Topics
Integers
- Natural Numbers
- Whole Numbers
- Negative and Positive Numbers
- Integers
- Representation of Integers on the Number Line
- Ordering of Integers
- Addition of Integers
- Subtraction of Integers
- Properties of Addition and Subtraction of Integers
- Multiplication of a Positive and a Negative Integers
- Multiplication of Two Negative Integers
- Product of Three Or More Negative Integers
- Closure Property of Multiplication of Integers
- Commutative Property of Multiplication of Integers
- Multiplication of Integers with Zero
- Multiplicative Identity of Integers
- Associative Property of Multiplication of Integers
- Distributive Property of Multiplication of Integers
- Making Multiplication Easier of Integers
- Division of Integers
- Properties of Division of Integers
Fractions and Decimals
- Concept of Fraction
- Types of Fractions
- Concept of Proper and Improper Fractions
- Concept of Mixed Fractions
- Concept of Equivalent Fractions
- Like and Unlike Fraction
- Comparing Fractions
- Addition of Fraction
- Subtraction of Fraction
- Multiplication of a Fraction by a Whole Number
- Using Operator 'Of' with Multiplication and Division
- Multiplication of Fraction
- Division of Fractions
- Concept of Reciprocals or Multiplicative Inverses
- Problems Based on Fraction
- The Decimal Number System
- Comparing Decimal Numbers
- Addition of Decimal Fraction
- Subtraction of Decimal Numbers
- Multiplication of Decimal Numbers
- Division of Decimal Numbers
- Problems Based on Decimal Numbers
Data Handling
Simple Equations
Lines and Angles
The Triangle and Its Properties
- Basic Concepts of Triangles
- Classification of Triangles based on Sides
- Classification of Triangles based on Angles
- Median of a Triangle
- Altitudes of a Triangle
- Exterior Angle of a Triangle and Its Property
- Some Special Types of Triangles - Equilateral and Isosceles Triangles
- Basic Properties of a Triangle
- Right-angled Triangles and Pythagoras Property
Comparing Quantities
- Ratio
- Concept of Equivalent Ratios
- Proportion
- Unitary Method
- Basic Concept of Percentage
- Estimation in Percentages
- Interpreting Percentages
- Conversion between Percentage and Fraction or Decimal
- Ratios to Percents
- Increase Or Decrease as Percent
- Basic Concepts of Profit and Loss
- Profit or Loss as a Percentage
- Calculation of Interest
Congruence of Triangles
- Similarity and Congruency of Figures
- Congruence Among Line Segments
- Congruence of Angles
- Congruence of Triangles
- Criteria for Congruence of Triangles
- Criteria for Similarity of Triangles
- SAS Congruence Criterion
- ASA Congruence Criterion
- RHS Congruence Criterion
- Exceptional Criteria for Congruence of Triangles
Rational Numbers
- Rational Numbers
- Equivalent Rational Number
- Positive and Negative Rational Numbers
- Rational Numbers on a Number Line
- Rational Numbers in Standard Form
- Comparison of Rational Numbers
- Rational Numbers Between Two Rational Numbers
- Addition of Rational Number
- Subtraction of Rational Number
- Multiplication of Rational Numbers
- Division of Rational Numbers
Perimeter and Area
- Basic Concepts in Mensuration
- Concept of Perimeter
- Perimeter of a Rectangle
- Perimeter of Squares
- Perimeter of Triangle
- Perimeter of Polygon
- Concept of Area
- Area of Square
- Area of Rectangle
- Triangles as Parts of Rectangles and Square
- Generalising for Other Congruent Parts of Rectangles
- Area of a Parallelogram
- Area of a Triangle
- Circumference of a Circle
- Area of Circle
- Conversion of Units
- Problems based on Perimeter
- Problems based on Area
Practical Geometry
- Construction of a Line Parallel to a Given Line, Through a Point Not on the Line
- Construction of Triangles
- Constructing a Triangle When the Length of Its Three Sides Are Known (SSS Criterion)
- Constructing a Triangle When the Lengths of Two Sides and the Measure of the Angle Between Them Are Known. (SAS Criterion)
- Constructing a Triangle When the Measures of Two of Its Angles and the Length of the Side Included Between Them is Given. (ASA Criterion)
- Constructing a Right-angled Triangle When the Length of One Leg and Its Hypotenuse Are Given (RHS Criterion)
Algebraic Expressions
Exponents and Powers
- Concept of Exponents
- Multiplying Powers with the Same Base
- Dividing Powers with the Same Base
- Taking Power of a Power
- Multiplying Powers with Different Base and Same Exponents
- Dividing Powers with Different Base and Same Exponents
- Numbers with Exponent Zero, One, Negative Exponents
- Miscellaneous Examples Using the Laws of Exponents
- Decimal Number System Using Exponents and Powers
- Crores
Symmetry
Visualizing Solid Shapes
- Features and Formulas
- Example 1
- Example 2
- Example 3
- Example 4
- Example 5
- Key Points Summary
Features and Formulas
| Feature | Perimeter | Area |
|---|---|---|
| Measures | Distance around the boundary | Space inside the shape |
| Units | Linear units (cm, m, km) | Square units (cm², m², km²) |
| Formula (Rectangle) | 2(l + b) | l × b |
| Formula (Square) | 4s | s² |
| Real Example | Length of fence needed | Amount of carpet needed |
Example 1
Find the area of a rectangle whose
(i) Length = 12 cm and breadth = 6 cm
=> Area of the rectangle = length × breadth
= 12 cm × 6 cm = 72 cm²
(ii) Length = 1.4 m and breadth = 0.5 m
∴ Area of the rectangle = length × breadth
= 1.4 m × 0.5 m = 0.70 m²
Find the area of a square whose each side is
(i) 3 m (ii) 2·4 cm
(i) Area of the square = (side)²
= (3 m)² = 3 m × 3 m = 9 m²
(ii) Area of the square = (side)²
= (2.4 cm)² = 2.4 cm × 2.4 cm = 5.76 cm²
Example 2
Question: What will happen to the area of a square field when each side is
(i) doubled (ii) trebled, or (iii) halved?
Solution:
Let the original side = s
So, original area A = s²
(i) Side is doubled
-
New side = 2s
-
New area = (2s)2 = 2s × 2s = 4s2 = 4A
-
So, new area = 4 times original area.
(ii) Side is trebled
-
New side = 3s
-
New area = (3s)2 = 3s × 3s = 9s2 = 9A
-
So, new area = 9 times original area.
(iii) Side is halved
-
New side = `s/2`
-
New area = (`s / 2` )² = `s/2` × `s/2` = `1/4` × s² = `1/4` × A
-
So, new area = `1/4` of the original area (one-fourth).
Example 3
Question: The perimeter of a rectangular field is 300 m and its length is 90 m. Find:
- its breadth
- its area
- cost of plowing the field at the rate of ₹80 per square meter.
Let length = 90m, perimeter = m.
Formula: P = 2(l + b)
Solution:
(i) Breadth
-
Perimeter of a rectangle = 2 × (length + breadth)
-
300 = 2(l + b)
-
300 = 2(90 + b)
-
Divide both sides by 2:
150 = 90 + b -
b = 150 − 90 = m
Breadth = 60 m
ii) Area of the field
-
Area of the rectangular field = length × breadth
-
Area = 90 m × 60 m
-
(iii) Cost of ploughing
-
Cost of ploughing the field = area of the field × rate of ploughing
-
Rate = ₹80 per m²
-
Area = 5400 m²
-
Cost = 5400 × 80
Cost of ploughing = ₹4,32,000
Example 4
Question: The floor of a big hall is in the shape of a square of side 8 m. The floor is to be covered by square tiles each of side 40 cm. Find the number of tiles required.
Solution:
Step 1: Find area of the floor
Floor is a square of side 8 m.
Area of floor = side2 = 8 × 8 = 64 m2
Step 2: Convert tile side into metres
Tile side = 40 cm
40 cm = `40/100` m = 0.4 m
Step 3: Find area of one tile
Tile is a square of side 0.4 m.
Area of one tile = 0.4 × 0.4 = 0.16 m2
Step 4: Find the number of tiles
Number of tiles = `Area of floor/Area of one tile` = `64/0.16`
Now divide:
64 ÷ 0.16 = 400
400 tiles are required to cover the floor.
Example 5
Question: Four square flower beds, each with a side of 4 m, are in four corners on a piece of land 12 m long and 10 m wide. Find the area of the remaining part of the land.
Solution:
Length (l) = 12 m,Width (w) = 10 m, side of each square bed s = 4 m
Step 1: Area of the whole land
Area of the whole land = l × w = 12 m × 10 m = 120 m².
Step 2: Area of one square flower bed
Area of one flower bed = s × s = 4 m × 4 m = 16 m².
Step 3: Total area of four beds
Area of one flower bed = s × s = 4 m × 16 m = 64 m².
Step 4: The remaining area of the land
Remaining area = Area of land − Total area of beds = 120 m² – 64 m² = 56 m².
Key Points Summary
Perimeter = total distance around a shape's boundary (measured in linear units like m or cm)
Area = total space inside a shape (measured in square units like m² or cm²)
Rectangle Perimeter = 2(length + breadth)
Rectangle Area = length × breadth
Square Perimeter = 4 × side
Square Area = side²
Unit Conversion Tips:
-
1 m = 100 cm, so 1 m² = 10,000 cm²
-
Always use consistent units before calculating
-
Express area in square units and perimeter in linear units
Example Question 3
Two crossroads, each of width 5 m, run at right angles through the center of a rectangular park of length 70 m and breadth 45 m and parallel to its sides. Find the area of the roads. Also, find the cost of constructing the roads at a rate of Rs. 105 per m2.
Area of the crossroads is the area of shaded portion, i.e., the area of the rectangle PQRS and the area of the rectangle EFGH. But while doing this, the area of the square KLMN is taken twice, which is to be subtracted.
Now,
PQ = 5 m and PS = 45 m
EH = 5 m and EF = 70 m
KL = 5 m and KN = 5 m
Area of the path = Area of the rectangle PQRS + area of the rectangle EFGH – Area of the square KLMN
= PS × PQ + EF × EH – KL × KN
= (45 × 5 + 70 × 5 − 5 × 5) m2
= (225 + 350 − 25) m2 = 550 m2
Cost of constructing the path = Rs. 105 × 550 = Rs. 57,750.
Example Question 1
A path 5 m wide runs along inside a square park of side 100 m. Find the area of the path. Also, find the cost of cementing it at the rate of Rs. 250 per 10 m2.
Let ABCD be the square park of side 100 m. The shaded region represents the path 5 m wide.
PQ = 100 – (5 + 5) = 90 m
Area of square ABCD = (side)2 = (100)2 m2 = 10000 m2
Area of square PQRS = (side)2 = (90)2 m2 = 8100 m2.
Therefore, area of the path = (10000 − 8100) m2 = 1900 m2
Cost of cementing 10 m2 =Rs. 250
Therefore, cost of cementing 1 m2 = Rs. `250/10`
So, cost of cementing 1900 m2 = Rs. `250/10 xx 1900` = Rs. 47,500.
Example Question 2
A rectangular park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path.
Let ABCD represent the rectangular park and the shaded region represents the path 2.5 m wide.
To find the area of the path, we need to find (Area of rectanglePQRS – Area of rectangle ABCD).
We have,
PQ = (45 + 2.5 + 2.5) m = 50 m
PS = (30 + 2.5 + 2.5) m = 35 m
Area of the rectangle ABCD = l × b = 45 × 30 m2 = 1350 m2
Area of the rectangle PQRS = l × b = 50 × 35 m2 = 1750 m2
Area of the path = Area of the rectangle PQRS − Area of the rectangle ABCD
= (1750 − 1350) m2 = 400 m2.
