#### Topics

##### Number Systems

##### Real Numbers

##### Algebra

##### Polynomials

##### 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

##### 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
- Angle Bisector
- Similarity of Triangles
- Ratio of Sides of Triangle

##### 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

##### 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

We have already learnt about the surface area and volumes of figures, but in this chapter we will given different combinations of solids. We are known with solids like cylinder, sphere, cone, cuboid, rectangle, square, etc. In this chapter we are suppose to find surface areas of different combinations of solids.

Now before we proceed, we must understand what is curved surface area. The solids having curved surface are cone, cylinder. Curved surface area is the outer portion, the area of flat ends of such solids will not be included while calculating Curved surface area. But while calculating Total surface area we must include the area of flat ends of the solids, provided that the solid should not be hollow from inside.

Example- 2 cubes each of volume `64 cm^3` are joined end to end. Find the surface area of the resulting cuboid.

Solution- `Volume of cube= 64cm^3`

`"edge"^3= 64cm^3`

`"edge of cube"` = `3 sqrt 64` cm

edge= 4 cm

By adding this two cube we get a cuboid

So we get dimensions of cuboid as

As we have two cube the lenght will be doubled

therefore, lenght = 2(4)= 8cm

breadth= 4cm

height= 4cm

Surface area of cuboid= 2 (lb+bh+hl)

= `2 (8 xx 4)+(4 xx 4)+(4 xx 8

)`

=` 2 (32+16+32)`

= `2 (80)`

Surface area of cuboid= `160 cm^2`

#### Video Tutorials

#### Shaalaa.com | Surface Area and Volume part 2 (Surface Area)

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