#### Topics

##### Number Systems

##### Algebra

##### Geometry

##### Trigonometry

##### Statistics and Probability

##### Coordinate Geometry

##### Mensuration

##### Internal Assessment

##### Real Numbers

##### 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

##### 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

##### Heights and Distances

##### Trigonometric Identities

##### Introduction to Trigonometry

##### Probability

##### Statistics

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

##### Areas Related to Circles

##### Surface Areas and Volumes

#### definition

A frustum may be formed from a right circular cone by cutting off the tip of the cone with a cut perpendicular to the height, forming a lower base and an upper base that are circular and parallel.

#### notes

In ΔAFE and ΔAGC

ΔAFE ≅ ΔAGC (Angle-Angle rule)

`"AF"/"AG"= "FE"/"GC"= "AE"/"AC"`

`(h_1-h)/(h_1)= r_2/r_1= (l_1-l)/l_1 `

`(r_2)/(r_1)= (l_1-l)/(l_1)`

`r_2l_1= r_1(l_1-l)`

`r_2l_1= r_1l_1- r_1l`

`r_1l= r_1l_1- r_2l_1`

`r_1l= l_1 (r_1-r_2)`

`l_1= (r_1l)/ (r_1-r_2)` ...........eq1

`(h_1-h)/h_1= r_2/r_1= (l_1-l)/l_1`

`(h_1-h)/h_1= r_2/r_1`

`r_1(h_1-h)= h_1r_2`

`r_1h_1- r_1h= h_1r_2`

`r_1h_1- r_2h_1= r_1h`

`h_1(r_1-r_2)= r_1h`

`h_1= (r_1h)/(r_1-r_2)` ............eq2

1) Curved surface area of frustum= Curved surface area of ABC- Curved surface area of ADE

= `pi r_1l_1- pi r_2 (l_1-l)`

= `pi r_1l_1- pi r_2l_1 + pi r_2l`

= `pi l_1 (r_1- r_2) + pi r_2l`

= `pi r_1/(r_1-r_2) (r_1-r_2) +pi r_2l` (from eq1)

= `pi r_1l + pi r_2l`

Curved surface area of frustum = `pi l(r_1+r_2)`

2) Total surface area of frustum= Curved surface area of frustum + area of the circle above+ area of the circle below

Total surface area of frustum = `pi l (r_1+r_2) + pi r_1^2+ pi r_2^2`

3) Volume of frustum= Volume of cone ABC- Volume of cone ADE

= `1/3 pi r_1^2h_1- 1/3 pi r_2^2 (h_1-h)`

=`1/3 pi r_1^2h_1- 1/3 pi r_2^2h_1+ 1/3 pi r_2^2h`

= `1/3 pi h_1 (r_1^2- r_2^2) +1/3 pi r_2^2h`

= `1/3 pi (r_1h)/(r_1-r_2) (r_1^2-r_2^2)+ 1/3 pi r_2^2h` (from eq2)

= `1/3 pi (r_1h)/(r_1-r_2) (r_1-r_2) (r_1-r_2)+ 1/3 pi r_2^2h`

= `1/3 pi r_1h (r_1+r_2)+ 1/3 pi r_2^2h`

= `1/3 pi h [r_1 (r_1+r_2)+ r_2^2]`

Volume of frustum= `1/3 pi h (r_1^2+ r_2^2+ r_1r_2)`

Example- Hanumappa and his wife Gangamma are busy making jaggery out of sugarcane juice. They have processed the sugarcane juice to make the molasses, which is poured into moulds in the shape of a frustum of a cone having the diameters of its two circular faces as 30 cm and 35 cm and the vertical height of the mould is 14 cm. If each `cm^3` of molasses has mass about 1.2 g, find the mass of the molasses that can be poured into each mould. `(Take pi= 22/7)`

Solution : Since the mould is in the shape of a frustum of a cone, the quantity (volume) of molasses that can be poured into it = `pi/3 h(r_1^2+ r_2^2+ r_1r_2)`

where `r_1` is the radius of the larger base and `r_2` is the radius of the smaller base

=`1/3 xx 22/7 xx 14 [(35/2)^2+ (30/2)^2= (35/2 xx 30/2)] cm^3`

It is given that `1 cm^3` of molasses has mass `1.2g.` So, the mass of the molasses that can be poured into each mould = `(11641.7 xx 1.2) g`

= `13970.04 g = 13.97 kg = 14 kg ` (approx.)