies a total space of `300 m^3`, and there are 20 workers, each of whom occupy about `0.08 m^3` space on an average. Then, how much air is in the shed? `(Take pi= 22/7)`

Solution- Volume of air in shed= Volume of shed- Space occupied by workers and machinery

Volume of shed= Volume of cuboid+ Volume of half cylinder

`= (8 xx 7 xx 15)+ 1/2 pi r^2h`

= `(8 xx 7 xx 15)+ 1/2 xx 22/7 xx (7/2)^2 xx 15`

Volume of shed =`1128.75 cm^3`

Volume of air in shed= Volume of shed- Space occupied by workers and machinery

`= 1128.75- 300+ (20 xx 0.08)`

Volume of air in shed= `827.15 m^3`

Example 2- A juice seller was serving his customers using glasses. The inner diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 10 cm, find the apparent capacity of the glass and its actual capacity. (Use π = 3.14.)

Solution: Capacity of the glass= Volume of glass- Volume of hemisphere

= `pi r^2h -2/3 pi r^3`

= `[3.14 (2.5)^2 xx 10] -[2/3 xx 3.14 xx (2.5)^3]`

`"Capacity of the glass"= 163.54 cm^3`