Topics
Number Systems
Real Numbers
Algebra
Polynomials
Pair of Linear Equations in Two Variables
- Introduction to linear equations in two variables
- Graphical Method
- 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
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
Arithmetic Progressions
Coordinate Geometry
Lines (In Two-dimensions)
Constructions
- Division of a Line Segment
- Construction of Tangents to a Circle
- Constructions Examples and Solutions
Geometry
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
- Concept of Angle Bisector
- Similarity of Triangles
- Ratio of Sides of Triangle
Circles
Trigonometry
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
Trigonometric Identities
Some Applications of Trigonometry
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
- 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
Statistics and Probability
Statistics
Probability
Internal Assessment
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`
