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AP State Syllabus AP Board 8th Class Biology Solutions Chapter 3 Story of Microorganisms 1 Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 3rd Lesson Story of Microorganisms 1

8th Class Biology 3rd Lesson Story of Microorganisms 1 Textbook Questions and Answers

Improve Your Learning

Question 1.
Which organisms are interlinked between living and non-living organisms? Why do you think so?
Answer:
Viruses are an interesting type of microorganisms. They usually made up of crystalized proteins. They behave like nonliving things when they are outside of a living cell. But they behave like living organisms when they are inside host living cells and reproduce just like bacteria. Hence viruses can also call as connecting links between living and nonliving things.

Question 2.
What are microorganisms? Where do you find them?
Answer:
We can see several organisms in our surroundings but we cannot see many of them with our unaided eyes. They can be seen only with the help of microscope only. They are called microorganisms. They can found in air, water, soil and all living organisms.

Question 3.
What type of microorganisms we can observe in pond water?
Answer:
Usually pond water contains bacteria, phytoplanktons, algal members, fungi, rotifers, hydra etc. Collect some pond water with greenish scrapings on a slide and we can observe different algal members like Spirogyra, Chara and Chlamydomonas through the microscope.

Question 4.
Whether microorganisms are useful or harmful. How? Explain.
Answer:

1. Microorganisms are useful and some are harmful.
2. Some microorganisms are useful in formation process, medicine preparation and increase soil fertility.
3. Some microorganisms are harmful by causing diseases and spoling food items.

Question 5.
How are the human actions causing the death of useful bacteria and fungi? What will happen if it continuous?
Answer:

1. Soil is highly rich in microorganisms such a bacteria, fungi, protozoa, micro arthropods.
2. The top eight inches of soil of one acre many contain as much as five and half tons of fungi and bacteria.
3. This is very much useful for growing crops.
4. Excess use of pesticides kills these bacteria.
5. Thus human actions causing death of useful bacteria and fungi.
6. If it happens continue, then it causes to ecological imbalance.

Question 6.
Why the cooked food spoil soon but not uncooked food ? Give your reasons.
Answer:

1. Cooking of food items makes the proteins in the food materials coagulate.
2. It also degrades the protective surface of the food.
3. Thus the cooked foods can be easily inhabited by microoganisms.
4. So they can be spoiled in less time than the uncooked food.

Question 7.
What questions would you like to ask your teacher to know about different shapes of bacteria ?
Answer:

1. Where can we find bacteria?
2. How can we see bacteria?
3. What do we call the round shaped bacteria?
4. What do we call the spring shaped bacteria?
5. What is the name of coma shape bacteria?
6. How many types of bacteria do we find in nature?
7. What is the shape of Lactobacillus bacteria? How is it useful?
8. How is septicemia bacteria harmful?
9. Which type of bacteria is responsible for food poisoning?
10. Which bacteria is present in root nodules of leguminous plants? How do they useful?
11. Name the bacteria that causes leprosy.
12. Which type of bacteria is responsible for tuberculosis?
13. What is the shape of Bacillus thuringiensis bacteria how is it useful to plants?
14. What is the shape of staphylococci bacteria? In what way it affect the health of people?

Question 8.
What will happen if you add buttermilk to chilled milk?
Answer:

1. Lacto bacillus bacteria is responsible for the formation of curd.
2. When we add buttermilk to luke warm milk it takes 2 or 3 hours time to form curd.
3. But if we add buttermilk to chilled milk it takes more time or curd would not form.
4. Curdling indicates that the increase in number of bacteria in milk.
5. In chilled condition the number of bacteria do not increase in number there by curd would not be formed.

Question 9.
How do you observe Lactobacillus bacterium?
Answer:
Take one or two drops of buttermilk on a slide and spread it. Heat the slide slightly on a lamp (3-4 sec¬onds). Add a few drops of crystal violet. Leave it for 30-60 seconds and wash the slide gently with water.
Observe the slide under the Compound Microscope to see the Lactobacillus bacterium.

Question 10.
Visit any bakery or milk chilling center near your school with the help of your teacher or parents. Learn about some techniques to culture and usage of some Microorganisms and prepare a note on them.
Answer:
The Milk Collection Station is a specially designed, integrated unit, which combines the several functions of a milk collection centre. It measures the weight, fat content and gives the price of the milk brought in by the each producer. The equipment is particularly useful for the milk cooperatives / milk collection centres as it can also maintain a summary of milk supplied. This state of the art equipment operates both on battery and mains and is able to process and record 120-150 milk collection per hour. An Electronic Milk weighing Unit, the Electronic Milk Tester and Data Processor Unit are main components of the system. The membership code of individual mem¬bers is entered automatically by member identity card / manually by an electronic key-board.

Question 11.
Observe some permanent slides of microorganisms in your school lab with the help of microscope. Draw its picture.
Answer:

Question 12.
Prepare a model of any microorganism. And write a note on them.
Answer:

Question 13.
Why should we clean our hands with soap before eating ?
Answer:

1. We touch the objects.
2. Microbes are present on them.
3. When we touch them, they will inhabit our hands.
4. Washing our hands with soap kills all the microbes.
5. And makes our hands clean and hygenic.
6. So we should clean our hands with soap before eating.

8th Class Biology 3rd Lesson Story of Microorganisms 1 Activities

Activity – 2

Question 1.
Identify the fungi present, in rotten vegetables.
Answer:
Take some rotten part of vegetable or black spoiled part of bread or coconut with help of a needle on a slide, Put a drop of water. Place a cover slip on it and we can see the following microorganisms through microscope.

Activity – 6

Question 2.
Observe different soil microorganisms through microscope and draw rough sketches.
Answer:
Collect some soil from the field in a beaker or in a glass. Add some water to it and stir it. Wait for some time to allow the soil particles to settle down. Take a drop of water on a slide and we can observe the following microorganisms.

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 2 Cell: The Basic Unit of Life Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 2nd Lesson Cell: The Basic Unit of Life

8th Class Biology 2nd Lesson Cell: The Basic Unit of Life Textbook Questions and Answers

Improve Your Learning

Question 1.
Who discovered the cell for the first time?
Answer:
It was the year 1665 Robert Hooke, a British scientist observed thin slices of cork under a simple magnifying device which he had made himself. He observed that the cork resembled the structure of a honey comb consisting of many empty spaces or empty box like structures. He thought it was made up of very small cavities, Robert Hooke called these cavities “Cell”.

Question 2.
Name the factors on which shape of the cells depend.
Answer:
The shape and size of the cells vary considerably but all of these cells ultimately determined by the specific function of the cells.
e.g.: Amoeba is changing its shape for specific functions like collection of food and locomotion.
The shape of the cell may vary for giving definite structure to the organism, e.g,: Epidermal cells.

Question 3.
Distinguish between unicellular and multi cellular organisms.
Answer:

Unicellular	Multicellular
1) An organism composed of just one cell.	1) An organism composed of more than cell.
2) Many of nature’s simplest creatures called “unicellular organisms”.	2) More biologically advanced creatures, called “multicellular organisms”.
3) Cell is an individual form no gathering to perform tasks, but they live together.	3) Different kinds of cells are joined together to perform specialized tasks.
4) The single cell a unicellular organism possesses, the smaller its body, e.g.: Amoeba, Chlamydomonas.	4) The more cells a multicellular organism possesses, the larger its body, e.g.: Fish, Neem tree.

Question 4.
How will you prepare slide without drying quickly?
Answer:
Preparation of slide is a technique to observe microscopic structures. Microscopic slide is prepared on a 2 mm thick. Thin flat plant material directly placed in a drop of water on the glass slide. A drop of glycerin is added to the water to keep the material for longer time. Glycerine saves the material from drying quickly.

Question 5.
Deekshith said that, “we can’t see cells with unaided eye.” Is the statement true or false? Explain.
Answer:
We can’t see cell with naked eye is true. All living things formed by microscopic cells which are visible through microscope only but the egg of birds is visible without microscope. The size of the cells in living organism may be as small as the millionth of a meter.
Most of the cells either in unicellular and multicellular are small in size to perform all the life processes perfectly. The smallest cell 0.1 to 0.5 micrometers found in bacteria. Some of the cells can be seen with naked eye. The largest cell 17 cm x 18 cm egg of Ostrich.

Question 6.
Correct the statement and if necessary rewrite. (OR)
What is cell wall and what are its functions?
a) Cell wall is essential in plant cells.
b) Nucleus controls cell activity.
c) Unicellular organisms perform all life processes like respiration, excretion, growth and reproduction.
d) To observe nucleus and organelles clearly, staining is not necessary.
Answer:
a) Cell wall is essential in plant cells.
Plant cell walls are essential for plant life and also have numerous industrial applications, ranging from wood to nutraceuticals.
The cell wall is the tough, usually flexible but sometimes fairly rigid layer that surrounds some types of cells. It is located outside the cell membrane and provides these cells with structural support and protection, in addition to acting as a filtering mechanism. A major function of the cell wall is to act as a pressure vessel, preventing over-expansion when water enters the cell. Cell wall is found in plants, bacteria, fungi, algae, archaea. Animals and protozoa do not have cell walls. “Plant cell wall is an essential component of biotic stress response mechanisms.”

b) Nucleus controls cell activity.
By containing the instructions for protein products in the DNA of the nucleus. All “control” work in the cell is carried out by proteins, such as enzymes, though DNA codes for other structural material, only protein has metabolic and behavioural control in the organism’s cells. Thus, the nucleus is the cell’s control center.

c) Unicellular organism perform all life processes like respiration, excretion, growth and reproduction.
All living organisms perform some basic life processes like respiration, excretion, etc., to sustain its life and improve its race. Unicellular organisms also perform all life processes.

d) To observe nucleus and organelles clearly, staining is not necessary.
To observe nucleus and organelles clearly, staining is necessary. Staining is a technique to get attached color to different parts of a cell. This helps to highlight particular areas in the cell.

Question 7.
Describe the structure of Nucleus.
Answer:
The nucleus is the largest cellular organelle in animals. In mammalian cells, the average diameter of the nucleus is approximately 6 micrometers (pm), which occupies about 10% of the total cell volume.

Question 8.
Explain the functions of Nucleus.
Answer:
Functions:

1. The main function of the cell nucleus is to control gene expression and mediate the replication of DNA during the cell cycle.
2. The nucleus provides a site for genetic transcription that is segregated from the location of translation in the cytoplasm, allowing levels of gene regulation that are not available to prokaryotes.

Question 9.
What is difference between cells in onion peel and cells in Spinach?
Answer:
Cells in onion peel arranged systematically with prominent nucleus. Cells in spinach are in different sizes and shapes without nucleus to perform nutrition.

Question 10.
Label parts of diagrammes given below. And identify which one is plant cell and which one is animal cell.

Answer:
A. Nucleus
B. Cytoplasm
C. Cell membrane
D. Vacuole
E. Nucleus
F. Cell wall
G. Cell membrane
H. Vocuole
I. Vacuole

Question 11.
What questions will you pose to know diversity in cells?
Answer:

1. Are all the cells similar in shape and size?
2. Do you find nuclei in all the cells?
3. How many different types of cells could you see?
4. What are the different shapes of the cells?
5. Do all the cells one of the same in length?

Question 12.
If you want to know about unicellular and multicellular organisms, what questions will you pose?
Answer:

1. What do you mean by unicelluar organism?
2. What do you mean by multicellular organism?
3. Give examples for unicellular and multicellular organisms.
4. What are the differences between unicellular and multicellular organism?

Question 13.
Get some floating slime from a puddle, pick a very small amount of slime and put it on a slide. Separate out one fiber and look at it through the microscope. Draw the diagram what you have observed.
Answer:

Question 14.
Collect different kinds of leaves from your surroundings and observe the shapes of the epidermal cells under microscope. Make a table which contains serial number, name of the leaf, shape of the leaf, shape of the epidermal cells. Do not forget to write specific findings below the table.
Answer:

Question 15.
Make sketches of animal and plant cells which you observed under microscope.
Answer:

Question 16.
Ameer said “Bigger onion has larger cells when compared to the cells of smaller onions”! Do you agree with his statement or not? Explain why.
Answer:
The sizes of the cells in living organisms are too small to be seen with naked eye. The size of the cell is related to its function. The cell has to perform similar function in all living organisms.
The size of the onion depends upon the number of cells and not the size of the cell. Cells are of different shape, size and number.
Hence, I don’t agree with Ameer. “Bigger onions have larger cells when compared to the cells of smaller onions”.

Question 17.
How do you appreciate the fact that a huge elephant, man and trees are made of cells, which are very small and we can look at them through microscope?
Answer:
The size of the cells in living organisms may be as small as the millionth of a meter (micron) or may be as large as a few centimeters. Majority of the cells are too small to be seen with naked eye.
The size of the cell is related to its function. For example, nerve cell both in man and animals are long and branched. They perform the same functions that of transferring message.
The size of the organism is depending upon the number of cells and not the size of the cell. Cells are of different shape, size and number.

Question 18.
Deepak said, “A plant can’t stand erect without cell wall”. Support this statement.
Answer:
Deepak said, “A plant can’t stand erect without cell wall.” We support this statement with the following reasons.
i) Plant cells differ from those of animals in having an additional layer around the cell membrane.
ii) We called if as ‘cell wall’.
iii) Cell wall gives strength and rigidity to planks.
So a plant can’t stand erect without cell wall.

8th Class Biology 2nd Lesson Cell: The Basic Unit of Life InText Questions and Answers

Question 1.
Make different questions to know cells and cell organelles.
Answer:

1. What are the structures present in the cells?
2. Why cells are considered to be structural and functional unit of life?
3. What is the need of cell wall in plant cells?
4. Did we see the cells with naked eye?

Question 2.
Prepare different questions to know the discovery of cell.
Answer:

1. In which year cell discovered?
2. Name the scientist who observed cells.
3. What type techniques used to observe the cell?
4. Is there any special devices used to study the cell?
5. Name the different devices used for discovery of cell.
6. Can we see living cells under the microscope?

Question 3.
Prepare permanent slides of Onion cell, cheek cell and compare practically.
Answer:
Take inner layer of onion and cheek cells. Stain them with saffranin or methylene blue and keep coverslip.
Observe both the slides under microscope. Cell membrane and cell wall is present in both the cells. Dark stained nucleus is presented in the centre of the cell.
Comparison between onion and cheek cells:

Onion cells	Cheek cells
Cells arranged compactly	Cells arranged loosely
Boundary of onion is cell membrane	Boundary of cheek cell is cell wall
Cytoplasm and nucleus is present	Cytoplasm surrounds the nucleus
Cell organelles present in cytoplasm	Cell organelles present in cytoplasm.

Question 4.
Explain diversity in leaf cells practically.
Answer:
Take a section of Neem leaf on the slide and put a drop of water, cover it with a coverslip and observe it under the microscope. We can see different types of tissues present in the leaf. The section of leaf shows the following features.
Three groups of tissues arranged in leaf.
The first group epidermis where the cells are barrel shaped and covered by waxy material for protection.
The second group mesophyll tissue where the cells contain chloroplasts for nutrition. Cells arranged loosely with air spaces and stomata for exchange of gases.
The third group vascular tissue where the cells are thick walled to transport water and food.

Question 5.
Make a sketch of blood cells.
Answer:

Question 6.
Draw a neat diagram of Clilamydomonas.
Answer:

Question 7.
Make a sketch of Amoeba. (OR)
a) Draw a labelled diagram of Amoeba.
b) What are pseudopodia?
Answer:
a)

b) The projections on the body of amoeba which help in locomotion and collecting food.

Question 8.
Have you listened to the words of the cell? Guess how big a cell is? Is the number and sizes of cells in both man and elephant the same? Are the cells of an elephant bigger than that of a man?
Answer:
The size of the cells in living organisms may be as small as the millionth of a meter (micron) or may be as large as a few centimeters. Majority of the cells are too small to be seen with naked eye.
The size of the cell is related to its function. For example, nerve cell in both man and elephant are long and branched. They perform the same functions that of transferring message.
The size of the organism is depending upon the number of cells and not the size of the cell. Cells are of different shape, size and number. Hence size of cells in both elephant and man are same. The number of cells are more in elephant than man.

8th Class Biology 2nd Lesson Cell: The Basic Unit of Life Activities

Activity – 2

Question 1.
Prepare a slide of an onion peel and find out the special characters.
Answer:
Peel an onion and cut out a small fleshy portion from the bulb. Break this into two small parts and try to separate them. We notice a thin film like material holding the pieces together. Take out small portion and spread it evenly on a slide. Cover it with a cover slip and observe it under microscope.
Draw the figure what you have observed.
Cells are arranged side by side without any gaps.

Activity – 4

Question 2.
Observation of the Nucleus in onion peel cells. (OR)
Sahitya is trying to observe the nucleus in the cells of onion peel. Explain the procedure to be followed for the experiment.
Answer:
Peel a membrane an onion now keep this membrane on a slide and add 1-2 drops of the stain (saffranin, methylene blue or red ink). Cover this with a cover slip and leave it for about five minutes. Then add water drop-wise from one side of the cover slip while soaking the extra water with a filter paper from the other side. This will help in washing away the extra stain. Now observe this slide under a microscope. The blue spot observed within the cell is the nucleus.

Activity – 5

Question 3.
Observe the following pictures and answer the questions given below.

i) What are the structures present in the cells ?
Answer:
Cell wall in onion cell and cell membrane in cheek cell.
Cytoplasm and nucleus are common in both the cells.

ii) Did you see a tiny dark stained thing in all the cells ?
Answer:
Yes there is a tiny dark stained nucleus is common in both the cells.

iii) Are they located in the center of the cell in both the cells ?
Answer:
Nucleus located in center of both the cells.

iv) What is the difference between boundary of onion cell and cheek cell ?
Answer:
Cell wall is the boundary of onion and cell membrane is the boundary of cheek cell.

Activity – 6

Question 4.
Collect leaves stems and roots of different plants from the field and take sections to study different types of cells and tissues present in leaf and stems practically.
Answer:
Take a section of grass leaf on the slide and put a drop of water, cover it with a cover slip and observe it under the microscope. We can see different types of tissues present in the leaf.
The section of root or stem shows the following features.
Four groups of cells can be observed.
First group is outermost layer called epidermis.
It protects stem or root externally.
Major portion of stem or root has second group of cells.
This group synthesizes food and preserve food.
Third group of cells transport water and food.
Fourth group of cells placed centrally.

Question 5.
a) Observe the following cells and collect permanent slides.

b) Fill the following table with help of your teacher.

Name of the cell	Shape of the cell	Name of the parts observed in it
RBC	Biconcave	Blood tissue
Smooth Muscle Cell	Rod	Muscles
Nerve Cell	Tree	Brain, spinal cord and nerves
Bone Cell	Star	All bones
White Blood cell	Amoeboid	Blood tissue

i) Are there any similarities in shape of the cells?
Answer:
No similarity is found in the shape and size of the above cells.

ii) Do you find nuclei in all the cells?
Answer:
In human RBC nucleus is absent. Muscle, Nerve, Bone and White Blood Cells consist nucleus.

iii) Do you know, which cell is the longest in all animals?
Answer:
Nerve cell is the longest in all animals.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers InText Questions and Answers.

8th Class Maths 15th Lesson Playing with Numbers InText Questions and Answers

Do this

Question 1.
Write the place value of numbers underlined.     (Pg. No: 312)
i) 29879   ii) 10344   iii) 98725
Answer:
i) 29879
Place value of 8 = 8 × 100 = 800
Place value of 2 = 2 × 10,000 = 20,000
ii) 10344
Place value of 4 = 4 × 1 = 4
Place value of 3 = 3 × 100 = 300
iii) 98725
Place value of 5 = 5 × 1 = 5
Place value of 8 = 8 × 1000 = 8,000

Question 2.
Write the following numbers in expanded form,        (Pg. No: 313)
i) 65    ii)    74    iii) 153    iv) 612
Answer:
Number Expanded form
i) 65 = 60 + 5 = (6 × 10¹) + (5 × 10⁰)
ii) 74 = 70 + 4 = (7 × 10¹) + (4 × 10⁰)
iii) 153 = 100 + 50 + 3 = (1 × 10²) + (5 × 10¹) + (3 × 10⁰)
iv) 612 = 600 + 10 + 2 = (6 × 10²) + (1 × 10¹) + (2 × 10⁰)

Question 3.
Write the following in standard notation.       (Pg. No: 313)
i) 10 × 9 + 4     ii) 100 × 7 + 10 × 4 + 3
Answer:
Expanded form General form
i) 10 × 9 + 4 = 90 + 4 = 94
ii) 100 × 7 + 10 × 4 + 3 = 700 + 40 + 3 = 743

Question 4.
Fill in the blanks.       (Pg. No: 313)
Answer:
i) 100 × 3 + 10 × ——— + 7 = 357 (5)
ii) 100 × 4 + 10 × 5 + 1 = ——— (451)
iii) 100 × ——— + 10 × 3 + 7 = 737 (7)
iv) 100 × ——— + 10 × q + r = \(\overline{\mathrm{pqr}}\) (p)
v) 100 × x + 10 × y + z = ——— (\(\overline{\mathrm{xyz}}\))
Do you know?

Question 5.
The number 8281807978777675747372717069686766656463626160595857565554535251504948474645444342414039383736353433323130292827262524232221201918 1716151413121110987654321 is written by starting at 82 and writing backwards to 1 and see that it is a prime number.        (Pg. No: 313)
Answer:
No.of digits in the given number are 155.

Question 6.
Write all the factors of the following numbers.       (Pg. No: 314)
a) 24    b) 15   c) 21   d) 27   e) 12   f) 20   g) 18   h) 23   i) 36
Answer:
a) Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
b) Factors of 15 = 1, 3, 5, 15
c) Factors of 21 = 1, 3, 7, 21
d) Factors of 27 = 1, 3, 9, 27
e) Factors of 12 = 1, 2, 3, 4, 6, 12
f) Factors of 20 = 1, 2, 4, 5, 10, 20
g) Factors of 18 = 1, 2, 3, 6, 9, 18
h) Factors of 23 = 1, 23
i) Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36

Question 7.
Write first five multiples of given numbers     (Pg. No: 314)
a) 5   b) 8   c) 9
Answer:
a) Multiples of 5 = 5, 10, 15, 20, 25
b) Multiples of 8 = 8, 16, 24, 32, 40
c) Multiples of 9 = 9, 18, 27, 36, 45

Question 8.
Factorize the following numbers into prime factors.    (Pg. No: 314)
a) 72    b) 158   c) 243
Answer:
a) 72 = 2 × 2 × 2 × 3 × 3
b) 158 = 2 × 7 × 9
c) 243 = 7 × 7 × 7

Question 9.
Check whether the following given numbers are divisible by 10 or not.   (Pg. No: 315)
a) 3860   b) 234   c) 1200   d) 10³   e) 10 + 280 + 20
Answer:
a) 3860, c) 1200, d) 103, e) 10 + 280 + 20 are divisible by ’10’.
[∵ the units digit of above numbers is ‘0’]
b) 234, is not divisible by 10.
[∵ its unit digit is 4]

Question 10.
Check whether the given numbers are divisible by 10 or not.    (Pg. No: 315)
a) 10¹⁰   b) 2¹⁰   c) 10³ + 10¹
Answer:
a) 10¹⁰ = 10000000000
b) 2¹⁰ = 1024
c) 10³ + 10¹ = 1000 + 10 = 1010
∴ a) 10¹⁰, c) 10³ + 10¹ are divisible by ’10’.
[∵ Their units digits are ‘0’.]
b) 1024 is not divisible by 10.
[∵ Its units digit is 4.]

Question 11.
Check whether the given numbers are divisible by 5 or not      (Pg. No: 315)
a) 205   b) 4560    c) 402    d) 105    e) 235785
Answer:
a) 205, b) 4560, d) 105, e) 235785 are divisible by 5.
[∵ The units digit of the above numbers are either 0 (or) 5.]
c) 402 is not divisible by 5.
[∵ Its units digit is 2.]

Question 12.
Check whether the given numbers which are divisible by 3 or 9 or by both,      (Pg. No: 318)
a) 3663    b) 186    c) 342    d) 18871    e) 120    f) 3789    g) 4542    h) 5779782
Answer:

Question 13.
Check whether the given numbers are divisible by 6 or not.      (Pg. No: 320)
a) 1632    b) 456     c) 1008     d) 789     e) 369    f) 258
Answer:

Question 14.
Check whether the given numbers are divisible by 6 or not.     (Pg. No: 320)
a) 458 + 676    b) 6³    c) 6² + 6³    d) 2² × 3²
Answer:

Question 15.
Can you arrange the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 in an order so that the number formed by first two digits is divisible by 2, the number formed by first three digits is divisible by 3, the number formed by first four digits is divisible by 4 and so on upto nine digits?
Solution: The order 123654987 looks promising check and verify.     (Pg. No : 320)
Answer:

∴ This number can’t continue upto ‘9’.
→ 123654987

∴ The given number 123654987 is not divisible by all the numbers like 2, 3, 4, 5,……… 9.

Question 16.
Check whether the given numbers are divisible by 4 or 8 or by both 4 and 8.
a) 464    b) 782     c) 3688    d) 100     e) 1000    f) 387856    g)4⁴     h) 8³ (Pg. No: 321)
Answer:
If a number is divisible by 4 then the last two digits of the number must be divisible by 4.
If the last 3 digits of a number is divisible by 8 then it is divisible by 8.

Question 17.
Check whether the given numbers are divisible by 7. (Pg. No: 322)
a) 322     b) 588     c) 952     d) 553    e) 448
Answer:

All the given numbers are divisible by ‘7’.

Question 18.
Check whether the given numbers are divisible by 11.    (Pg. No: 323)
i) 4867216      ii) 12221     iii) 100001
Answer:
If the difference between the sum of digits of odd places and even places is divisible by 11, then entire number is divisible by 11.

Question 19.
Take different pairs of numbers and check the above four rules.      (Pg. No: 325)
Answer:
a) Consider a factor of 36, say 9.
Factors of 9 are 1,3,9.
∴ 36 is divisible by 1, 3, 9.
∴ 36 is also divisible by all the factors of 9.
b) Let us consider a number 60. It is divisible by 5 and 6. It is also divisible by 5 x 6 = 30 Where 5, 6 are co-primes.
c) Take two numbers 25, 30. These numbers are both divisible by 5.
The number 25 + 30 = 55 is also divisible by 5.
d) Take two numbers 36, 54. These numbers are both divisible by 9.
Their difference i.e., 54 – 36 = 18 is also divisible by 9.

Question 20.
144 is divisible by 12. Is it divisible by the factors of 12? Verify.     (Pg. No : 325)
Answer:
Factors of 12 = 1, 2, 3, 4, 6, 12.
If 12 is a factor of 144 then 144 is divisible by all the factors of 12.

Question 21.
Check whether 2³ + 2⁴ + 2⁵ is divisible by 2. Explain.       (Pg. No : 325)
Answer:
2³ + 2⁴ + 2⁵ = 8 + 16 + 32 = 56 is an even.
∴ 56 is divisible by 2.

Question 22.
Check whether 33 – 32 is divisible by 3. Explain    (Pg. No : 325)
Answer:
33 – 32 = 27 – 9 = 18 → 1 + 8 = 9 ⇒ \(\frac{9}{3}\) (R = 0)
∴ It is divisible by ‘3’.

Question 23.
Check the result if the numbers chosen were       (Pg. No : 328)
i) 37    ii) 60    iii) 18   iv) 89
Answer:
i) If the digits are interchanged in 37 then it becomes as 73.
∴ 37 + 73 = 110 → \(\frac{110}{11}\) (R = 0)
It is divisible by ’11’.

Question 24.
In a cricket team there are 11 players. The selection board purchased 10x + y T-shirts to players. They again purchased ‘10y + x’ T-shirts and total T-shirts were distributed to players equally. How many T-shirts will be left over after they distributed equally to 11 players ? How many each one will get?     (Pg. No : 328)
Answer:
No.of players in the team = 11
No.of T- shirts are purchased at first = 10x + y
No. of T – shirts are purchased for the 2nd time = 10y + x
Sum of the T – shirts = (10x + y) + (10y + x)
= 11x + 11y = ll(x + y)
∴ 11(x + y) T – shirts are distributed among 11 players then each will get ll(x + y)
\(\frac{11(x + y)}{11}\) = x + y
Remaining T – shirts = Purchased T – shirts – 11 (No.of T-shirts got by each)
= 11(x + y) – 11(x + y)
= 0

Question 25.
In a basket there are ‘10a + b’ fruits (a ≠ 0 and a > b). Among them ‘10b + a’ fruits are rotten. The remaining fruits distributed to 9 persons equally. How many fruits are left over after equal distribution? How many fruits would each child get?      (Pg. No: 328)
Answer:
No. of fruits in a basket = 10a + b
No. of fruits are rotten = 10b + a
Remaining fruits to be distributed = (10a + b) – (10b + a)
= 10a + b – 10b – a
= 9a – 9b = 9(a – b)
∴ 9(a – b) fruits are distributed among ‘9’ Children
then each will get = 9(a – b) ÷ 9 = \(\frac{9(a – b)}{9}\) = (a – b)
No. of fruits left over after distribution
= Total no. of fruits distributed – No.of fruits got by each
= 9(a – b) – 9(a – b) = 0

Question 26.
Check in the above activity with the following numbers.      (Pg. No: 329)
i) 657    ii) 473     iii) 167    iv) 135
Answer:

Question 27.
If 21358AB is divisible by 99, find die values of A and B.     (Pg. No: 331)
Answer:
If 21358AB is divisible by 99, then it is divisible by 9 and 11.
If 21358AB is divisible by 9 then the sum of the digits is divisible by 9.
2 + 1 + 3 + 5 + 8 + A + B = 9 × 3 say
⇒ 19 + A + B = 27
⇒ A + B = 27-19 = 8
A + B = 8 …… (1)
If 21358AB is divisible by ‘ll’ then the difference of sum of even and odd digits will be divisible by’ll’.
2 1 3 5 8 A B
∴ (2 + 3 + 8 + B) – (1 + 5 + A) = 11 × 1 say
⇒ 13 + B – 6 – A = 11
⇒ B – A = 11 – 7 = 4 ……. (2)
From (1) & (2) A = 2, B = 6
∴ The required number is 21358AB = 2135826 which is divisible by 99.

Question 28.
Find the values of A and B of file number 4AB8 (A, B are digits) which is divisible by 2, 3, 4, 6, 8 and 9.       (Pg. No: 331)
Answer:
Given number is 4AB8.
4AB8 → \(\frac{8}{2}\) (R = 0) so, it is divisible by ‘2’.
4AB8 → If it is divisible by ‘3’, sum of all the digits should be a multiple of ‘3’. .
∴ 4 + A + B + 8 = 3 or 6 or 9 or 12 or 15 …….

4AB8 → If it is divisible by 9, sum of all the digits should be a multiple of ‘9’.
∴ 4 + A + B + 8 = 9 or 18 or 27 or 36
⇒ A + B + 12 = 9 ∣18∣ 27∣ 36 ……. (3)
From (1) & (3)
A + B + 12 = 9 or 18 say
If A + B + 12 = 9
A + B = 9 – 12 = -3
It is impossible
If A + B + 12 = 18
A + B = 18 – 12 = 6
∴ A + B = 6
If A = 4 & B = 2
4AB8 = 4428
4AB8 → 4428 → \(\frac{428}{8}\) (R ≠ 0)
∴ A = 4 & B = 2 are not possible.
If A = 2& B = 4
4AB8 → 4248 → \(\frac{248}{8}\) (R = 0)
∴ A = 2 and B = 4

Question 29.
By using the above method check whether 7810364 is divisible by 4 or not.        (Pg. No: 333)
Answer:
Given number = 7810364

Sum of product of place values and remainders of place values = 0 + 0 + 0 + 0 + 0 + 12 + 4
→ \(\frac{16}{4}\) (R = 0)
∴ 7810364 is divisible by ‘4’.

Question 30.
By using the above method check whether 963451 is divisible by 6 or not.     (Pg. No: 333)
Answer:
The given number = 963451

Sum of product of place values and remainders of place values
= 36 + 24 + 12 + 16 + 20 + 1 → \(\frac{109}{6}\) (R ≠ 0)
∴ 963451 is not divisible by ‘6’.

Try these

Question 1.
In the division 56 Z ÷ 10 leaves remainder 6, what might be the value of Z.     (Pg. No: 315)
Answer:
Let 56Z, Z = 0, 1,2, 3, 4, ….. , 9 say.
To obtain remainder ‘6’ when divided by 10, Z = 6
\(\frac{566}{10}\) = \(\frac{560+6}{10}\)
Remainder is 6.

∴ Z = 6

Question 2.
If 4B ÷ 5 leaves remainder 1, what might be the value of B?    (Pg. No : 316)
Answer:
If 4B is divided by 5 then remainder should be ‘1’,
∴ B = {0, 1, 2, ….. , 9}
i.e., 40, 41, 42, …… , 49
From the above numbers we have to take 41 and 46.
41 and 46 are divided by 5 and leaves the remainder 1.
∴ B = {1, 6}

Question 3.
If 76C ÷ 5 leaves remainder 2, what might be the value of C?     (Pg. No: 316)
Answer:
To get remainder 2, when 76C is divided by 5 take C = {0, 1,……, 9}.
If C = 2, 7 then
76C = 762 or 767 are divided by 5 leaves the remainder 2.

Question 4.
“If a number is divisible by 10, it is also divisible by 5.” Is the statement true? Give reasons.     (Pg. No : 316)
Answer:
The given statement is true.
∵ When a number is divisible by ’10’, then its units digit should be ‘0’.
Similarly the units digit of a number is 5 or 0, then it is divisible by 5.
∴ The number which is divisible by 10 is also divisible by 5.

Question 5.
“If a number is divisible by 5, it is also divisible by 10.” Is the statement is true or false? Give reasons.     (Pg. No : 316)
Answer:
The given statement is false.
∵ If a number is divisible by 5, then its units digit must be ‘5’ or ‘0’. But in case of 10, it . must be ‘0’ only.
∴ The number which is divisible by ‘5’ is need not be divisible by ’10’.

Question 6.
Check whether the given numbers are divisible by 4 or 8 or by both 4 and 8.     (Pg. No : 321)
a) 4² × 8² b) 10³ c) 10⁵ + 10⁴ + 10³ d) 4³ + 4² + 4¹ – 2²
Answer:

Question 7.
Take a four digit general number, make the divisibility rule for ‘7’.     (Pg. No : 322)
Answer:
Let the 4 – digited number be ‘abcd’ say.
The remainders when divided by ‘7’,

∴ If (6a + 2b + 3c + d) is divisible by 7 then the 4 – digited number be divisible by ‘7’.

Question 8.
Check your rule with the number 3192 which is a multiple of 7.      (Pg. No : 322)
Answer:
The given number is 3192
⇒ a = 3, b = 1, c = 9, d = 2
6a + 2b + 3c + d = 6 × 3 + 2 × l + 3 × 9 + 2.
= 18 + 2 + 27 + 2
= 49 → \(\frac{49}{7}\) (R = 0)
∴ 3192 is divisible by 7’according to my law.

Question 9.
1) Verify whether 789789 is divisible by 11 or not.     (Pg. No: 323)
2) Verify whether 348348348348 is divisible by 11 or not.
3) Take an even palindrome i.e. 135531 check whether this number is divisible by 11 or not.
4) Verify whether 1234321 is divisible by 11 or not.
Answer:

Question 10.
Check whether 1576 × 1577 × 1578 is divisible by 3 or not.     (Pg. No : 325)
Answer:
The given number is 1576 × 1577 × 1578.
The product of any 3 consecutive numbers is divisible by ‘3’.
Ex : 4 × 5 × 6 = 120 → \(\frac{120}{3}\) (R = 0)
∴ 1576 × 1577 × 1578 is divisible by ’3’.

Question 11.
Check the above method applicable for the divisibility of 11 by taking 10-digit number.     (Pg. No : 326)
Answer:
The largest 10 – digited number = 9,99,99,99,999
D C B A
∴ 9/999/999/999
⇒ B + D = 9 + 999 = 1008
A + C = 999 + 999 = 1998
∴ (A + C) – (B + D) = 990 → \(\frac{990}{11}\) (R = 0)
∴ The largest 10 – digited number should be divisible by 11 according to this method.

Question 12.
Take a three digit number and make the new numbers by replacing its digits as (ABC, BCA, CAB). Now add these three numbers. For what numbers the sum of these three numbers is divisible?      (Pg. No : 329)
Answer:

Question 13.
If YE × ME = TTT find the numerical value of Y + E + M + T.
[Hint: TTT = 100T + 10T + T = T(111) = T(37 × 3)]      (Pg. No: 332)
Answer:
TTT = 100T + 10T + T
= T(111)
= T(37 × 3)
∴ YE × ME = T(37 × 3)
∴ T ={1, 2, 3, ….., 9}
But T = {3, 6, 9} are multiples of 3.
T(37 × 3) = 3(111), 6(111), 9(111) are divisible by 3.
∴ YE × ME = 333∣666∣999
YE × ME = 999 = 27 × 37
∴ Y = 2, M = 3, E = 7, T = 3
∴ Y + E + M + T = 2 + 3 + 7 + 3 = 15

Question 14.
If cost of 88 articles is A733B, find the values of A and B.       (Pg. No: 334)
Answer:
If A733B is divisible by 88 then it is divisible by 8 × 11.
Divisibility of 11:
⇒ A733B → (A + 3 + B) – (7 + 3) = 0
⇒ A + B = 7 ……. (1)
Divisibility of 8:
⇒ A733B ⇒ \(\frac{33B}{8}\)
∴ \(\frac{336}{8}\) (R = 0) (If B = 6 then it is divisible by 8)
∴ B = 6 ……. (2)
From (1), (2)
∴ A = 1, B = 6

Question 15.
Check whether 456456456456 is divisible by 7, 11 and 13.     (Pg. No: 334)
Answer:
∴ The given number = 456456456456
456456456456 = 456 (1001001001)
= 456 × (7 × 11 × 13) × (1000001)
∴ 456456456456 is divisible by 7, 11 and 13.

Think, Discuss and Write

Question 1.
Find the digit in the units place of a number if it is divided by 5 and 2 leaves the remainders 3 and 1 respectively.   (Pg. No: 316)
Answer:
If a number is divided by 5 and 2 leaves the remainders 3 and 1 respectively, then its units digit be 3.
Ex: \(\frac{13}{5}\) ⇒ (R = 3), \(\frac{13}{2}\) ⇒ (R = 1)
\(\frac{23}{5}\) ⇒ (R = 3), \(\frac{23}{2}\) ⇒ (R = 1)
∴ The unit’s digit of a required number be 3.

Question 2.
Take a two digit number reverse the digits and get another number. Subtract smaller number from bigger number. Is the difference of those two numbers is always divisible by 9?      (Pg. No : 328)
Answer:

Question 3.
1) Can we conclude 10²ⁿ – 1 is divisible by both 9 and 11? Explain.     (Pg. No: 333)
2) Is 10²ⁿ⁺¹ – 1 is divisible by 11 or not? Explain.
Answer:

Question 4.
Verify a⁵ + b⁵ is divisible by (a + b) by taking different natural numbers for ‘a’ and ‘b’.    (Pg. No : 334)
Answer:

∴ a⁵ + b⁵ is divisible by (a + b).
∴ (a⁵ + b⁵) is divisible by (a + b) for all the values of a, b.

Question 5.
Can we conclude (a²ⁿ⁺¹ + b²ⁿ⁺¹) is divisible by (a + b)?      (Pg. No : 334)
Answer:

a²ⁿ⁺¹ + b²ⁿ⁺¹ is divisible by (a + b) for all the values of ‘n’.

AP State Syllabus 8th Class Maths Solutions 14th Lesson Surface Areas and Volume (Cube-Cuboid) InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 14 Surface Areas and Volume (Cube-Cuboid) InText Questions and Answers.

8th Class Maths 14th Lesson Surface Areas and Volume (Cube-Cuboid) InText Questions and Answers

Do this

Question 1.
Find the total surface area of the following cuboid.      [Page No. 298]

Answer:
i) l = 4 cm, b = 4 cm, h = 10 cm.
The total surface area of a cuboid = 2 (lb + bh + lh)
= 2 (4 × 4 + 4 × 10 + 4 × 10) = 2(16 + 40 + 40)
= 2 × 96
= 192 Sq. cms.

ii) l = 6 cm, b = 4 cm, h = 2 cm.
The total surface area of a cuboid = 2 (lb + bh + lh)
= 2(6 × 4 + 4 × 2 + 6 × 2)
= 2 (24 + 8 + 12)
= 2 × 44
= 88 sq. cms.

Question 2.
Let us find the volume of a cuboid whose length, breadth and height are 6 cm, 4 cm and 5 cm respectively.      [Page No. 287]

Let place 1 cubic centimeter blocks along the length of the cuboid. How many blocks can we place along the length? 6 blocks, as the length of the cuboid is 6 cm.
How many blocks can we place along its breadth? 4 blocks, as the breadth of the cuboid is 4 cm. So there are 6 × 4 blocks can be placed in a layer.
How many layers of blocks can be placed in the cuboid? 5 layers, as the height of the cuboid is 5 cm. Each layer has 6 × 4 blocks. So, all the 5 layers will have 6 × 4 × 5 blocks i.e. length × breadth × height.
This discussion leads us to the formula for the volume of a cuboid.
Volume of a cuboid = length × breadth × height      [Page No. 305]
Answer:
The dimensions of a cuboid are 6 cm, 4 cm, 5 cm respectively.
∴ Volume (V) = lbh
= 6 × 4 × 5.
= 120 cm³

Question 3.
Arrange 64 unit cubes in as many ways as you can to form a cuboid. Find the surface area of each arrangement. Can solid cuboid of same volume have same surface area? [Page No. 306]
Answer:
No. of cuboids are formed using 64 unit cubes
64 = 1 × 64 ……. (1)
= 2 × 32 …….. (2)
= 4 × 16 …….. (3)
1) l = 64 cm, b = 1 cm, h = 1 cm.
The total surface area of a cuboid, A = 2 (lb + bh + lh)
= 2 (64 × 1 + 1 × 1 + 1 × 64)
= 2 (64 + 1 + 64)
= 2 × 129
= 258 Sq. cm.

2) l = 32 cm, b = 2 cm, h = 1 cm.
A = 2 (lb + bh + lh)
= 2 (32 × 2 + 2 × 1 + 32 × 1)
= 2 (64 + 2 + 32)
= 2 × 98 = 196 Sq. cm.

3) l = 16 cm, b = 4 cm, h = 1 cm.
A = 2 (lb + bh + lh)
= 2 (16 × 4 + 4 × 1 + 16 × 1)
= 2 (64 + 4 + 16)
= 2 × 84 = 168 Sq. cm.

No, the volume of a cuboid is not same as the surface area of a cuboid.

Try These

Question 1.
Find the surface area of cube ‘A’ and lateral surface area of cube ‘B’.      [Page No. 300]

Answer:
a = 10 cm.
The total surface area of a figure ‘A’ = 6a²
= 6 × (10)²
= 6 × 100 = 600 Sq. cm.
Lateral surface area of a figure ‘B’ – 4a²
= 4 × (8)² [∵ a = 8 cm.]
= 4 × 64 = 256 Sq. cm.

Question 2.
Two cubes each with side ‘b’ are joined to form a cuboid as shown in the given fig. What is the total surface area of this cuboid?      [Page No. 300]

Answer:

Total surface area of a cuboid = 2 (lb + bh + lh)
= 2 (2b × b + b × b + 2b × b)
= 2 (2b² + b² + 2b²)
= 2(5b²) = 10b² Sq. cm.

Question 3.
How will you arrange 12 cubes of equal lengths to form a cuboid of smallest surface area?      [Page No. 300]

Answer:
We can’t obtain the least total surface area by arranging 12 cubes by side by side.

∴ A = 2 (lb + bh + lh)
= 2 (12 × 1 + 1 × 1 + 12 × 1)
= 2 (12 + 1 + 12)
= 2 × 25 = 50 Sq. cm.
We can obtain the least total surface area by arranging 3 cubes by 4 cubes.
∴ A = 2 (lb + bh + lh)
= 2 (3 × 1 + 1 × 4 + 3 × 4) (∵ l = 3; b = 1; h = 4)
= 2 (3 + 4 + 12)
= 2 × 19
= 38 Sq. cm.

Question 4.
The surface area of a cube of 4 × 4 × 4 dimensions is painted. The cube is cut into 64 equal cubes. How many cubes have
(a) 1 face painted? (b) 2 faces painted? (c) 3 faces painted? (d) no face painted?       [Page No. 300]
Answer:
If the 4 × 4 × 4 cube is divided into 64 equal cubes then the length of its each side = 1 unit.

[∵ \(\frac{4 \times 4 \times 4}{64}\) = 1]
a) No.of cubes (a = 4) have painted 1 face = 6(a – 2)² = 6(4 – 2)² = 6 × 4 = 24
b) No.of cubes have painted 2 faces = 12(a – 2) = 12(4 – 2) = 24
c) No.of cubes have painted 3 faces = 4 × a = 4 × 2 = 8
d) No.of cubes have painted no faces = (a – 2)³ = (4 – 2)³ = (2)³ = 8

Think, Discuss and Write

Question 1.
Can we say that the total surface area of cuboid = lateral surface area + 2 × area of base.      [Page No. 299]
Answer:
Total surface area of a cuboid = L.S.A + 2 × Area of base
= 2h (l + b) + 2 × lb
= 2lh + 2bh + 2lb
= 2 (lb + bh + lh)
We can conclude that total surface area of a cuboid = L.S.A + 2 × Area of base

Question 2.
If we change the position of cuboid from Fig. (i) to Fig. (ii) do the lateral surface areas become equal?     [Page No. 299]

Answer:
There will be no change in the L.S.A of a cuboid if its positions are changed.

Question 3.
Draw a figure of cuboid whose dimensions are l, b, h are equal. Derive the formula for LSA and TSA.       [Page No. 299]
Answer:

Lateral surface area of a cuboid
= 4 × (areas of 4 faces)
= 2 (l × h) + 2 × (b × h) (1 + 2 + 3 + 4 faces)
= 2h(l + b) sq.units (1 = 3, 4 = 2)
∴ Total surface area of a cuboid
= 4 × (Area of 4 faces) + (Areas of upper & lower faces)
= 2h (l + b) + 2 (lb)
= 2lh + 2bh + 27b
= 2 (lb + bh + lh) sq.units.

AP State Syllabus 8th Class Maths Solutions 13th Lesson Visualizing 3-D in 2-D InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions and Answers.

8th Class Maths 13th Lesson Visualizing 3-D in 2-D InText Questions and Answers

Do This

Question 1.
Name some 3 – Dimensional objects.   [Page No. 282]
Answer:

1. Cube
2. Cylinder
3. Sphere
4. Cuboid
5. Cone

Question 2.
Give some examples of 2 – Dimensional objects.     [Page No. 282]
Answer:

1. Square
2. Rectangle
3. Line segment
4. Circle
5. Triangle

Question 3.
Draw a kite in your notebook. Is it 2 – D or 3 – D object?      [Page No. 282]
Answer:

Kite is a 2 – D object.

Question 4.
Identify some objects which are in cube or cuboid shape.      [Page No. 282]
Answer:
Shapes of Cube                           Shapes of Cuboid
a) Chalk piece box                       a) Duster
b) Dice                                         b) Cell phone (Cuboidal in shape)
c) Cube shaped cake                   c) Plasma T.V.

Question 5.
How many dimensions that a circle and sphere have?      [Page No. 282]
Answer:
Circle has 2 dimensions.
Sphere has 3 dimensions.

Question 6.
Identify the faces, edges and vertices of given figures.      [Page No. 288]

Answer:

Question 7.
Write the names of the prisms given below:     [Page No. 290]

Answer:
(i) Cube
(ii) Triangular prism
(iii) Pentagonal prism
(iv) Hexagonal prism
(v) Rectangular prism

Question 8.
Write the names of the pyramids given below:       [Page No. 290]

Answer:
(i) Square pyramid
(ii) Pentagonal pyramid
(iii) Hexagonal pyramid

Question 9.
Fill the table:      [Page No. 290]

Answer:

Question 10.
Explain the difference between prism and pyramid.      [Page No. 290]
Answer:
Upper and lower sides of a prism are equal in number. But, in a pyramid the base is a plane and all the edges are coincide in a single point on the top.

Try These

Question 1.
Name three things which are the examples of polyhedron. [Page No. 287]
Answer:

Question 2.
Name three things which are the examples of non-polyhedron. [Page No. 287]
Answer:

Think, Discuss and Write

Question 1.
How to find area and perimeter of top view and bottom view of the given figure.       [Page No. 283]

Answer:
Let the side of each face be ‘1’ unit say.
Shapes of different positions (I)                      Their areas (II)
1. Front view                                                    A = (1 × 1) + (1 × 1) + (1 × 1) = 3 Sq. Units
2. Top view                                                      A = (1 + 1 + 1) × (1 + 1) = 3 × 2 = 6 Sq. Units
3. Bottom view                                                A = (1 + 1 + 1) × (1 + 1) = 3 × 2 = 6 Sq. Units
Perimeters (III)
1. —————>                                             1 + 1 + 1 = 3 Units
2. —————>                                             2(l + b) = 2 (3 + 2) = 2 × 5 = 10 Units
3. —————>                                             2(l + b) = 2 (3 + 2) = 2 × 5 = 10 Units

AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation InText Questions and Answers.

8th Class Maths 12th Lesson Factorisation InText Questions and Answers

Do this

Question 1.
Express the given numbers in the form of product of primes. [Page No. 267]
(i) 48      (ii) 72      (iii) 96
(i) 48
Answer:
48 = 2 × 2 × 2 × 2 × 3

ii) 72
Answer:
72 = 2 × 2 × 2 × 3 × 3

iii) 96
Answer:
96 = 2 × 2 × 2 × 2 × 2 × 3

Question 2.
Find the factors of following:      [Page No. 268]
(i) 8x²yz     (ii) 2xy (x + y)       (iii) 3x + y³z
Answer:
i) 8x²yz = 2 × 2 × 2 × x × x × y × z
ii) 2xy (x + y) = 2 × x × y × (x + y)
iii) 3x + y³z = (3 × x) + (y × y × y × z)

Question 3.
Factorise:      [Page No. 270]
(i) 9a² – 6a
(ii) 15a³b – 35ab³
(iii) 7lm – 21lmn
Answer:
(i) 9a² – 6a = 3 × 3 × a × a – 2 × 3 × a
= 3 × a (3a – 2)
∴ 9a² – 6a = 3a (3a – 2)

ii) 15a³b – 35ab³
= 3 × 5 × a × a × a × b – 7 × 5 × a × b × b × b
= 5 × a × b [3 × a × a – 7 × b × b]
= 5ab [3a² – 7b²]

iii) 7lm – 21lmn
= 7 × l × m – 7 × 3 × m × n × l
= 7 × l × m [1 – 3n]
= 7lm [1 – 3n]

Question 4.
Factorise:
i) 5xy + 5x + 4y + 4
ii) 3ab + 3a + 2b + 2 [Pg. No. 271]
Answer:
i) 5xy + 5x + 4y + 4
= (5xy + 5x) + (4y + 4)
= 5x(y + 1) + 4(y + 1)
= (y + 1) (5x + 4)

ii) 3ab + 3a + 2b + 2
= [3 × a × b + 3 × a] + [2 × b + 2]
= 3 × a [b + 1] + 2 [b + 1]
= (b + 1) (3a + 2)

Think, Discuss and Write

While solving some problems containing algebraic expressions in different operations, some students solved as given below. Gan you identify the errors made by them? Write correct answers.     [Page No. 279]
Question 1.
Srilekha solved the given equation as shown below.
3x + 4x + x + 2x = 90
9x = 90 Therefore x = 10 What could say about the correctness of the solution?
Can you identify where Srilekha has gone wrong?
Answer:
Srilekha’s solution is wrong,
∵ 3x + 4x + x + 2x = 90
10x = 90
x = \(\frac{90}{10}\)
∴ x = 9

Question 2.
Abraham did the following.      [Page No. 280]
For x = -4, 7x = 7 – 4 = -3.
Answer:
Abraham’s solution is wrong.
∴ If x = -4
⇒ 7x = 7(-4) = -28

Question 3.
John and Reshma have done the multiplication of an algebraic expression by the following methods: verify whose multiplication is correct.      [Page No. 280]

Answer:

∴ John’s solutions are wrong and Reshma’s solutions are correct.

Question 4.
Harmeet does the division as (a + 5) ÷ 5 = a + 1 His friend Srikar done the same (a + 5) ÷ 5 = a/5 + 1 and his friend Rosy did it this way (a + 5) ÷ 5 = a Can you guess who has done it correctly? Justify!     [Page No. 280]
Answer:
The solutions of Harmeet, Rosy are wrong.
(a + 5) ÷ 5 = \(\frac{a+5}{5}\)
= \(\frac{a}{5}\) + \(\frac{5}{5}\)
= \(\left(\frac{a}{5}+1\right)\)
∴ Srikar had done it correctly.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions InText Questions and Answers.

8th Class Maths 11th Lesson Algebraic Expressions InText Questions and Answers

Do this

Question 1.
Find the number of terms in following algebraic expressions.
5xy², 5xy³ – 9x, 3xy + 4y – 8, 9x² + 2x + pq + q.    [Page No. 248]
Answer:

Question 2.
Take different values for x and find values of 3x + 5.     [Page No. 248]
Answer:
If x = 1 then 3x + 5 = 3(1) + 5 = 3 + 5 = 8
If x = 2 then 3x + 5 = 3(2) + 5 = 6 + 5 = 11
If x = 3 then 3x + 5 = 3(3) + 5 = 9 + 5 = 14

Question 3.
Find the like terms in the following: ax²y, 2x, 5y², -9x², -6x, 7xy, 18y².    [Pg. No. 249]
Answer:
Like terms are (2x, – 6x) (5y², 18y²).

Question 4.
Write 3 like terms for 5pq².     [Pg. No. 249]
Answer:
Like terms of 5pq² are – 3pq², pq², \(\frac{\mathrm{pq}^{2}}{2}\)etc.,

Question 5.
If A = 2y² + 3x – x², B = 3x² – y² and C = 5x² – 3xy then find          [Pg. No. 250]
(i) A + B (ii) A – B (iii) B + C (iv) B – C (v) A + B + C (vi) A + B – C
Answer:
A = 2y² + 3x – x², B = 3x² – y², C = 5x² – 3xy
i) A + B = (2y² + 3x – x²) + (3x² – y²)
= (2y² – y²) + 3x + (3x² – x²)
∴ A + B = y² + 3x + 2x² = 2x² + 3x + y²

ii) A – B = (2y² + 3x – x²) – (3x² – y²)
= 2y² + 3x – x² – 3x² + y²
∴ A – B = 3y² + 3x – 4x²

iii) B + C = (3x² – y²) + (5x² – 3xy)
= 3x² + 5x² – y² – 3xy
∴ B + C = 8x² – y² – 3xy

iv) B – C = (3x² – y²) – (5x² – 3xy)
= 3x² – y² – 5x² + 3xy
∴ B – C = – 2x² – y² + 3xy

v) A + B + C = A + (B + C)
= (2y² + 3x – x²) + (8x² – y² – 3xy)
= (8x² – x²) + (2y² – y²) + 3x – 3xy
∴ A + B + C = 7x² + y² + 3x – 3xy

vi) A + B – C = A + (B – C)
= (2y² + 3x – x²) + (-2x² – y² + 3xy)
= (2y² – y²) + (-x² – 2x²) + 3x + 3xy :
∴ A + B – C = y² – 3x² + 3x + 3xy

Question 6.
Complete the table:       [Page No. 253]

Answer:

Question 7.
Check whether you always get a monomial when two monomials are multiplied.        [Page No. 253]
Answer:
Yes, the product of two monomials is always a monomial.
Ex: 2xy × 5y = 10xy is a monomial.

Question 8.
Product of two monomials is a monomial? Check.     [Pg. No. 253]
Answer:
Yes, the product of two monomials is a monomial.
∵ 2x × y = 2xy

Question 9.
Find the product: (i) 3x(4ax + 8by) (ii) 4a²b(a – 3b) (iii) (p + Sq²) pq (iv) (m³ + n³) 5mn²       [Pg. No. 255]
Answer:
i) 3x (4ax + 8by) = 3x × 4ax + 3x × 8by
= 12ax² + 24bxy

ii) 4a²b (a – 3b) = 4a²b × a – 4a²b × 3b
= 4a³b – 12a²b²

iii) (p + 3q²) pq = p × pq + 3q² × pq
= p²q + 3pq³

iv) (m³ + n³) 5mn² = m³ × 5mn² + n³ × 5mn²
= 5m⁴n² + 5mn⁵

Question 10.
Find the number of maximum terms in the product of a monomial and a binomial?       [Pg. No. 255]
Answer:
The no.of terms in the product of a monomial and a binomial are two (2).

Question 11.
Find the product:       [Pg. No. 257]
(i) (a – b) (2a + 4b)
(ii) (3x + 2y) (3y – 4x)
(iii) (2m – l)(2l – m)
(iv) (k + 3m)(3m – k)
Answer:
i) (a – b) (2a + 4b) = a(2a + 4b) – b(2a + 4b)
= (a × 2a + a × 4b) – (b × 2a + b × 4b)
= 2a² + 4ab – (2ab + 4b²)
= 2a² + 4ab – 2ab – 4b²
= 2a² + 2ab – 4b²

ii) (3x + 2y) (3y – 4x) = 3x(3y – 4x) + 2y(3y – 4x)
= 9xy – 12x² + 6y² – 8xy
= xy – 12x² + 6y²

iii) (2m – l) (2l – m) = 2m(2l – m) – l(2l – m)
= 2m × 2l – 2m × m – l × 2l + l × m
= 4lm – 2m² – 2l² + lm
= 5lm – 2m² – 2l²

iv) (k + 3m) (3m – k) = k(3m – k) + 3m(3m – k)
= k × 3m – k × k + 3m × 3m – 3m × k
= 3km – k² + 9m² – 3km
= 9m² – k²

Question 12.
How many number of terms will be there in the product df two binomials?        [Page No. 257]
Answer:
No. of terms in the product of two binomials are 4.
Ex: (a + b) (c + d) = ac + ad + be + bd

Question 13.
Verify the following are identities by taking a, b, c as positive integers.    [Pg. No. 260]
(i) (a – b)² = a² – 2ab + b²
(ii) (a + b) (a – b) = a² – b²
(iii) (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Answer:
i) (a – b)² = a² – 2ab + b²
a = 3, b = 1
⇒ (3 – 1)² = (3)² – 2 × 3 × 1 + 1²
⇒ (2)² = 9 – 6 + 1
⇒ 4 = 4
∴ (i) is an identity,

ii) (a + b) (a – b) = a² – b²
a = 2, b = 1
⇒ (2 + 1) (2 – 1) = (2)² – (1)²
⇒ 3 × 1 = 4 – 1
⇒ 3 = 3
∴ (ii) is an identity.

iii) (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
a = 1, b = 2, c = 0
⇒ (1 + 2 + 0)² = 1² + 2² + 0² + 2 × 1 × 2 + 2 × 2 × 0 + 2 × 0 × 1
⇒ (3)² = 1 + 4 + 0 + 4 + 0 + 0
⇒ 9 = 1 + 4 + 4
⇒ 9 = 9
∴ (iii) is an identity.

Question 14.
Now take x = 2, a = 1 and b = 3, verify the identity (x + a) (x + b) s x + (a + b)x + ab.        [Pg. No. 260]
i) What do you observe? Is LHS = RHS?
ii) Take different values for x, a and b for verification of the above identity.
iii) Is it always LHS = RHS for all values of a and b?
Answer:
i) (x + a) (x + b) = x² + (a + b)x + ab
x = 2, a = 1, b = 3 then
⇒ (2 + 1) (2 + 3) = 2² + (1 + 3)² + 1 × 3
⇒ 3 × 5 = 4 + 4x² + 3
⇒ 15 = 4 + 8 + 3 ⇒ 15 = 15
∴ LHS = RHS

ii) x = 0, a = 1, b = 2 then
⇒ (0 + 1) (0 + 2) = 0² + (1 + 2) 0 + 1 × 2
⇒ 1 × 2 = 0 + 0 + 2
⇒ 2 = 2
∴ LHS = RHS for different values of x, a, b.

iii) LHS = RHS for all the values of a, b.

Question 15.
Consider (x + p) (x + q) = x + (p + q)x + pq.
(i) Put q instead of ‘p’ what do you observe?
(ii) Put p instead of ‘q’ what do you observe?
(iii) What identities you observed in your results?            [Pg. No. 261]
Answer:
i) (x + p) (x + q) = x² + (p + q)x + pq …… (1)
Substitute q instead of p in (1).
⇒ (x + q) (x + q) = x² + (q + q)x + q × q
⇒ (x + q)² = x² + 2qx + q²

ii) Substitute ‘p’ instead of q in (1).
⇒ (x + p) (x + p) = x² + (p + p)x + p × p
⇒ (x + p) = x² + 2px + p²

iii) ∴ I observe the following identities.
(x + q)² = x² + 2qx + q²
(x + p)² = x² + 2px + p²

Question 16.
Find: (i) (5m + 7n)²
(ii) (6kl + 7mn)²
(iii) (5a² + 6b²)²
(iv) 302²
(v) 807²
(vi) 704²
(vii) Verify the identity: (a – b)² = a² – 2ab + b², where a = 3m and b = 5n.         [Pg. No. 261]
Answer:
i) (5m + 7n)² is in the form of (a + b)².
(a + b)² = a² + 2ab + b² [a = 5m, b = 7n]
(5m + 7n)² = (5m)² + 2 × 5m × 7n + (7n)²
= (5m × 5m) + 70 mn + 7n × 7n
= 25m² + 70mn + 49n²

ii) (6kl + 7mn)²
We know that (a + b)² = a² + 2ab + b²
∴ (6kl + 7mn)² = (6kl)² + 2 × 6kl × 7mn + (7mn)²
= 36 k²l² + 84 klmn + 49 m²n²

iii) (5a² + 6b²)²
a = 5a², b = 6b²
(5a² + 6b²)² = (5a²)² + 2 × 5a² × 6b² + (6b²)²
= (5a² × 5a²) + 60a²b² + (6b² × 6b²)
= 25a⁴ + 60a²b² + 36b⁴

iv) (302)² = (300 + 2)²
a = 300, b = 2
∴ (300 + 2)² = (300)² + 2 × 300 × 2 + (2)²
= (300 × 300) + 1200 + (2 × 2)
= 90,000 + 1200 + 4
= 91,204

v) (807)² = (800 + 7)²
a = 800, b = 7
∴ (800 + 7)² = (800)² + 2 × 800 × 7 + (7)²
= (800 × 800) + 11,200 + (7 × 7)
= 6,40,000 + 11,200 + 49
= 6,51,249

vi) (704)² = (700 + 4)²
a = 700, b = 4
∴ (700 + 4)² = (700)² + 2 × 700 × 4 + 4²
= (700 × 700) + 5600 +(4 × 4)
= 4,90,000 + 5600 + 16
= 4,95,616

vii) (a – b)² = a² – 2ab + b² …… (1)
Substitute a = 3m, b = 5n in (1).
LHS = (3m – 5n)² = (3m)² – 2 × 3m × 5n + (5n)²
= 9m² – 30mn + 25n²
RHS = (3m)² – 2 × 3m × 5n + (5n)²
= 9m² – 30mn + 25n²
∴ LHS = RHS

Question 17.
Find:
(i)(9m – 2n)²
(ii) (6pq – 7rs)²
(iii) (5x² – 6y²)²
(iv) 292²
(v) 897²
(vi) 794²        [Pg. No. 262]
Answer:
i) (9m – 2n)² is in the form of (a – b)².
(a – b)² = a² – 2ab + b²
(9m – 2n)² = (9m)² – 2 × 9m × 2n + (2n)²
= (9m × 9m) – 36mn + (2n × 2n)
= 81m² – 36mn + 4n²

ii) (6pq – 7rs)²
a = 6pq, b = 7rs
(6pq – 7rs)² = (6pq)² – 2 × 6pq × 7rs + (7rs)²
= (6pq × 6pq) – 84pqrs + (7rs × 7rs)
= 36p²q² – 84pqrs + 49r²s²

iii) (5x² – 6y²)² = (5x²)² – 2 × 5x² × 6y² + (6y²)²
= (5x² × 5x²) – 60x²y² + (6y² × 6y²)
= 25x⁴ – 60x²y² + 36y⁴

iv) (292)² = (300 – 8)²
a = 300, b = 8
∴ (300 – 8)² = (300)² – 2 × 300 × 8 + (8)² = (300 × 300) – 4800 + (8 × 8)
= 90,000 – 4800 + 64
= 90,064 – 4800
= 85,264

v) (897)² = (900 – 3)²
= (900)² – 2 × 900 × 3 + (3)²
= 8,10,000 – 5400 + 9
= 8,10,009 – 5400
= 8,04,609

vi) (794)² = (800 – 6)²
= (800)² – 2 × 800 × 6 + (6)²
= 6,40,000 – 9600 + 36
= 6,40,036 – 9600
= 6,30,436

Question 18.
Find:
(i) (6m + 7n) (6m – 7n)
(ii) (5a + 10b) (5a – 10b)
(iii) (3x² + 4y²) (3x² – 4y²)
(iv) 106 × 94
(v) 592 × 608
(vi) 92² – 8²
(vi) 984² – 16²      [Pg. No. 262]
Answer:
i) (6m + 7n) (6m -,7n) is in the form of (a + b) (a – b). (a + b) (a – b) = a² – b²,
here a = 6m, b = 7n
(6m + 7n) (6m – 7n) = (6m)² – (7n)²
= 6m × 6m – 7n × 7n
= 36m² – 49n²

ii) (5a + 10b) (5a – 10b) = (5a)² – (10b)² [∵ (a + b) (a – b) = a² – b²]
= 5a × 5a – 10b × 10b
= 25a² – 100b²

iii) (3x² + 4y²) (3x² – 4y²)
= (3x²)² – (4y²)²
= 3x² × 3x² – 4y² × 4y²
= 9x⁴-16y⁴ [∵ (a + b) (a – b) = a² – b²]

iv) 106 × 94 = (100 + 6) (100 – 6)
= 100² – 6² = 100 × 100 – 6 × 6 [∵ (a + b) (a- b) = a²– b²]
= 10,000 – 36
= 9,964

v) 592 × 608 = (600 – 8) (600 + 8)
= (600)² – (8)²
= 600 × 600 – 8 × 8
= 3,60,000 – 64
= 3,59,936

vi) 92² – 8² is in the form of a² – b² = (a + b) (a – b).
92² – 8² = (92 + 8)(92 – 8)
= 100 × 84
= 8400

vii) 9842 – 162 = (984 + 16) (984 – 16)
= (1000) (968) [∵ (a + b)(a – b) = a² – b²]
= 9,68,000

Try These

Question 1.
Write an algebraic expression using speed and time; simple interest to be paid, using principal and the rate of simple interest.    [Pg. No. 251]
Answer:
Distance = speed × time
d = s × t

Question 2.
Can you think of two more such situations, where we can express in algebraic expressions?     [Pg. No. 251]
Answer:
Algebraic expressions are used in the following situations:
i) Area of a triangle = \(\frac{1}{2}\) × base × height = \(\frac{1}{2}\) bh
ii) Perimeter of a rectangle = 2(length + breadth) = 2(l + b)

Think, Discuss and Write

Question 1.
Sheela says the sum of 2pq and 4pq is 8p²q² is she right? Give your explanation.      [Pg. No. 249]
Answer:
The sum of 2pq and 4pq = 2pq + 4pq = 6pq
According to Sheela’s solution it is 8p²q².
6pq ≠ 8p²q²
Sheela’s solution is wrong.

Question 2.
Rehman added 4x and 7y and got 1 lxy. Do you agree with Rehman?     [Pg. No. 249]
Answer:
The sum of 4x and 7y
= (4x) + (7y)
= 4x + 7y ≠ 11xy
I do not agree with Rehman’s solution.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions InText Questions and Answers.

8th Class Maths 10th Lesson Direct and Inverse Proportions InText Questions and Answers

Do this

Question 1.
Write five more such situations where change in one quantity leads to change in another quantity.     [Page No. 231]
Answer:
The change in one quantity leads to change in another quantity will see in the following situations.

1. If speed increases then time decreases.
2. In a family, the number of persons are increased then their consumption will also increases.
3. If water consumption increases then water levels decreases.
4. If the capacity of worker’s increases then time decreases.
5. If thickness of a wire increases then its resistance decreases.

Question 2.
Write three situations where you see direct proportion.      [Page No. 233]
Answer:

1. The relation between number of students to number of teachers.
2. Number of buffaloes to their consumption of grass.
3. Number of workers to length of wall.

Question 3.
Let us consider different squares of sides 2, 3, 4 and 5 cm. Find the areas of the squares and fill the table.

What do you observe? Do you find any change in the area of the square with a change in its side? Further, find the ratio between the area of a square to the length of its side. Is the ratio same? Obviously not.
∴ This variation is not a direct proportion.     [Page No. 233]
Answer:

From the above table the ratios are not equal.
∴ So the change is not in direct proportion.
If the measure of side of a square will be change then its area also be changed.

Question 4.
The following are rectangles of equal breadth on a graph paper. Find the area for each rectangle and fill in the table.


Is the area directly proportional to length?      [Page No. 233]

Answer:
Yes, the area is directly proportional to its length.

Question 5.
Take a graph paper make same rectangles of same length and different width. Find the area for each. What can you conclude about the breadth and area?        [Page No. 233]
Answer:

Area of first rectangle (A₁) = 3 × 1 = 3 sq. cm.
Area of second rectangle (A₂) = 3 × 2 = 6 sq. cm.
∴ The relation between the areas of rectangle and breadths is in direct proportion.
[∵ \(\frac{1}{3}\) = \(\frac{2}{6}\)]

Question 6.
Measure the distance in the given map and using that calculate actual distance between (i) Vijayawada and Visakhapatnam, (ii) Tirupati and Warangal. (Scale is given)        [Page No. 235]

Answer:
i) The distance between Vijayawada and Visakhapatnam = 2 cm
According to the sum
1 cm = 300 km then 2 cm = ?
1 …… 300
2 …… ? (x)
⇒ x = 2 × 300 = 600 km
The distance between the above two cities is 600 km.
ii) The distance between Tirupathi and Warangal = 3 cm
But given that 1 cm = 300 km
3 cm = ? (x)
x = 3 × 300 = 900 km
∴ The distance between Tirupathi and Warangal = 900 km.

Question 7.
Write three situations where you see inverse proportion.      [Page No. 238]
Answer:
i) Time – work capacity
ii) Speed – distance
iii) Time – speed

Question 8.
To make rectangles of different dimensions on a squared paper using 12 adjacent squares. Calculate length and breadth of each of the rectangles so formed. Note down the values in the following table.

What do you observe? As length increases, breadth decreases and vice-versa (for constant area).
Are length and breadth inversely proportional to each other?      [Page No. 238]
Answer:
In a rectangle if length is increases then breadth is decreases and vise-versa.
∴ Length and breadth of a rectangle are in inverse proportion.

Think, discuss and write

Question 1.
Can we say that every variation is a proportion.
A book consists of 100 pages. How do the number of pages read and the number of pages left over in the book vary?

What happened to the number of left over pages, when completed pages are gradually increasing? Are they vary inversely? Explain.         [Page No. 239]
Answer:
In every situation number of pages read and number of pages left over in the book are in inverse proportion.
If number of pages read are increases then number of pages left are decreases.

AP State Syllabus 8th Class Maths Solutions 9th Lesson Area of Plane Figures InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 9 Area of Plane Figures InText Questions and Answers.

8th Class Maths 9th Lesson Area of Plane Figures InText Questions and Answers

Do this

Question 1.
Find the area of the following figures:     [Page No. 200]
i)

Answer:
Area of a parallelogram = b × h = 7 × 4 = 28 sq.cm.

ii)

Answer:
Area of a triangle = \(\frac{1}{2}\) bh = \(\frac{1}{2}\) × 7 × 4
= 14 sq.cm.

iii)

Answer:
Area of a triangle = \(\frac{1}{2}\) bh = \(\frac{1}{2}\) × 5 × 4
= 10 sq.cm.

iv)

Answer:
Area of rhombus = \(\frac{1}{2}\) d₁d₂
= \(\frac{1}{2}\) × (4+4) × (3+3)
[∴ d₁ = 4 + 4 = 8, d₂ = 3 + 3 = 6]
= \(\frac{1}{2}\) × 8 × 6
= 24 cm²

v)

Answer:
Area of a rectangle = l × b
= 20 × 14 = 280 sq.cm

vi)

Answer:
Area of a square = s²
= s × s
= 5 × 5 = 25 cm²

Question 2.
The measurements of some plane figures are given in the table below. However, they are incomplete. Find the missing information.     [Page No. 200]
Answer:

Question 3.
Find the area of the following trapezium.      [Page No. 204]
fig (i)

Answer:
Area of a trapezium

fig (ii)

Answer:
Area of a trapezium

Question 4.
Area of a trapezium is 16 cm². Length of one parallel side is 5 cm and distance between two parallel sides is 4 cm. Find the length of the other parallel side. Try to draw this trapezium on a graph paper and check the area.
[Page No. 204]
Answer:
Given that
Area of a trapezium = 16 sq.cm
Length of one of the parallel sides is a = 5 cm; h = 4 cm
Length of 2nd parallel side (b) = ?
A = \(\frac{1}{2}\)h(a + b)

Graph Sheet:

Area of parallelogram ABCD = 12 sq.cm + (S + P) + (Q + R) + (W + T) + (V + U)
= 12 + 1 + 1 + 1 + 1
= 12 + 4
= 16 sq.cm

Question 5.
ABCD is a parallelogram whose area is 100 sq.cm. P is any point insile the parallelogram (see fig.) find tie area of △APB + △CPD.       [Page No. 204]

Answer:
Area of parallelogram ABCD = 100 sq.cm
From the given figure,
ar (△APB) + ar (△CPD) = ar (△PD) + ar (△BPC)

Question 6.
The following details are noted in meters in the field book of a surveyor. Find the area of the fields.     [Page No. 213]
i)

Answer:

From the above figure
i) A, B, C, D, E are the vertices of pentagonal field,
ii) AD is the diagonal.
iii) Now the area of the field = Areas of 4 triangles and a trapezium.
PQ = AQ – AP = 50 – 30 = 20
QD = AD – AQ = 140 – 50 = 90
RD = AD – AR = 140 – 80 = 60
Area of △APB:

Area of trapezium PBCQ:

Area of △QCD:

Area of △DER:

Area of △ERA:

∴ Area of the field = ar △APB + ar trapezium PBCQ + ar △QCD + ar △DER + ar △ERA
= 450 + 800 + 2250 + 1500 + 2000 = 7000 sq. units
ii)

Answer:

From the above figure
i) A, B, C, D, E are the vertices of a pentagonal field.
ii) AC is the diagonal.
iii) The area of a field is equal to areas of 4 triangles and a trapezium.
QC = AC – AQ = 160 – 90 = 70
RC = AC – AR = 160 – 130 = 30
PR = AR – AP = 130 – 60 = 70
Area of △AQB:

Area of △QBC :

Area of △DRC :

Area of trapezium EPRD:

Area of △EPA :

∴ Area of the field = ar △AQB + ar △QBC + ar △DRC + ar trapezium EPRD + ar △EPA
= 2700 + 2100 + 450 + 2450 + 1200 = 8900 sq. units

Try these

Question 1.
We know that parallelogram is also a quadrilateral. Let us split such a quadrilateral into two triangles. Find their areas and subsequently that of the parallelogram. Does this process in turn with the formula that you already know?   [Page No. 209]

Answer:
Area of a parallelogram ABCD

Area of parallelogram ABCD
= base x height
= bh sq. units
(OR)

Area of parallelogram ABCD
= ar △ABC + ar △ACD
= \(\frac{1}{2}\) BC × h₁ + \(\frac{1}{2}\) AD × h₂
= \(\frac{1}{2}\) bh + \(\frac{1}{2}\) bh [∵ h₁ = h₂]
= bh sq. units.
∴ This process in turn with already known formula.

Question 2.
Find the area of following quadrilaterals.      [Page No. 213]
i)

Answer:
d = 6 cm, h₁ = 3 cm, h₂ = 5 cm
Area of a quadrilateral
= \(\frac{1}{2}\)d(h₁ + h₂)
= \(\frac{1}{2}\) × 6 (3 + 5) = 3(8) = 24 cm²

ii)

Answer:
d₁ = 7 cm; d₂ = 6 cm
Area of a rhombus A = \(\frac{1}{2}\) d₁d₂
= \(\frac{1}{2}\) × 7 × 6
= 7 × 3 = 21 cm²

iii)

Answer:
Area of a parallelogram (A) = bh
(∵ The given fig. is a parallelogram in which two opposite sides are parallel)
Area of a parallelogram = 2 ar AADC
= 2 × \(\frac{1}{2}\) × 8 × 2 = 16 Sq. cm.
[∵ Area of a parallelogram = ar △ADC + ar △ABC. But ar △ABC = ar △ADC]

Question 3.
i) Divide the following polygon into parts (triangles and trapezium) to find out its area.     [Page No. 214]

Answer:
FI is a diagonal of polygon EFGHI.
If perpendiculars GA, HB are drawn on the diagonal FI, then the given figure pentagon is divided into 4 parts.
∴ Area of a pentagon EFGHI = ar △AFG + ar AGHB + ar △BHI + ar △EFI.

NQ is a diagonal of polygon MNOPQR. Here the polygon is divided into two parts.
∴ Area of a hexagon MNOPQR = ar NOPQ + ar MNQR.

ii) Polygon ABCDE is divided into parts as shown in the figure. Find the area.     [Page No. 215].

If AD = 8 cm, AH = 6 cm, AF = 3 cm and perpendiculars BF = 2 cm, GH = 3 cm and EG = 2.5 cm.
Answer:
Area of polygon ABCDE = ar △AFB + ar FBCH + ar △HCD + ar △AED

So, the area of polygon ABCDE = 3 + 7.5 + 3 + 10 = 23.5 sq.cm

iii) Find the area of polygon MNOPQR if MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm, MA = 2 cm.   [Page No. 215].

NA, OD, QC and RB are perpendiculars to diagonal MP.
Answer:
Area of MNOPQR
= ar △MAN + ar ADON + ar △DOP + ar △CQP + ar BCQR + ar △MBR
Hence CP = MP – MC = 9 – 6 = 3 cm
BC = MC – MB = 6 – 4 = 2 cm
AB = MB – MA = 4 – 2 = 2 cm
DP = MP – MD = 9 – 7 = 2 cm
AD = MD – MA = 7 – 2 = 5 cm

= 2.5 + (2.5 × 5.5) + 3 + 3 + 4.5 + (2 × 2.5)
= 2.5 + 13.75 + 3 + 3 + 4.5 + 5
= 31.75 sq.cms

Think, discuss and write

Question 1.
A parallelogram is divided into two congruent triangles by drawing a diagonal across it. Can we divide a trapezium into two congruent triangles?    [Page No. 213]
Answer:

No, we cannot divide a trapezium into two congruent triangles.
∵ From the adjacent figure,
△ABC ≆ △ADC

AP State Syllabus 8th Class Maths Solutions 8th Lesson Exploring Geometrical Figures InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions and Answers.

8th Class Maths 8th Lesson Exploring Geometrical Figures InText Questions and Answers

Do this

Question 1.
Identify which of the following pairs of figures are congruent.     [Page No. 184]

Answer:
The congruent figures are (1, 10), (2, 6, 8), (3, 7), (12, 14), (9, 11), (4, 13).

Question 2.
Look at the following pairs of figures and find whether they are congruent. Give reasons. Name them.    [Page No. 185]

Answer:
i) △ABC, △PQR
∠A = ∠Q Angle
There is no information about other angles (or) sides.
But if we overlap each other, they coincide.
∴ △ABC ≅ △PQR

ii) From △PLM, △QNM
PL = QN (S)
LM = MN (S)
PM = QM (S)
By S.S.S congruency, these two triangles are congruent.
∴ △PLM ≅ △QNM

iii) From △LMN, △PQR
NL ≠ PQ,LM ≠ QR, NM ≠ RP [∵ The corresponding angles are not given]
∴ △LMN ≆ △PQR

iv) From fig. ABCD is a parallelogram and LMNO is a rectangle.
In any case a rectangle and a parallelogram are not congruent.
∴ ▱ ABCD ≆ □ DLMNO

v) Both the circles are having same radii,
i.e., r₁ = r₂ = 2 units
∴ The given circles are congruent to each other.

Question 3.
Identify the out line figures which are similar to those given first.    [Page No. 186]

Answer:
The similar figures are (a) (ii), (b) (ii).

Question 4.
Draw a triangle on a graph sheet and draw its dilation with scale factor 3. Are those two figures are similar?      [Page No. 191]
Answer:
Step – 1: Draw a △ PQR and choose the center of dilation C which is not on the triangle. Join every vertex of the triangle from C and produce.

Step – 2: By using compasses, mark three points P’, Q’ and R’ on the projections
so that
CP’ = k(CP) = 3CP
CQ’ = 3 CQ
CR’ = 3 CR

Step- 3: Join P’Q’,Q’R’and R’P’.
Notice that △P’Q’R’ ~ △PQR

Question 5.
Try to extend the projection for any other diagram and draw squares with scale factor 4, 5. What do you observe? [Page No. 191]
Answer:
Sometimes we need to enlarge 10 the figures say for example while making cutouts, and sometimes we reduce the figures during designing. Here in every case the figures must be similar to the original. This means we need to draw enlarged or reduced similar figures in daily life. This method of drawing enlarged or reduced similar figure is called ‘Dilation’.

Observe the following dilation ABCD, it is a square drawn on a graph sheet.
Every vertex A, B, C, D are joined from the sign ‘O’ and produced to 4 times the length upto A, B, C and D respectively. Then A, B, C, Dare joined to form a square which 4 times has enlarged sides of ABCD. Here, 0 is called centre of dilation and
\(\frac{OA’}{OA}\) = \(\frac{4}{1}\) = 4 is called scale factor.

Question 6.
Draw all possible lines of symmetry for the following figures.     [Page No. 193]

Answer:

Try these

Question 1.
Stretch your hand, holding a scale in your hand vertically and try to cover your school building by the scale (Adjust your distance from the building). Draw the figure and estimate height of the school building.      [Page No. 189]
Answer:
Illustration: A girl stretched her arm towards a school building, holding a scale vertically in her arm by standing at a certain distance from the school building. She found that the scale exactly covers the school building as in figure. If we compare this illustration with the previous example, we can say that

By measuring the length of the scale, length of her arm and distance of the school building, we can estimate the height of the school building.

Question 2.
Identify which of the following have point symmetry.     [Page No. 196]
1.

2. Which of the above figures are having symmetry ?
3. What can you say about the relation between line symmetry and point symmetry?
Answer:
1. The figures which have point symmetry are (i), (ii), (iii), (v).
2. (i), (iii), (v).
3. Number of lines of symmetry = Order of point symmetry.

Think, discuss and write

Question 1.
What is the relation between order of rotation and number of axes of symmetry of a geometrical figure?     [Page No. 195]
Answer:
The line which cuts symmetric figures exactly into two halves is called line of symmetry. The figure is rotated around its central point so that it appears two or more times as original. The number of times for which it appears the same is called the order of rotation.

From the above table number of lines of symmetry = Number of order of rotation.

Question 2.
How many axes of symmetry does a regular polygon has? Is there any relation between number of sides and order of rotation of a regular polygon?      [Page No. 195]
Answer:
Number of sides of a regular polygon are n. Then its lines of symmetry are also n.

AP State Syllabus 8th Class Maths Solutions 7th Lesson Frequency Distribution Tables and Graphs InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions and Answers.

8th Class Maths 7th Lesson Frequency Distribution Tables and Graphs InText Questions and Answers

Do this

Question 1.
Here are the heights of some of Indian cricketers. Find the median height of the team.   [Page No. 154]

Answer:
The ascending order of heights is 5’3″, 5’5″, 57″, 5’8″, 5’9″, 571″, 571″, 6’0″, 6’0″, 6’7″
Number of players = 10 (is an even)
Median = Mean of \(\left(\frac{\mathrm{n}}{2}\right)\) and \(\left(\frac{n}{2}+1\right)\) terms = Mean of \(\left(\frac{10}{2}\right)\) and \(\left(\frac{10}{2}+1\right)\) terms = Mean of 5, 6 terms =

Question 2.
Ages of 90 people in an apartment are given in the adjacent grouped frequency distribution.

i) How many Class Intervals are there in the table?
ii) How many people are there in the Class Interval 21 – 30?
iii) Which age group people are more in that apartment?
iv) Can we say that both people the last age group (61-70) are of 61, 70 or any other age?    [Page No. 158]
Answer:
i) 7    ii) 17     iii) 31 – 40     iv) Yes, they are 62, 63, ……, 69

Question 3.
Long jump made by 30 students of a class are tabulated as

I. Are the given class intervals inclusive or exclusive?
II. How many students are in second class interval?
III. How many students jumped a distance of 3.01 m or more?
IV. To which class interval does the student who jumped a distance of 4.005 m belongs?    [Page No. 160]
Answer:
I. Inclusive
II. 7
III. 15 + 3 + 1 = 19
IV. 401 – 500

Question 4.
Calculate the boundaries of the class intervals in the above table.     [Page No. 160]
Answer:
Boundaries:
100.5 – 200.5
2005 – 300.5
300.5 – 400.5
400.5 – 500.5
500.5 – 600.5

Question 5.
What is the length of each class interval in the above table?     [Page No. 160]
Answer:
100

Question 6.
Construct the frequency polygons of the following frequency distributions.       [Page No. 174]
i) Runs scored by students of a class in a cricket friendly match.

ii) Sale of tickets for drama in an auditorium.

Answer:
i)

Steps of construction: Runs scored (Mid values of C.I.)
Step – 1: Calculate the mid points of every class interval given in the data.
Step – 2: Draw a histogram for this data and mark the mid points of the tops of the rectangles, (like B, C, D, E, F respectively).
Step – 3: Join the mid points successively.
Step – 4: Assume a class interval before the first class and another after the last class. Also calculate their mid values (A and G) and mark on the axis. (Here, the first class is 10 – 20. So, to find the class preceding 10 – 20, we extend the horizontal axis in the negative direction and find the mid point of the imaginary class interval 60 – 70.
Step – 5: Join the first end point B to A and last end point F to G which completes the frequency polygon.
Frequency polygon can also be drawn independently without drawing histogram. For this, we require the midpoints of the class interval of the data.

ii)

Steps of construction:
Step -1: Calculate the mid points of every class interval given in the data.
Step – 2: Draw a histogram for this data and mark the mid points of the tops of the rectangles (like B, C, D, E, F respectively).
Step – 3: Join the mid points successively.
Step – 4: Assume a class interval before the first class and another after the last class.
Step – 5: To find the class preceding 2.5 – 7.5, we extend the horizontal axis in the negative direction and find the mid point of the imaginary class interval 32.5 – 37.5 like A, G.
Step – 6: Join A to B and G to F.
∴ The required ABCDEFG polygon is formed.

Try these

Question 1.
Give any three examples of data which are in situations or in numbers.      [Page No. 148]
Answer:
1) The data of 35 students who like different games:

2) The data of 35 students who like different colours:

3) The data of 35 students who like different fruits:

Question 2.
Prepare a table of estimated mean, deviations of the above cases. Observe the average of deviations with the difference of estimated mean and actual mean. What do you infer?
[Hint: Compare with average deviations]      [Page No. 151]
Answer:

Mean = \(\frac{\Sigma x_{i}}{N}\) = \(\frac{80}{5}\) = 16
Mean of the deviations = \(\frac{-5}{5}\) = -1
Mean = Assumed mean + Mean of deviations = 17 + (-1) = 16
∴ Assumed mean, original arithmetic mean are equal.

Question 3.
Estimate the arithmetic mean of the following data.      [Page No. 153]
i) 17, 25, 28, 35, 40
ii) 5, 6, 7, 8, 8, 10 10, 10, 12, 12, 13, 19, 19, 19, 20
Verify your answers by actual calculations.
Answer:
i) 17, 25, 28, 35, 40
Assumed Mean = 35

A.M in general method = \(\frac{\text { Sum of the observations }}{\text { No. of the observations }}\)

ii) 5, 6, 7, 8, 8, 10, 10, 10, 12, 12, 13, 19, 19, 19, 20
Assumed Mean = 10

Question 4.
Find the median of the data 24, 65, 85, 12, 45, 35, 15.       [Page No. 155]
Answer:
The ascending order of the data is 12, 15, 24, 35, 45, 65, 85
Number of observations (n) = 7 (odd)
∴ Median = \(\frac{n+1}{2}\) = \(\frac{7+1}{2}\) = 4th term
∴ Median = 35

Question 5.
If flie median of x, 2x, 4x is 12, then find mean of the data.       [Page No. 155]
Answer:
Given observations are x, 2x, 4x
∴ Median = 2x
According to the sum
2x = 12 ⇒ x = 6
2x = 2 × 6 = 12
4x = 4 × 6 = 24
∴ The mean of 6, 12, 24 = \(\frac{6+12+24}{3}\) = \(\frac{42}{3}\) = 14

Question 6.
If the median of the data 24, 29, 34, 38, x is 29 then the value of ‘x’ is
i) x > 38   ii) x < 29   iii) x lies in between 29 and 34   iv) none       [Page No. 155]
Answer:
Median of 24, 29, 34, 38, x is 29.
n = 5 is an odd.,
∴ Median = \(\frac{n+1}{2}\) = \(\frac{5+1}{2}\) = 3rd term
If x is less than 29, then only 29 should be a 3rd term.
∴ x < 29

Question 7.
Less than cumulative frequency is related to …….     [Page No. 165]
Answer:
Upper boundaries

Question 8.
Greater than cumulative frequency is related to ……..      [Page No. 165]
Answer:
Lower boundaries

Question 9.
Write the Less than and Greater than cumulative frequencies for the following data.       [Page No. 165]

Answer:

Question 10.
What is total frequency and less than cumulative frequency of the last class above problem? What do you infer?    [Page No. 165]
Answer:
The sum of the frequencies in the above distribution table = 30
Less than C.F of the last C.I = 30
∴ Sum of the observations = Less than C.F of last C.I.

Question 11.
Observe the adjacent histogram and answer the following questions.     [Page No. 169]

i) What information is being represented in the histogram?
ii) Which group contains maximum number of students?
iii) How many students watch TV for 5 hours or more?
iv) How many students are surveyed in total?
Answer:
i) The histogram represents students who watch the T.V.’s .(Duration of watching T.V).
ii) 4th class interval contains maximum number of students.
iii) 35 + 15 + 5 = 55
iv) Number of students are surveyed = 10 + 15 + 20 + 35 + 15 + 5 = 100

Think, discuss and write

Question 1.
Is there any change in mode, if one or two or more observations, equal to mode are included in the data?    [Page No. 155]
Answer:
If one or two or more observations equal to mode are included there will be no change in the mode.
Ex: The mode of 5, 6, 7, 8, 7, 9 is 7.
If 3, 7’s are added to above observations there will be no change in the mode.

Question 2.
Make a frequency distribution of the following series.
1, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7.    [Page No. 161]
Answer:
The range of the observations = Highest value – Least value
∴ Range = 7 – 1 = 6
If number of classes = 7 then
Class Interval = \(\frac{\text { Range }}{\text { No. of classes }}\) = \(\frac{6}{7}\) = 0.8 (approx.)

Question 3.
Construct a frequency distribution for the following series of numbers.
2, 3, 4, 6, 7, 8, 9, 9, 11, 12, 12, 13, 13, 13, 14, 14, 14, 15, 16, 17, 18, 18, 19, 20, 20, 21, 22, 24, 24, 25. (Hint: Use inclusive classes)      [Page No. 161]
Answer:
Range = Maximum value – Minimum value = 25 – 2 = 23
Class Interval = \(\frac{\text { Range }}{\text { No. of classes }}\) = \(\frac{23}{5}\) = 4.6 = 5 (approx.) [∵ No. of classes = 5]

Question 4.
What are the differences between the above two frequency distribution tables?      [Page No. 161]
Answer:
The class intervals of first frequency distribution table are exclusive class intervals. The C.I’s of 2nd frequency distribution table are inclusive class intervals.

Question 5.
From which of the frequency distributions we can write the raw data again?      [Page No. 161]
Answer:
Classes

Question 6.
All the bars (or rectangles) in a bar graph have     [Page No. 168]
a) same length b) same width c) same area d) equal value
Answer:
b) same width

Question 7.
Does the length of each bar depend on the lengths of other bars in the graphs?     [Page No. 168]
Answer:
No

Question 8.
Does the variation in the value of a bar affect the values of other bars in the same graph?      [Page No. 168]
Answer:
No

Question 9.
Where do we use vertical bar graphs and horizontal bar graphs?     [Page No. 168]
Answer:
Vertical and horizontal bar graphs are used to present the equal widths corresponding to the given frequencies.

Question 10.
Class boundaries are taken on the X-axis. Why not class limits?      [Page No. 172]
Answer:
The difference between upper and lower boundaries gives the class interval i.e., we take class boundaries on X-axis.

Question 11.
Which value decides the width of each rectangle in the histogram?      [Page No. 172]
Answer:
Class Interval

Question 12.
What does the sum of heights of all rectangles represent?     [Page No. 172]
Answer:
Sum of the frequencies.

Question 13.
How do we complete the polygon when there is no class preceding the first class?       [Page No. 173]
Answer:
The frequency of preceding class should be taken as ‘0’ (zero) then it should be joined.

Question 14.
The area of histogram of a data and its frequency polygon are same. Reason how?       [Page No. 173]
Answer:
Because both the figures are constructed on the basis of mid values of class intervals.

Question 15.
Is it necessary to draw histogram for drawing a frequency polygon?       [Page No. 173]
Answer:
No need.

Question 16.
Shall we draw a frequency polygon for frequency distribution of discrete series?       [Page No. 173]
Answer:
No, we can’t.

Question 17.
Histogram represents frequency over a class interval. Can it represent the frequency at a particular point value?            [Page No. 175]
Answer:
Yes, the histogram represents the frequency at a particular point value. Since the length of a histogram represents the value of its corresponding frequency (length of the frequency).

Question 18.
Can a frequency polygon give an idea of frequency of observations at a particular point?       [Page No. 175]
Answer:
Yes, we can identify the frequency of observation with a frequency polygon at a particular point. Since the height of the polygon is equal to frequency of polygon.

AP State Syllabus 8th Class Maths Solutions 6th Lesson Square Roots and Cube Roots InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 6 Square Roots and Cube Roots InText Questions and Answers.

8th Class Maths 6th Lesson Square Roots and Cube Roots InText Questions and Answers

Do this

Question 1.
Find the perfect squares between (i) 100 and 150 (ii) 150 and 200      [Page No. 124]
Answer:
i) The perfect squares between 100 and 150 are = 121, 144
ii) Perfect squares between 150 and 200 = 169, 196

Question 2.
Is 56 a perfect square? Give reasons.      [Page No. 124]
Answer:
Product of primes of 56 = 8 × 7 = (2 × 2) × 2 × 7
56 is not a perfect square. Since it can’t be written as product of two same numbers.

Question 3.
How many non perfect square numbers are there between 9² and 10²?      [Page No. 128]
Answer:
No. of non perfect square numbers between 9² and 10² are
= 2 × base of first number = 2 × 9 = 18
They are 82, 83, ……. 99.

Question 4.
How many non perfect square numbers are there between 15² and 16²?     [Page No. 128]
Answer:
No. of non perfect square numbers between 15 and 16 are = 2 × base of first number = 2 × 15 = 30
They are 226, 227, ……. 255,

Question 5.
Check whether the following numbers form pythagorean triplet.     [Page No. 129]
(i) 2, 3, 4
(ii) 6, 8, 10
(iii) 9, 10, 11
(iv) 8,15, 17
Answer:

Question 6.
Take a pythagorean triplet. Write their multiples. Check whether these multiples form a pythagorean triplet.      [Page No. 129]
Answer:
3, 4, 5 are pythagorean triplets.

From 6,8,10
⇒ 10² = 8² + 6²
⇒ 100 = 64 + 36
⇒ 100 = 100 (T)
From 9, 12, 5
⇒ 15² = 9² + 12²
⇒ 225 = 81 + 144
⇒ 225 = 225 (T)
∴ The multiples of pythagorean triplets are also pythagorean triplets.

Question 7.
By subtraction of successive odd numbers And whether the following numbers are perfect squares or not.        [Page No. 131]
(i) 55 (ii) 90 (iii) 121
Answer:
(i) √55
Step 1 → 55 – 1 = 54 (1st odd number be subtracted)
Step 2 → 54 – 3 = 51 (2nd odd number be subtracted)
Step 3 → 51 – 5 = 46 (3rd odd number be subtracted)
Step 4 → 46 – 7 = 39 (4th odd number be subtracted)
Step 5 → 39 – 9 = 30 (5th odd number be subtracted)
Step 6 → 30 – 11 = 19 (6th odd number be subtracted)
Step 7 → 19 – 13 = 6 (7th odd number be subtracted)
∴ 55 is not a perfect square number.
(∵ difference of consecutive odd numbers is not equal to ‘0’)

ii) √90
Step 1 → 90 – 1 =89 (1st odd number be subtracted)
Step 2 → 89 – 3 = 86 (2nd odd number be subtracted)
Step 3 → 86 – 5 = 81 (3rd odd number be subtracted)
Step 4 → 81 – 7 = 74 (4th odd number be subtracted)
Step 5 → 74 – 9 = 65 (5th odd number be subtracted)
Step 6 → 65 – 11 = 54 (6th odd number be subtracted)
Step 7 → 54 – 13 = 41 (7th odd number be subtracted)
Step 8 → 41 – 15 = 26 (8th odd number be subtracted)
Step 9 → 26 – 17 = 9 (9th odd number be subtracted)
∴ 90 is not a perfect square number.
(∵ difference of consecutive odd numbers is not equal to ‘0’)

iii) √121
Step 1 → 121 – 1 = 120 (1st odd number is subtracted)
Step 2 → 120 – 3 = 117 (2nd odd number is subtracted)
Step 3 → 117 – 5 = 112 (3rd odd number is subtracted)
Step 4 → 112 – 7 = 105 (4th odd number is subtracted)
Step 5 → 105 – 9 = 96 (5th odd number is subtracted)
Step 6 → 96 – 11 = 85 (6th odd number is subtracted)
Step 7 → 85 – 13 = 72 (7th odd number is subtracted)
Step 8 → 72 – 15 = 57 (8th odd number is subtracted)
Step 9 → 57 – 17 = 40 (9th odd number is subtracted)
Step 10 → 40 – 19 = 21 (10th odd number is subtracted)
Step 11 → 21 – 21 = 0 (11th odd number is subtracted)
∴ At the 11th step, the difference of consecutive odd numbers is ‘0’
121 is a perfect square number.
∴ √121 = \(\sqrt{11 \times 11}\) = 11 (∵ It ends at 11th step)

Question 8.
Which of the following are perfect cubes?     [Page No. 143]
(i) 243    (ii) 400    (iii) 500   (iv) 512     (v) 729
Answer:

∴ 512 and 729 are perfect cubes.

Try These

Question 1.
Guess and give reason which of the following numbers are perfect squares. Verify from the above table. (Refer table in Text Page no: 124)         [Page No. 124]
(i) 84   (ii) 108   (iii) 271   (iv) 240    (v) 529
Answer:
(i), (ii), (iii), (iv) are not perfect squares.
(v) 529 = 23 × 23
∴ 529 is a perfect square number.

Question 2.
Which of the following have one in its units place?     [Page No. 125]
(i) 126²    (ii) 179²    (iii) 281²     (iv) 363²
Answer:

Number	Square of units digit	Units digit of a number
i) 1262	(6)2 = 36	6
ii) 1792	(9)2 = 81	1
iii) 2812	(1)2 = 1	1
iv) 3632	(3)2 = 9	9

Question 3.
Which of the following have 6 in the units place?
(i) 116²    (ii) 228²    (iii) 324²    (iv) 363²        [Page No. 125]
Answer:
i) 1162 ⇒ (6)² = 36 units digit = 6
ii) 2282 ⇒ (8)² = 64 units digit = 4
iii) 3242 ⇒ (4)² = 16 units digit = 6
iv) 3632 ⇒ (3)² = 9 units digit = 9
∴ Numbers which are having ‘6’ in its unit’s digit are: (i) 116² (iii) 324²

Question 4.
Guess, how many digits are there in the squares of i) 72   ii) 103    iii) 1000        [Page No. 125]
Answer:

Question 5.

27 lies between 20 and 30
27² lies between 20² and 30²
Now find what would be 27² from the following perfect squares.      [Page No. 125]
(i)329      (ii) 525     (iii) 529    (iv) 729
Answer:
The value of (27)² = 27 × 27 = 729

Question 6.
Rehan says there are 37 non square numbers between 9² and 11². Is he right? Give your reason.       [Page No. 128 ]
Answer:
No. of (integers) non perfect square numbers between 9² and 11²
= 82, 83, ……. 100 …… 120 = 39
But 100 is a perfect square number.
∴ Required non perfect square numbers are = 39 – 1 = 38
∴ No, his assumption is wrong.

Question 7.
Is 81 a perfect cube?      [Page No. 140]
Answer:
81 = 3 × 3 × 3 × 3 = 3⁴
No, 81 is not a perfect cube.
[∵ 81 can’t be written as product of 3 same numbers.]

Question 8.
Is 125 a perfect cube?       [Page No.140]
Answer:
125 = 5 × 5 × 5 = (5)³
Yes, 125 is a perfect cube.
[∵ It can be written as product of 3 same numbers]

Question 9.
Find the digit in units place of each of the following numbers.      [Page No. 141]
(i) 75³   (ii) 123³    (iii) 157³    (iv) 198³    (v) 206³
Answer:

Number	Cube of a units digit	Units digit
i) 753	53= 125	5
ii) 1233	33 = 27	7
iii) 1573	73 = 343	3
iv) 1983	83 = 512	2
v) 2063	63 = 216	6

Think, Discuss and Write

Question 1.
Vaishnavi claims that the square of even numbers are even and that of odd are odd. Do you agree with her? Justify.  [Page No. 125]
Answer:
The square of an even number is an even
∵ The product of two even numbers is always an even.
Ex: (4)² = 4 × 4 = 16 is ah even.
The square of an odd number is an odd.
∵ The product of two odd numbers is an odd number.
Ex: 11² = 11 × 11 = 121 is an odd.

Question 2.
Observe and complete the table:      [Page No. 125]

Answer:

Question 3.
How many perfect cube numbers are present between 1 and 100,1 and 500,1 and 1000?     [Page No. 140]
Answer:
Perfect cube numbers between 1 and 100 = 8, 27, 64
Perfect cube numbers between 1 and 500 = 8, 27, 64, 125, 216, 343
Perfect cube numbers between 1 and 1000 = 8, 27, 64, 125, 216, 343, 512, 729

Question 4.
How many perfect cubes are there between 500 and 1000?      [Page No. 140]
Answer:
Perfect cubes between 500 and 1000 = 512 and 729