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AP State Syllabus AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers Ex 2.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 2nd Lesson Whole Numbers Exercise 2.2

Question 1.
Find the sum by suitable rearrangement.
i) 238 + 695 + 162
ii) 154 + 197 + 46 + 203
Answer:
i) 238 + 695 + 162 = 238 + 162 + 695 (Commutative property)
= (238 + 162) + 695 (Associative property)
= 400 + 695
= 1095

ii) 154 + 197 + 46 + 203 = 154 + 46 + 197 + 203 (Commutative property)
= (154 + 46) + (197 + 203) (Associative property)
= 200 + 400
= 600

Question 2.
Find the product by suitable rearrangement.
i) 25 × 1963 × 4
ii) 20 × 255 × 50 × 6
Answer:
i) 25 × 1963 × 4 = 25 × (1963 × 4)
= 25 × (4 × 1963) (Commutative property)
= (25 × 4) × 1963 (Associative property)
= 100 × 1963 = 196300

ii) 20 × 255 × 50 × 6 = 20 × 50 × 255 × 6 (Commutative property)
= (20 × 50) × (255 × 6) (Associative property)
= 1000 × 1530
= 15,30,000

Question 3.
Find the product using suitable properties.
1)205 × 1989 ii) 1991 × 1005
Answer:
i) 205 × 1989 = (200 + 5) × 1989
(Distributive property of multiplication over addition)
= (200 × 1989) + (5 × 1989)
= 397800 + 9945
= 407745

ii) 1991 × 1005
= 1991 × (1000 + 5)
Distributive property of multiplication over addition.
= (1991 × 1000)+ (1991 × 5)
= 1991000 + 9955
= 2000955

Question 4.
A milk vendor supplies 56 liters of milk in the morning and 44 liters of milk in the evening to a hostel. If the milk costs Rs. 50 per liter, how much money he gets per day?
Answer:
Capacity of milk supplied in the morning = 56 l
Capacity of milk supplied in the evening = 44 l
Total capacity of milk supplied in one day = (56 + 44)l
Cost of one liter milk = Rs. 50
Cost of (56 + 44) liters milk =(56 + 44) × 50
= 100 × 50
= Rs. 5000
∴ Money got by vendor per day = Rs. 5000

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 2 Whole Numbers Ex 2.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 2nd Lesson Whole Numbers Exercise 2.1

Question 1.
How many whole numbers are there in between 27 and 46?
Answer:
Number of whole numbers upto 27 is 28 (from zero to 27)
Number of whole numbers upto 45 is 46 (excluding 46)
Number of whole numbers between 27 and 46 = 46 – 28 = 18
They are 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45.

Question 2.
Find the following using number line.
i) 6 + 7 + 7
ii) 18 – 9
iii) 5 × 3
Answer:
i) 6 + 7 + 7
Answer:
Draw the number line starting from zero (0).

Starting from 6, we make 7 jumps to the right of 6 on the number line. Then we reach 13. Again make 7 jumps to the right of 13. Then we reach 20.
So, 6 + 7 + 7 = 20

ii) 18 – 9
Answer:

Draw the number line starting from zero (0).
Start from 18. We make 9 jumps to the left of 18 on the number line.
Then we reach 9.
So, 18 – 9 = 9

iii) 5 × 3
Answer:

Draw the number line starting from zero (0).
Start from 0 and make 3 jumps to the right of the zero on the number line.
Now, treat 3 jumps as one step.
So, to make 5 steps (i.e., 3, 6, 9, 12 and 15) move on the right side, we read 15 on the number line.
So, 5 × 3 = 15

Question 3.
In each pair, state which whole number on the number line is on the right of the other number.
i) 895, 239
Answer:
As 239 < 895, we conclude that 895 is on the RHS of 239.

895 is right of 239 on the number line.

ii) 1001, 10001
Answer:
As 1001 < 10001, we conclude that 10001 is on the RHS of 1001.

10001 is right of 1001 on the number line.

iii) 15678, 4013
Answer:
As 4013 < 15678, we can say that 15678 is on the RHS of 4013.

15,678 is right of 4,013 on the number line.

Question 4.
Mark the smallest whole number on the number line.
Answer:
We know that zero is the smallest whole number mark it on the number line.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us Unit Exercise Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 1st Lesson Numbers All Around us Unit Exercise

Question 1.
Write each of the following in numeral form.
i) Hundred crores hundred thousands and hundred.
Answer:
Indian system: 100,01,00,100.

ii) Twenty billion four hundred ninety seven million pinety six thousands four hundred seventy two.
Answer:
20,497,096,472

Question 2.
Write each of the following in words in both Hindu-Arabic and International system.
i) 8275678960
ii) 5724500327
iii) 1234567890
Answer:
i) 8275678960
Hindu – Arabic system: 827,56,78,960
Eight hundred twenty seven crores fifty six lakhs seventy eight thousand nine hundred and sixty.
International system: 8,275,678,960
Eight billion two hundred seventy five million six hundred seventy eight thousand nine hundred and sixty.

ii) Hindu-Arabic system: 572,45,00,327
Five hundred seventy two crores forty five lakhs three hundred and twenty seven. International system: 5,724,500,327
Five billion seven hundred twenty four million five hundred thousand three hundred and twenty seven,

iii) 1234567890
Hindu-Arabic system: 123,45,67,890
One hundred twenty three crores forty five lakhs sixty seven thousand eight hundred and ninety.
International system: 1,234,567,890
One billion two hundred thirty four million five hundred sixty seven thousand eight hundred and ninety.

Question 3.
Find the difference between the place values of the two eight’s in 98978056.
Answer:
Place values of 8 in Hindu-Arabic system of the given number are 8,000 and 80,00,000
Difference = 80,00,000 – 8,000 = 79,92,000
Place values of 8 in International system of the given number are 8000 and 8,000,000
Difference = 8,000,000 – 8,000 = 79,92,000

Question 4.
How many 6 digit numbers are there in all?
Answer:
Number of 6 digit numbers = greatest 6 digit number – greatest 5 digit number
= 9,99,999 – 99,999 = 9,00,000.

Question 5.
How many thousands make one million?
Answer:
1000 thousands can make one million.
1 Million = 1000 Thousands.

Question 6.
Collect ‘5’ mobile numbers and arrange them in ascending and descending order.
Answer:
Let the 5 mobile numbers are: 9247568320, 9849197602, 8125646682, 6305481954, 7702177046
Ascending order: 6305481954, 7702177046, 8125646682, 9247568320, 9849197602
Descending older: 9849197602, 9247568320, 8125646682, 7702177046, 6305481954

Question 7.
Pravali has one sister and one brother. Pravali’s father earned one million rupees and wanted to distribute the amount equally. Estimate approximate amount each will get in lakhs and verify with actual division.
Answer:
One million = 10,00,000 = 10 lakhs
Pravali’s father distributed 10 lakhs amount to his 3 children equally.
So, the share of each children = 10 lakhs ÷ 3 = Rs. 3,33,333
= Rs. 3,00,000 (approximately)

Question 8.
Government wants to distribute Rs. 13,500 to each farmer. There are 2,27,856 farmers in a district. Calculate the total amount needed for that district. (First estimate, then calculate)
Answer:

Question 9.
Explain terms Cusec, T.M.C, Metric tonne, Kilometer.
Answer:
a) Cusec: A unit of flow equal to 1 cubic foot per second.
1 Cusec = 0.028316 cubic feet per second = 28.316 litre per second.
Cusec is the unit to measure the liquids in large numbers quantity.

b) TMC: TMC is the unit to measure the water in large quantity.
TMC means Thousand Million Cubic feet.
1 TMC = 0.28316000000 litre
= 28.316 billion litre
= 2831.6 crores litre

c) Metric tonne: Metric tonne is the unit of weight.
Metric tonne = 1000 kg = 10 quintals
We should use this unit in measuring crops, paddy, dall, etc.

d) Kilometer: Kilometre is the unit of length.
1 kilometer = 1000 meters.
We should use this unit is measuring distance between villages, towns, cities,…. etc.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us Ex 1.4 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 1st Lesson Numbers All Around us Exercise 1.4

Question 1.
Write some daily life situation where we can use large numbers.
Answer:
We should use large number in our daily life in
a) Counting money at banks, etc.
b) Population of the city or state or country or world.
c) Austronautical distances, etc.

Question 2.
A box of medicine contains 3,00,000 tablets each weighing 15mg. What is the total weight of all the tablets in the box in grams and in kilograms?
Answer:
Weight of a tablet = 15 mg
Weight of 3,00,000 tablets = 300000 × 15
Weight of one box of tablets = 45,00,000 mg
we know 1000 mg = 1 gram
To convert mg into grams we have to divide grams by 1000 mg = 45,00,000 ÷ 1000
Weight of one box of tablets in grams = 4500 grams.
We know 1000g = 1kg
To convert ‘g’ into kilograms we have to divide kilograms by 1000g = \(\frac{4500}{1000}\) = 4.5 kg

Question 3.
Damodhar wants to buy onions in Kurnool market. Each onion bag weighs 45 kg. He loaded 326 onion bags with 45kg in a lorry. Find the total weight of onions in kilograms and quintals.
Answer:
Weight of one bag onions = 45 kg
Weight of 326 bag onions = 326 × 45 = 14,670 kg
We know, 100 kg = 1 quintal.
To convert kgs into quintal we have to divide kgs by 100
= 14670 ÷ 100 = 146.7 quintals.

Question 4.
The population of 4 South Indian States according to 2011 Census:
Andhra Pradesh: 8,46,65,533, Karnataka: 6,11,30,704, Tamil Nadu: 7,21,38,958 and Kerala: 3,33,87,677. What is the total population of South Indian States?
Answer:
Population of Andhra Pradesh = 8,46,65,533
Karnataka = 6,11,30,704
TamilNadu = 7,21,38,958
Kerala = 3,33,87,677
Total population of 4 states = 25,13,22,872

Question 5.
A famous cricket player has so far scored 28,754 runs in International matches. He wishes to complete 50,000 runs in his career. How many more runs does he need?
Answer:
Number of runs wishes to complete = 50,000
Number of runs scored by the player = 28,754
Number of runs needed = 50,000 – 28,754 = 21,246 runs

Question 6.
In an election, the successful candidate registered 1,32,356 votes and his nearest rival secured 42,246 votes. Find the majority of successful candidate.
Answer:
Number of votes secured by the winner = 1,32,356
Number of votes secured by the rival = 42,246
Number of more votes secured by the winner = 1,32,356 – 42,246 = 90,110
Majority of the winner = 90,110 votes.

Question 7.
Write the greatest and smallest six digit numbers formed by all the digits 6, 4, 0, 8, 7, 9 and find the sum and difference of those numbers.
Answer:
Given digits are 6, 4, 0, 8, 7, 9
The greatest number formed by the digits = 9,87,640
The smallest number formed by the digits = 4,06,789
Sum of the numbers = 9,87,640 + 4,06,789 = 13,94,429
Difference between numbers = 987640 – 406789 = 580851

Question 8.
Haritha has Rs. 1,00,000 with her. She placed an order for purchasing 124 ceiling fans at Rs. 726 each. How much money will remain with her after the purchase?
Answer:
Cost of each fan = Rs. 726
Cost of 124 fans = 726 × 124 = Rs. 90,024
Money with Haritha = Rs. 1,00,000
Cost of 124 fans = Rs. 90,024
Remaining money after purchasing = Rs. 9,976

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us Ex 1.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 1st Lesson Numbers All Around us Exercise 1.3

Question 1.
Write each of the following numbers in digits by using International place value chart. Also write them in expanded form.
i) Nine million seven hundred thousand and six hundred five.
Answer:
9,700,605: 9,000,000 + 700,000 + 600 + 5

ii) Seven hundred million eight hundred seventy two thousand and four hundred seven.
Answer:
700,872,407: 700,000,000 + 800, 000 + 70,000 + 2000 + 400 + 7

Question 2.
Rewrite each of the following numerals with proper commas in the International system of numeration and write the numbers in word form.
i) 717858
ii) 3250672
iii) 75623562
iv) 956237676
Answer:
i) 717,858: Seven hundred seventeen thousand eight hundred and fifty eight.
ii) 3,250,672: Three million two hundred fifty thousand six hundred and seventy two.
iii) 75,623,562: Seventy five million six hundred twenty three thousand five hundred and sixty two.
iv) 956,237,676: Nine hundred fifty six million two hundred thirty seven thousand six hundred and seventy six.

Question 3.
Write the following number names in both Indian and International systems.
i) 6756327
ii) 45607087
iii) 8560707236
Answer:
i) 6756327
Indian system:
67,56,327: Sixty seven lakh fifty six thousand three hundred and twenty seven.
International system:
6,756,327: Six million seven hundred fifty six thousand three hundred and twenty seven.

ii) 45607087 Indian system:
4,56,07,087: Four crores fifty six lakhs seven thousand eighty seven.
International system:
45,607,087: Forty five million six hundred seven thousand and eighty seven.

iii) 8560707236 Indian system:
856,07,07,236: Eight hundred fifty six crores seven lakh seven thousand two hundred and thirty six.
International system:
8,560,707,236: Eight billion five hundred sixty million seven hundred seven thousand two hundred and thirty six.

Question 4.
Express the following numbers in other system.

Answer:

Question 5.
Write the following numbers in International system (Word Form).
i) Twenty Nine crore thirty five lakh forty six thousand seven hundred and fifty three.
Answer:
293,546,753
Word form: Two hundred ninety three million five hundred forty six thousand seven hundred and fifty three.

ii) Thousand crore ninety nine lakh and forty three.
Answer:
10,009,900,043
Word form: Ten billions nine million nine hundred thousand and forty three.

Question 6.
Write following numbers in Indian system (Word Form).
i) Nine billion twenty four million fifty thousand and seventy two.
Answer:
902,40,50,072
Word form: Nine hundred two crores forty lakhs fifty thousand and seventy two.

ii) Seven hundred billions six millions four thousand seven hundred and five.
Answer:
70000,60,04,705
Word form: Seventy thousand crores sixty lakhs four thousand seven hundred and five.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us Ex 1.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 1st Lesson Numbers All Around us Exercise 1.2

Question 1.
Write each of the following in numeral form.
i) Sixty crores seventy five lakhs ninety two thousands five hundred and two.
Answer:
60, 75, 92, 502

ii) Nine hundred forty four crores six lakhs fifty five thousand four hundred and eighty six.
Answer:
944, 06, 55, 486

iii) Ten crores ten thousand and ten.
Answer:
10,00,10,010

Question 2.
Insert commas in the correct positions to separate periods and write the following numbers in words.
i) 57657560
ii) 70560762
iii) 97256775613
Answer:
i) 5,76,57,560: Five crores seventy six lakhs fifty seven thousands five hundred and sixty.
ii) 7,05,60,762: Seven crores five lakhs sixty thousands seven hundred and sixty two.
iii) 9725,67,75,613: Nine thousand seven hundred and twenty five crores sixty seven lakhs seventy five thousand six hundred and thirteen.

Question 3.
Write the following in expanded form.
i) 756723
ii) 60567234
iii) 8500756762
Answer:
i) 756723
Expanded form: 7 × 1,00,000 + 5 × 10,000 + 6 × 1,000 + 7 × 100 + 2 × 10 + 3 × 1
: 7 lakhs + 5 ten thousands + 6 thousands + 7 hundreds + 2 tens + 3 ones
Word form: Seven lakh fifty six thousand seven hundred and twenty three.

ii) 60567234
Expanded form: 6 × 1,00,00,000 + 5 × 1,00,000 + 6 × 10,000 + 7 × 1,000 + 2 × 100 + 3 × 10 + 4 × 1
: 6 crores + 5 lakhs + 6 ten thousands + 7 thousands + 2 hundreds + 3 tens + 4 ones
Word form: Six crore five lakh sixty seven thousand two hundred and thirty four.

iii) 8500756762
Expanded form: 8 × 1,00,00,00,000 + 5 × 10,00,00,000 + 7 × 1,00,000 + 5 × 10,000 + 6 × 1000 + 7 × 100 + 6 × 10 + 2 × 1
: 8 hundred crores + 5 ten crores + 7 lakhs & 5 ten thousands + 6 thousands + 7 hundreds + 6 tens + 2 ones
Word form: Eight hundred and fifty crore seven lakh fifty six thousand seven hundred and sixty two.

Question 4.
Determine the difference between the place value and the face value of 6 in 86456792.
Answer:
Given number is 86456792. By putting commas to separate periods the given number can be written as 8,64,56,792.
i) Place value of ‘6’ in thousand place = 6 x 1000 = 6,000
Face value of 6 = 6
Difference = 6,000 – 6 = 5,994
ii) Place value of ‘6’ in ten lakhs palce = 6 x 10,00,000 = 60,00,000
Face value of 6 = 6
Difference = 60,00,000 – 6 = 59,99,994

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 1 Numbers All Around us Ex 1.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 1st Lesson Numbers All Around us Exercise 1.1

Question 1.
Identify the greatest and smallest among the following numbers.

Create Your Own Problem on Block No: 3 and fill the above table.
Answer:

Question 2.
Write the given numbers in ascending and descending order.

Answer:

Question 3.
Write the numbers in word form.

Create Your Own Problem on Block No:4 and fill the above table.
Answer:

Question 4.
Write the numbers in figures.

Create Your Own Problem on Block No:4 and fill the above table.
Answer:

Question 5.
Write 4 digit numbers is many as possible with 6, 0, 5, 7 digits.
Answer:
5067, 5076, 5706, 5760, 5670, 5607
6057, 6075, 6705, 6750, 6570, 6507
7056, 7065, 7605, 7650, 7506, 7560

Question 6.
Form the greatest and smallest numbers with given digits and find the difference without repetition.

Answer:

Question 7.
Observe the table and fill the empty boxes.

Answer:

Students can go through AP Board 6th Class Maths Notes Chapter 12 Data Handling to understand and remember the concepts easily.

AP State Board Syllabus 6th Class Maths Notes Chapter 12 Data Handling

→ Data: Information which is in the shape of numbers or words or pictures which help us in taking decisions is called data.
If data is expressed in numbers it is called numerical data.
Eg: The marks obtained by five students at an examination is 25, 32, 28,14 & 24.
If data is expressed in words it is called data in words.
Eg : The colours liked by some students are Red, black, pink, white, etc.

→ Frequency: Number of times a particular observation occurs in a given data is called its frequency.

→ Frequency distribution table: A table showing the frequency or count of various items is called a frequency distribution table.
A data can be arranged in a tabular form using tally marks. The data arranged in a tally table is easy understood and interpret.

→ Pictograph: If a data is arranged using pictures, then it is called a pictograph.

→ Bar graph: If a data is arranged using rectangles, then it is called a bar graph.
The rectangles can be either vertical or horizontal in a bar graph.

→ Scale: Scale is a convenient way to represent the data. It quantifies what a single unit represents in a given bar graph or pictograph.

Students can go through AP Board 6th Class Maths Notes Chapter 11 Perimeter and Area to understand and remember the concepts easily.

AP State Board Syllabus 6th Class Maths Notes Chapter 11 Perimeter and Area

→ Perimeter: The perimeter of a polygon is sum of all its sides.
The perimeter of an equilateral triangle is P = 3 × side
The perimeter of a rectangle P = 2 (length + breadth)
And its area A = length × breadth A = l × b
The perimeter of a square is P = 4 × side And its area A = side × side (or)
A = s × s The circumference of a circle C = 2πr where r is the radius of the circle.

→ Area: The region occupied by a plane figure is called its area.
To find the area of a complex figure, we divide the given shape into the combination of rectangles, squares and triangles where ever necessary.

Students can go through AP Board 6th Class Maths Notes Chapter 10 Practical Geometry to understand and remember the concepts easily.

AP State Board Syllabus 6th Class Maths Notes Chapter 10 Practical Geometry

→ The above are the components we see in a geometry box. They are pair of set squares, protractor, graduated ruler, compasses and the divider.

→ Graduated ruler: A scale is used to draw straight edges of given length. It is also used to measure the lengths of the given line segments. A scale is also called a graduated ruler.

→ Compass: A compass is used to draw a circle of given radius. Sometimes we draw only a part of a circle which is called an arc. A compass is used in the construction of a line segment of given length, a perpendicular to a given line, given angle and in many more geometrical shapes.

→ Divider: A divider is used to measure the lengths of straight line segments and curved lines. It is also used to compare the lengths of two line segments.

→ Set squares: The pair of set squares is used to draw the pair of parallel lines.

→ Perpendicular bisector: The perpendicular bisector of a given line segment is the line which divides the given line segment into two equal parts at right angles.

→ Angle bisector: Angle bisector is a ray which divides the given angle into two equal parts.

Students can go through AP Board 6th Class Maths Notes Chapter 9 2D-3D Shapes to understand and remember the concepts easily.

AP State Board Syllabus 6th Class Maths Notes Chapter 9 2D-3D Shapes

→ Space: A plane is a set of all points which extends in all directions endlessly in three dimensions.

→ Plane: A plane is a smooth surface which extends in all directions in two dimensions.
Eg: The surfaces of a table, the surface of a blackboard are examples for part of a plane.

→ Polygon: A simple closed figure formed by line segments is called a polygon.

→ Triangles:
A simple closed figure formed by three line segments is called a triangle.
The line segments \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{BC}}\) & \(\overline{\mathrm{CA}}\) are called the sides of the triangle.

A triangle contains three sides.
A triangle contains three interior angles ∠A, ∠B & ∠C and three vertices A, B & C

→ A triangle divides the plane on which it lies into three sets of points.
1. Interior points on the triangle
2. Points on the triangle
3. Points in the exterior of the triangle

→ Quadrilateral:
A simple closed figure bounded by four line segments is called a quadrilateral.

→ Circle: If we draw a boundary along the edge of a round shaped object, then we get the following shape. The shape is called a circle. The length of the curved edge is called circumference. “A” is called the centre of the circle.

→ Chord: A line segment joining any points ‘A’ & ‘B’ on the circumference of the circle is called a chord. DE and FG are chords of the circle.

→ Diameter: The longest chord passing through the centre of the circle is called a diameter. AC & DG are diameters.

→ Arc: A part of the circle is called an arc.

CIRCLE TERMINOLOGY

→ Circumference: The distance around a circle.

→ Radius: The distance from the centre of a circle to the circumference. Half the diameter.

→ Diameter: A straight line passing through the centre of a circle to touch both sides of the circumference. Twice
as long as the radius.

→ Chord: A straight line joining two points on the circumference of a circle. The diameter is a special kind of chord.

→ Arc: A section of the circumference.

→ Sector: A section of a circle, bounded by two radii and an arc.

→ Segment: A section of a circle, bounded by a chord and an arc.

→ Tangent: A straight line touching the circumference once at a given point.

→ A circle divides the plane on which it lies into three parts.

1. Interior points
2. Exterior points
3. Points on the circle

→ BASIC PARTS OF A CIRCLE

• Interior of a Circle
  points A, B, C
• ON the circle
  point D
• Exterior of a Circle
  points E, F, G

→ The region in the interior of a circle enclosed by its boundary is called circular region.

→ Symmetry: Some figures appear beautiful because of their symmetry. Such shapes can be divided into two identical parts along a straight line which is called line of symmetry. A symmetrical figure may have more than one line of symmetry.

→ English alphabet – lines of symmetry:

3-D shapes:
→ NAMES OF 3D SHAPES

→ 3D SHAPES IN REAL – LIFE

→ Cube: A cube is a 3-dimensional figure. It has 6-identical faces. Each face of a cube is a square. All its sides are equal.

→ Cuboid: A cuboid is three a dimensional figure having three measures length, breadth and height.

→ Cylinder: A cylinder has circular faces at its both ends. It has two measures namely radius of the base and height of the cylinder.

→ Cone: A cone is a 3-d figure having curved surface with a circular base.

→ Prism and a Pyramid:

A Prism is a 3-d shape with parallelograms as its lateral surfaces. A Pyramid is a 3-d figure with triangles as its lateral surfaces.
A Prism / Pyramid may have a triangle/ square/rectangle…as its base.
A Prism/Pyramid is named as per its base.

→ Euler’s formula:
The number of faces (F), vertices (V) and edges (E) of a polyhedron are related by this formula: F + V = E + 2

Looking at the box to the right, calculate the number of edges:
Faces = 6
Vertices = 8
F + V = E + 2
6 + 8 = E + 2
14 = E + 2
E = 12

Students can go through AP Board 6th Class Maths Notes Chapter 8 Basic Geometric Concepts to understand and remember the concepts easily.

AP State Board Syllabus 6th Class Maths Notes Chapter 8 Basic Geometric Concepts

→ Line: A line is a straight edge which extends endlessly on both sides. A line has no width.

Line is a set of infinite points extending in two opposite directions endlessly. A line is represented by any two points lying on it.
The above line is represented by \(\overline{\mathrm{AB}}\)
A line can also be represented by a lower case letter of English alphabet such as l.
A line has no end points.
A line may not be always straight. Sometimes it is curved as shown in the figure.

It is called a curved line.

→ Ray: A ray is a straight edge starting from a point and extends only in one direction endlessly.

A ray has no width.
A ray has only one end point. The end point is called the vertex of the ray. A ray is represented by the initial point and any arbitrary point in the direction in which it extends. Here the ray is \(\overline{\mathrm{AB}}\).

→ Line segment: A part of a line is called a line segment.
A line segment has two ends.

A line segment is represented by its two end points. Line segment \(\overline{\mathrm{AB}}\).
Recall that a small dot made by a shape edged pencil may be treated as a point.

→ History:
Geometry has a long and rich historical nature. The term ‘GEOMETRY’ is derived from the greek word ‘GEOMETRON’. ‘GEO’ means earth and ‘METRON’ means measurement. So, Geometry is the mathematics related to the earth’s measurement.
Early geometry was a collection Of empirically discovered principles concerning lengths, angles, areas and volumes which were developed to meet some practical need in surveying, construction, astronomy and various crafts.
In the ancient India Aryabhatta, Brahmagupta were some of the Indian Mathematicians who contributed their works in geometry.

→ Intersecting lines: Two lines meeting at a single point are called intersecting lines. The point is called the point of intersection. Two lines l & m intersecting at the point A.

→ Concurrent lines: Three or more lines passing through a single point are called concurrent lines and the point is called the point of concurrence.

Here the lines l, m & n are meeting at a single point O and hence are concurrent lines. The point of concurrence is O.

→ Parallel lines: Two lines are said to be parallel, if they never meet each other.

The distance between two parallel lines is constant throughout their length.
Here the lines l & m are parallel to one another.

→ Perpendicular lines: Two lines are said to be perpendicular, if they are straight to one another. Two adjacent sides of a paper are perpendicular to each other.

→ Measuring the length of a line segment: To measure the length of a given line segment we use a graduated scale or a divider in the instrument box.
Adjust the scale along the line segment and find the readings at the two end points. The difference of readings gives us the length of the line segment.
Adjust the two pointed legs of the divider on the two end points of the given line segment. Now measure the width between the legs on a graduated scale.
Measuring line segments in Centimeters

→ Angle: A figure formed by two rays with a common end point is called an angle.

The common end point is called the vertex of the angle. The two rays are called the two arms of the angle.

→ Sexagesimal system : A system related to number sixty is called sexagesimal system. We use this system in measuring angles.

Full angle is also called complete angle.
We use protractor to measure the angle.