Kerala Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 1 सुख-दुख

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Kerala State Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 1 सुख-दुख (कविता)

सुख-दुख पाठ्यपुस्तक के प्रश्न और उत्तर

सुख दुख कविता का सारांश Kerala Syllabus 8th प्रश्ना 1.
‘सुख-दुख की खेल मिचौनी खोले जीवन अपना मुख।’ इन पंक्तियों का आशय क्या है?
Sukh Dukh Summary In Hindi Kerala Syllabus 8th
उत्तर:
सुख और दुख एक सिक्के के दो पहलू हैं। जीवन में सुख और दुख का होना स्वाभाविक है। जीवन का असली मुख सुख-दुख के मिलन में खुलता है।

Sukh Dukh Poem Summary In Hindi Kerala Syllabus 8th प्रश्ना 2.
‘फिर घन में ओझल हो शशि फिर शशि में ओझल हो घन।’ यहाँ घन और शशि किन-किन के प्रतीक हैं?
दुख कविता का सारांश Kerala Syllabus 8th
उत्तर:
यहाँ घन दुख का प्रतीक है और शशि सुख का प्रतीक है।

सुख-दुख कविता का सारांश Kerala Syllabus 8th प्रश्ना 3.
‘अविरत सुख भी उत्पीड़न’ क्या आप इससे सहमत है? क्यों?
Sukh Dukh Poem Meaning In Malayalam Kerala Syllabus 8th
उत्तर:
मैं इस कथन से सहमत हूँ। क्योंकि जीवन का असली मुख सुख-दुख पूर्ण है। हमेशा सुख हो तो जीवन की असलियत की पहचान नहीं होगी। सदा सुखी रहनेवालों के जीवन में दुख के आने पर वे उसके सामना करने में असफल हो जाते हैं।

सुख-दुख Textbook Activities

Sukh Dukh Kavita Ka Saransh Kerala Syllabus 8th प्रश्ना 1.
कविता से शब्द युग्मों का चयन करके लिखें।
Kerala Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 1 सुख-दुख 4
उत्तर:
1. सुख-दुख
2. निशा-दिवा
3. सोता-जागता
4. साँझ-उषा
5. विरह-मिलन
6. हास-अश्रु

Sukh Dukh Poem Kerala Syllabus 8th प्रश्ना 2.
वर्गपहेली की पूर्ति करें।
Kerala Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 1 सुख-दुख 5
Sankranti Sukha Dukha Kerala Syllabus 8th
दाईं ओर:
Kerala Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 1 सुख-दुख 7
1. ‘मुख’ का समानार्थी शब्द।
5. इसका अर्थ है- ‘आकाश’।
6. सुख-दुख के मिलन से यह परिपूर्ण हो जाता है।
9. यह ‘निरंतर’ का समानार्थी है।

नीचे की ओर:
Kerala Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 1 सुख-दुख 8
2. सफलता पाने के लिए इसकी ज़रूरत है।
3. ‘मेघ’ के लिए इस कविता में प्रयुक्त शब्द।
4. कविता में ‘संसार’ का प्रतीकात्मक शब्द।
7. ‘उषा’ का अर्थ।
8. ‘सुख-दुख’ किस विधा की रचना है?
10. यह कभी नहीं बोलना चाहिए।
उत्तर:
दाईं ओर:
Kerala Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 1 सुख-दुख 9
1. आनन,
5. गगन,
6. जीवन,
9. अविरत
नीचे की ओर:
Kerala Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 1 सुख-दुख 10
2. मेहनत,
3. घन,
4. जग,
7. प्रभात,
8. कविता,
10. झूठ

सुख दुख कविता का आशय Kerala Syllabus 8th प्रश्ना 3.
सुख-दुख कविता का आशय लिखें।
Kerala Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 1 सुख-दुख 11
उत्तर:
सुख-दुख कविता श्री. सुमित्रानंदन पंत की है। इसमें कवि जीवन की असलियत को समझाने का प्रयास किया है। वे कहते हैं कि जीवन सुख-दुखपूर्ण है। सुख-दुख की खेलमिचौनी से जीवन अपना वास्तविक मुख खोल देता है। सुख-दुख के मधुर मिलन से जीवन पूर्ण हो जाता है। जीवन के दुख कभी सुख में बदलता है तो कभी सुख दुख में परिणित हो जाता है। संसार अति सुख और दुख से पीड़ित है। सुख-दुख मानव जीवन में बराबर रहें। निरंतर दुख और निरंतर सुख दोनों पीड़ा देनेवाले हैं। सुख-दुख रूपी दिन-रात में संसार का जीवन सोता और जागता है। इस संध्या और उषा के आँगन में विरह और मिलन का आलिंगन हो रहा है। मानव जीवन का मुख सदा हँसी और आँसू से भरा है।

सुख-दुख Summary in Malayalam and Translation

Sukh Dukh Poem Summary In Malayalam Kerala Syllabus 8th

सुख-दुख शब्दार्थ Word meanings

Sukh Dukh Question Answer Kerala Syllabus 8th

Kerala Syllabus 8th Standard Maths Solutions Chapter 2 Equations

You can Download Equations Questions and Answers, Activity, Notes, Kerala Syllabus 8th Standard Maths Solutions Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Maths Solutions Chapter 2 Equations

Equations Text Book Questions and Answers

Textbook Page No. 34

Kerala Syllabus 8th Standard Maths Solution Chapter 2 Question 1.
“Six more marks and I would’ve got full hundred marks in the maths test.” Rajan was sad. How much mark did he actually get? the num¬ber.
Solution:
The marks Rajan got is six less from 100
∴ 100 – 6 = 94

Kerala Syllabus Class 8 Maths Solutions Chapter 2 Question 2.
Mother gave Rs.6o to Lissy for buying hooks. She gave back the 13 rupees left. For how much money did she buy books?
Solution:
Money Lissy got from mother = Rs. 6o
Amount she returned = Rs. 13
Amount Lissy used to buy books = 60 – 13 = Rs. 47

Kerala Syllabus 8th Standard Maths Notes Chapter 2 Question 3.
Gopalan bought a bunch of bananas. 7 of them were rotten which he threw away. Now there are 46. How many bananas were there in the bench?
Solution:
No. of rotten bananas = 7
No. of bananas left =46
∴ Total number of bananas
= 46 + 7 = 53

Hss Live Guru 8th Maths Chapter 2  Question 4.
Vimala spent 163 rupees shopping and now she has 217 rupees. How much money did she have at first?
Solution:
Amount spent by Vimala for shopping = Rs. 163
Amount left with her after shopping Rs. 217
Total amount = 163 + 217 = Rs. 380

Hsslive Guru 8th Class Maths Chapter 2 Question 5.
264 added to a number makes it 452. What is the number?
Solution:
Number = 452 – 264 = 188

Kerala Syllabus 8th Standard Notes Maths Chapter 2 Question 6.
198 subtracted from a number makes it 163. What is the number.
Solution:
Number = 198 + 163 = 361

Textbook Page No. 35

8th Scert Maths Solutions Chapter 2 Question 1.
In a company the manager’s salary in five times that of a peon. The manager gets Rs, 40,000 a month. How much does a peon get a month?
Solution:
Salary of the manager = Rs.40,000.
\(\frac{1}{5}\) th salary of the manager in the salary of the peon.
∴ Salary of the peon = \(\frac{40000}{5}\)
= 8000

Kerala Syllabus 8th Standard Maths Textbook Solutions Question 2.
The travellers of a picnic split equally the 5200 rupees spent. Each gave 1300 rupees. How many travellers were there?
Solution:
Total amount spent = Rs. 5200
Share of one = Rs.1300
∴ Members in the group = \(\frac{5200}{1300}\) = 4

Kerala Syllabus 8th Maths Solutions Chapter 2 Question 3.
A number multiplied by 12 gives 756. What is the number?
Solution:
Number = \(\frac{756}{12}\) = 63

Class 8 Maths Chapter 2 Kerala Syllabus Question 4.
A number divided by 21 gives 756. What is the number
Solution:
Number 756 × 21 = 15,876

Textbook Page No. 37

8th Standard Maths Guide Kerala Syllabus Chapter 2 Question 1.
Anita and her friends bought pens. For five pens bought toge¬ther, they got a discount of three rupees and it cost them 32 rupees. Had they bought the pens separately, how much would each have to spend?
Solution:
Cost for 5 pens = Rs. 32
discount = Rs.3
Real cost = 32 + 3 = 35
∴ Real cost of 1 pen = \(\frac{35}{5}\) = 7
Kerala Syllabus 8th Standard Maths Solution Chapter 2

8th Class Maths Notes Kerala Syllabus Chapter 2 Question 2.
The perimeter of a rectangle is 25 metres and one of its side is 5m. How many metres is the other side?
Solution:
Perimeter of the rectangle = 25m
One side = 5m
Double of the sum of length and breadth is perimeter = \(\frac{25}{2}\) – 5 = 7.5
Kerala Syllabus Class 8 Maths Solutions Chapter 2

Maths Class 8 Kerala Syllabus Chapter 2 Question 3.
In each of the problems below, the result of doing some operations on a number is given. Find the number?
(i) Three added to double is 101
(ii) Two added to triple is 101
(iii) Three substracted from double is 101
(iv) Two substracted from triple is 101
Solution:
Kerala Syllabus 8th Standard Maths Notes Chapter 2

Hsslive 8th Class Maths Chapter 2 Question 4.
Half a number added to the number gived III. What is the number?
Solution:
1\(\frac{1}{2}\) of the numbers = 111
ie \(\frac{3}{2}\) of the number = 111
number = 111 × \(\frac{2}{3}\) = 74

Hss Live Guru 8 Maths Chapter 2 Question 5.
A piece of folk maths. A child asked a flock of birds, ‘How many are you”? A bird replied
“We and us again
with half of us
And half of that
with one more.
would make hundred” How many birds were there?
Solution:
We and us ⇒ double (2 times)
Hss Live Guru 8th Maths Chapter 2

Textbook Page No. 41

Hsslive Guru Maths 8th Standard Chapter 2 Question 1.
The perimeter of a rectangle is 80 metre and its length is one metre more than twice the breadth. What are its length and breadth?
Solution:
If x be the breadth, Then length
= 2x + 1
2 (x + 2x + 1) = 80
2(3x + 1) = 80
6x + 2 = 80
6x = 80 – 2
x = \(\frac{80-2}{6}\) = 13.
breadth = 13 metre
Length = 2 × 13 + 1 = 27 metre

Question 2.
From a point on a line another line is to be drawn such that the angle on one side is 50° more than the angle on the other side. How much is the smaller angle?
Solution:
Let one angle is x, the second angle = x + 50
Sum of two angles on a line = 180°
ie x + x + 50 = 180°
2x + 50 = 180°, 2x = 180 – 50
x = \(\frac{180-50}{2}\) = 65
The angles are 65°, 115°

Question 3.
The price of a book is 4 rupees more than the price of a pen. The price of a pencil is 2 rupees less than the price of the pen. The total price of 5 books, 2 pens and 3 pencils is 74 rupees. What is the price of each?
Solution:
Let the cost of a pen = x
cost of a book = x + 4
Cost of a pencil = x – 2
ie.5(x + 4) + 2x + 3(x – 2) = 74
5x + 20 + 2x + 3x – 6 = 74
10 x + 14 = 74, 10x = 74 – 14
x = \(\frac{74-14}{10}\) = 6
Cost of pen = Rs. 6
Cost of book = Rs. 10
Cost of pencil = Rs. 4

Question 4.
(i) The sum of three consecutive natural numbers is 36. What are the numbers
(ii) The sum of three consecutive even numbes is 36. What are the numbers?
(iii) Can the sum of three consecutive odd numbers be 36. Why?
(iv) The sum of three consecutive odd numbers is What are the numbers?
(v) The sum of three consecutive natural numbers is 33. What are the numbers?
Solution:
x + x + 1 + x + 2 = 36
3x + 3 = 36
x = \(\frac{36-3}{3}\) = 11
∴ Numbers are 11, 12, 13
x – 1 + x + x + 1 = 36
3x = 36
x = 12
Or
∴ Numbers are 11, 12, 13

(ii) (x – 2) + x + (x + 2) = 36
3x = 36
x = 12
∴ The numbers are 10, 12, 14

(iii) The sum of three odd numbers is odd.
∴ It is not possible to get 36 as the sum of three odd numbers.

(iv) (x – 2) + x + (x + 2) = 33
3x = 33
x = 11
∴ Numbers are 9, 11, 13

(v) (x – 1)+ x + (x + 1) = 33
3x = 33
x = 11
The numbers are 10, 11, 12

Question 5.
(i) In a calender, a square of four numbers is marked. The sum of the numbers is 80. What are the numbers?
(ii) A square of nine numbers is marked in a calendar. The sum of all there numbers is 90. What are the numbers?
Solution:
(i) Let,
x x + 1
x + 7 x + 8
four numbers in the square of 4 numbers.
Sum of the numbers = x + (x + 1) + (x + 7) + (x + 8) = 80
4x + 16 = 80, 4x = 80 – 16
Hsslive Guru 8th Class Maths Chapter 2

Textbook Page No. 44

Question 1.
Ticket rate for the science exhibition is rupees 10 for a child and 25 rupees for the adult. Rs. 740 was got from 50 persons. How many children among them?
Solution:
Let the numbers of children be x
Then the number of adult = 50 – x
∴ 10 x + 25 (50 – x) = 740
10 x + 1250 – 25 x = 740
1250 – 15 x = 740
15 x = 1250 – 740
x = \(\frac{1250-740}{15}\) = 34
Number of adults = 50 – 34 = 16

Question 2.
A class has the same numbers of girls and boys. Only 8 boys were absent on a particular day and then the number of girls was double the number of boys and girls?
Solution:
Let the number of boys =number of girls = x
2 (x – 8) = x
2x – 16 = x, 2x – x = 16
x – 16 = 0
∴ x = 16
∴ The number of boys = No. of girls = 16

Question 3.
Ajayan is ten years older than Vijayan. Next year Ajayan’s age would be double that of Vijayan. What are their ages now?
Solution:
Let the age of Vijayan = x
Age of Ajayan = x + 10
Age of Vijayan after an year = x + 1
Age of Ajayan after an year = x + 11
2 (x + 1) = x + 11
2x + 2 = x + 11
2x – x = 11 – 2
x = 9
∴ Age of Vijayan = 9
Age of Ajayan = 19

Question 4.
Five times a number is equal to three times the sum of the number and 4. What is the number?
Solution:
Let the number = x
Five times number = 5x
4 more than the number = x + 4
Three times of it = 3(x + 4)
ie. 5 x = 3 (x + 4)
5 x = 3x + 12
5 x – 3 x = 12
2 x = 12
x = 6

Question 5.
In a co-operative society, the number of men is thrice the number of women 29 women and 16 men more joined the society and now the number of men is double the number of women. How many women were there in the society at first?
Solution:
Let the number of women = x
No of men = 3 x
No. of women when 29 more were joined = x + 29
No. of men when 16 more were joined = 3x + 16
3 x + 16 = 2 (x + 29)
3x + 16 = 2x + 58
3x – 2x = 58 – 16
x = 42
No. of women = 42;
No. of men = 3 × 42 = 126

Equations Additional Questions & Answers

Question 1.
“If you give rupees 5 to me both of us have equal amounts with us”. Ajay told to Vineeth. Then Vineeth told to Ajay “ If you give rupees 5 to me I will have 5 times more money than yours. Find out the amount both of them have?
Solution:
Let Ajay has Rs.x with him and when Vineeth gives Rs. 5, both of them have x + 5 rupees with them. If Vineeth gets RS.5 he has 5 times more than Ajay.
x + 5 + 5 = 5x
4 x = 10
x = \(\frac{10}{4}=\frac{5}{2}\) = 2.5 Rupees
Amount with Ajay = Rs. 2.50
Amount with Vineeth = Rs.12.50

Question 2.
The perimeter of a triangle is 49 cm. One side is 7 cm more than the second side and 5 cm less than the third side. Find out the lengths of three sides?
Solution:
Let the second side = xcm
First side = x + 7 cm
Third side = x + 12 cm
Perimeter = x + x + 7 + x + 12 = 49
= 3x = 49 – 19 = 30
3x = 20, x = 10 cm
The sides are = 10 cm, 17 cm, 22 cm

Question 3.
The sum of two numbers is 9, 8 times the smaller number is 2 more than 6 times the bigger number. Write an equation to find the numbers?
Solution:
Let the smaller number = x
Bigger number = 9 – x
Equation is, 8 x = 6 (9 – x) + 2

Question 4.
Anitha and friends bought pens. When they bought a packet of 5 pens they got a discount of rupees 2. They paid rupees 18. Find the cost of each one buys separately?
Solution:
Total cost in including discount = Rs. 20
Cost of 5 pens = Rs.20
Cost of 1 pen = \(\frac{20}{5}\) = 4

Question 5.
The cost of a chair and a table is Rs. 1500. The cost of table is 4 times the cost of chair. Find the expense of each?
Solution:
Let the cost of chair = x
Cost of table = 4x
x + 4 x = 1500
5x = 1500
x = 300
cost of chair = Rs. 300
Cost of table = Rs. 1200

Question 6.
There were 25 questions in the examination written by Jafar. Each correct answer gets 2 marks. There is a loss of mark for each wrong answer. Jafar answered all the questions. He got 35 marks. Find^ut the number of correct answers?
Solution:
Let the number of correct answers = x
Total number of questions = 25
No. of wrong answers = 25 – x
Marks for correct answers = 2x
marks losses for wrong answers = 25
ie. 2x – (25 – x) = 25
2x – 25 + x = 35
3x = 60
x = \(\frac{60}{3}\) = 20, ∴ number of correct answers = 20.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 20 Static Electricity in Malayalam

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Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 20 Static Electricity in Malayalam

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Kerala Syllabus 8th Standard Basic Science Solutions Chapter 3 Let’s Regain Our Fields

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Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 3 Let’s Regain Our Fields

Let’s Regain our Fields Questions and Answers

Let’s Regain Our Fields Kerala Syllabus 8th Food Safety

Food Safety is a condition in which every individual are provided with food enough to lead a healthy life. It is very’ essential to ensure food safety to create a society in which there is no fear of poverty and health issues due to malnutrition. It is a challenge to ensure food safety in a condition in which the agricultural fields are declining. A culture that loves soil and agriculture has to be recreated. It is also important to regain the lost agricultural fields.

Indicators (Text Book Page No:35)

Lets Regain Our Fields Notes Kerala Syllabus 8th Question 1.
Didn’t you notice the illustration and the newspaper report?
Answer:
Population explosion, insufficient agricultural fields, negative attitude towards agriculture, unscientific vision, methods of agriculture, influence of consumer society, etc., are the causes of scarcity of food.

Lets Regain Our Fields Notes Pdf Kerala Syllabus 8th Question 2.
What is the concept indicated by the illustration?
Answer:
Modern techniques of agriculture play a major role to overcome the problems and crises mentioned. The techniques agriculture like polyhouse farming, open precision farming, integrated pest management, hydroponics. aeroponics etc are the contributions of science. The novel information and findings form remedy for food scarcity.

Let’s Regain Our Fields Pdf Kerala Syllabus 8th Question 3.
Discuss it with your friends using the given indicators. Write your inferences in the science diary.
Answer:
The major goal of food safety is to eliminate hunger and poverty from our country. Supplying food grains in cheaper rates is ultimately beneficial to the people of the lower strata.

Indicators (Text Book Page No:36)

Class 8 Biology Chapter 3 Kerala Syllabus Question 4.
What were the changes that occurred in the area of agricultural fields from the year 1971 to 2011?
Answer:
During the period 1971-2011, the area of agricultural fields considerably reduced. In 1971 the area was 8.75 lakh hectors and in 1991 it was reduced to 5.5 lakh hectors. It was reduced to 2.08 lakh hectors in the year 2011.

Let’s Regain Our Fields Notes Kerala Syllabus 8th Question 5.
What tendency could be observed in rice production and population growth during the period?
Answer:
In this period population increased from 2.13 crores to 3.34 crores. But the production of rice steeply reduced from 13.65 lakh tons to 5.69 lakh tons.

8th Class Biology Notes Pdf Kerala Syllabus Question 6.
Is this tendency desirable? Why?
Answer:
This tendency is not advisable. The increase in population and decrease in the area of crop field and the production of rice may lead the nation to poverty. The goal of food safety will not be attainable.

Hss Live Guru 8 Biology Kerala Syllabus Question 7.
What are the obstacles faced by farmers today?
Basic Science Question and Answer:
The Crises in the field of Agriculture

Farmers face a number of crises. They are mainly loss, cost of production, limited cultivable land, climate change, exploitation of brokers, environmental destruction and health issues, etc. Many of these problems can be overcome if a positive attitude towards agriculture develop. There are many possibilities to solve every problems.

i. Fertile soil

Around 20 different elements are required for the growth of plants. They are called essential elements.
eg: Carbon, Hydrogen, Oxygen, Nitrogen, Phosphorus, Potassium, Sulphur

These elements are naturally available in the soil by the decomposition .of microorganisms. The fertility of soil can be increased by proper manuring. Organisms like bacteria, fungi, algae, termite, earthworm, etc., help to increase the fertility of soil.

Indicators (Text Book Page No:38)

Hss Live Guru 8th Biology Kerala Syllabus Question 8.
What is the role of microorganisms in ensuring the natural availability of elements in the soil?
Answer:
The elements are naturally available in the soil by the decaying action of micro-organisms. When leguminous plants are grown in fields, the microorganisms that harbour in the root nodules fixes atmospheric nitrogen to the soil, bacteria, fungi, algae, termite, earthworm, etc., increase the fertility of the soil.

Hsslive Guru Biology 8th Kerala Syllabus Question 9.
What is the need of testing the soil?
Answer:
Soil analysis (soil testing) is done in order to detect the amount of elements present and their pH value of the soil.

Let’s Regain Our Fields Pdf Download Question 10.
Why does the application of fertilizers become essential for better crop yield?
Answer:
Fertility of the soil is increased by proper manuring. Manuring helps to make up the deficiency of elements in the soil that are required for plant growth. As a result better yield is obtained.

Lets Regain Our Fields Pdf Kerala Syllabus 8th Microbial Fertilizer

Microbial fertilizers are substances that contain micro-organisms that facilitate the fertility of the soil. Micro-organisms increase the amount factors, required for plant growth, in the soil. The bacteria like rhizobium, Azotobacter, azospirillum, and the aquatic plant azolla increase the amount of nitrogen in soil. Many things are to be taken care to retain the microorganisms in the soil. They are

  • Ensure the availability of organic fertilizer(biofertilizers) in the soil.
  • Sufficient water supply must be there.
  • Do not use chemical fertilizers and pesticides.

Consequences of Unscientific application of fertilizers

Indicators (Text Book Page No:39)

Kerala Syllabus 8th Standard Biology Notes Question 11.
What are the consequences of unscientific application of chemical fertilizers? Discuss on the basis of the following indicators.

  • composition of soil
  • microorganisms in soil
  • health issues
  • financial factors

Answer:
Unscientific fertigation loses the fertility and changes the natural texture of soil. Certain elements required for plant growth will exceed in the soil and certain others become insufficient. Chemical fertilizers do not provide all the elements required for plant growth in proper amount and proportion.

Excessive fertigation and the use of pesticides kill the microorganisms in the soil. Thus the natural fertility of the soil loses.

The chemical substances present in chemical fertilizers get accumulated in agricultural crops (through biological magnification). It causes many chronic diseases in organisms including man who consume this food.

Chemical fertilizers and pesticides are expensive. Fertilizers are manufactured by small scale industries as well, as multinational companies. The expenses is not affordable by the farmers. Thus unscientific application of fertilizers causes consequences of varied dimensions.

Hsslive Guru 8th Class Biology Kerala Syllabus Pest Control

Many methods are adapted today to control pests. Chemical pesticides are widely used. Chemical pesticides kill the pest as a whole. Use of chemical fertilizers cause a number environmental and health problems. High amount of chemical fertilizers are reported in ground water too.
Another possibility is to control pests using ultrasonic sound waves.
Using radiation the reproductive capacity of male pests can be lost and makes pest control effective.
Pheromone traps like devices also makes pest control effective. Pest control using the natural enemies of pest is highly effective and having no environment and health consequences.

Natural Enemies of Pest (Friendly Pests)

Organisms that eat pests, cause disease to them or parasitises on them are called their natural enemies.

8th Std Biology Notes Kerala Syllabus Integrated Pest Management

In this method, the use of chemical pesticides is highly reduced. Pest control is made possible by the combined use of biological pesticides, friendly pests, mechanical control methods, etc. It will not disturb the equilibrium of the ecosystem.
It do not destroy pests as a whole. But it prevents the multiplication of pests and the number of pest is cortrolled to prevent crop loss.

Advantages of Integrated Pest Management

  • Do not kill the pests as a whole
  • Use of chemical pesticides is highly reduced.
  • No environmental or health issues as friendly pests, mechanical control measures, Biological pesticides, etc. are used.
  • Does not disturb the equilibrium of ecosystem.

Waste Management

Indicators (Text Book Page No:41)

Hss Live Guru 8 Physics Kerala Syllabus Question 12.
Waste management and sustainable agriculture
Answer:
Live Stock management, Poultry farming, Pisciculture etc, help not only to earn income but also to the treatment of biological wastes. Composting, biogas production etc, are possible by using bio wastes. Cow dung is a very good biological manure and an essential component . for the production of biogas. By preparing cattle feed, fish food, poultry feed, etc., from bio waste more earnings can be done and biological waste management also possible.
Let's Regain Our Fields Kerala Syllabus 8th

Hsslive Guru 8th Basic Science Kerala Syllabus Certain methods of Agriculture

Many farming techniques that help to earn improved income by scientific approach are in practice.
eg: Rearing of cattle’s, Poultry farming, Sericulture, Pisciculture, Floriculture, Apiculture, Cuniculture, Mushroom culture, Horticulture, Medicinal plant cuitivation etc.
→ Live stock Management
Rearing of cattle’s for milk, meat and agricultural purposes.
→ Poultry farming
Rearing of birds for egg, meat etc.
→ Sericulture
Rearing of silk worms
→ Pisciculture
Rearing of fishes
→ Floriculture
Cultivation of flowers for commercial purposes.
→ Apiculture
Rearing of honeybees
→ Cuniculture
Rearing of rabbits
→ Mushroom culture
Cultivation of mushrooms .
→ Horti culture
Cultivation of fruits and vegetables.

Completion of Table(Text Book Page No:44)

Hss Live Guru Biology 8 Kerala Syllabus Question 13.
Complete the following table related to various agricultural sectors.
Lets Regain Our Fields Notes Kerala Syllabus 8th

Answer:

Areas Products Varieties
Pisciculture fish, fish liver oil Pearl spot, Rohu
Apiculture Honey, Wax Kolan, Mellifera Njodiyan
Mushroom culture Edible mushroom Milk mushroom, Button mushroom
Livestock management Milk, Meat jercy(Cow) Murrah (Buffalo) Jamnapari (Goat)
Cuniculture Meat Grey giant, White giant
Sericulture Silk thread Mulberry silkworm Tussar silkworm Muga silkworm
Poultry farming Egg, Meat Athulyaf(hen) Muscovy (duck) Bobwhite (Quail)

Modern Techniques of Agriculture
1. Polyhouse farming
Polyhouse is a special arrangement made by completely or partially covering transparent sheet like polythene. Humidity and temperature kept constant in polyhouse. So plant growth will be fast. Nutrients are dissolved in water and sprayed. Pest attack also will be less.

2. Open Precision farming
This is a technique in which the nature of soil in the agricultural land, quantity of elements in the soil, presence of water in the soil, etc., are studied accurately with the help of modern technology and suitable crops are selected for cultivation. The attack of weeds also is less because the soil is covered using polythene sheets. Water supply or irrigation can be limited.

3. Hydroponics
Plants are grown in nutrient solution.
4. Aeroponics
Plants are grown in such a way that their roots are grown towards air and nutrients are directly sprayed to the roots.

Indicators (Text Book Page No:46)

Hss Live Guru 8th Basic Science Kerala Syllabus Question 14.
How are modern agricultural practices helpful in reducing crop loss due to climate change?
Answer:
In arrangements like polyhouses temperature and humidity are maintained constant. Plant grows fast and gets better yield. Polyhouse farming helps to reduce crop less due to climatic changes.

Basic Science Class 8 Chapter 3 Kerala Syllabus Question 15.
What are the advantages of precision farming?
Answer:
This is a technique in which the nature of soil in the agricultural land, quantity of elements in the soil, presence of water in the soil etc., are studied accurately with the help of modem technology and suitable crops are selected for cultivation. The attack of weeds also is less because the soil is covered using polythene sheets. Water supply or irrigation can be limited.

Class 8 Science Notes Kerala Syllabus Question 16.
How does cultivation become possible without depending on soil?
Answer:
• Hydroponics and aeroponics are soilless cultivation methods.
• Plants are grown in nutrient solution.
• Aeroponic plants are grown in such a way that their roots are grown towards air and nutrients are directly sprayed to the roots.

Farmers groups

Nowadays farmers groups, that provide farmers an opportunity to sell and buy their products without brokers mediators are very lively. Online gathering of farmers also is widespread as the demand for organic product raised. These online groups help to find customers for the quality organic products and to get good price for the products.

Completion of Table(Text Book Page No:51)

Question 17.
Complete the following table, adding important ideas.
Lets Regain Our Fields Notes Pdf Kerala Syllabus 8th

Answer:

Challenges Remedies
Climatic change • Polyhouse farming
• Hydroponics
Environmental distruction and health hazards • Scientific manuring
• Integrated Pest manage ment. Biological waste Management
Cost of production • Microbial fertilizers
• Integrated farming
Crop loss • Polyhouse farming
• Open precision  farming
• Integrated Pest management
Limited space • Hydroponics
• Aeroponics
• Terrace farming Loss
Loss • Farmers societies

• Online gatherings

Let’s Regain Our Fields Let us Assess (Text Book Page No:52)

Question 1.
Cuniculture is related to
a. Keeping of honey bees
b. Rearing of rabbits
c. Cultivation of fruits and vegetables
d. Rearing of fish
Answer:
Rearing of Rabbits

Question 2.
High quality hybrid varieties provide high yield. Then, what is the need of native varieties? Record your response to this statement.
Answer:
The native varieties exist in a particular locality by acquiring natural resistance and adaptations to the climate, availability of food, nature of soil etc., though they have low productivity.
The native varieties have high resis-tance to diseases and environmental conditions. The extinction of native varieties leads to the depletion of our biodiversity. New varieties can be developed only from the native varieties.

 

Question 3.
Which is the most appropriate way to reduce crop loss due to pests?
a. Using high concentration pesticides
b. Protecting friendly pests.
c. Practicing integrated pest management
d. Applying organic pesticides only.
Answer:
C  Adopt integrated Pest control

Question 4.
‘Lower price during higher yield’. Suggest a practical solution to overcome this crisis faced by farmers.
Answer:
Collect and distribute resources through farmers societies.

Let’s Regain Our Fields Additional Questions and Answers

Question 1.
Find the odd one in each group.
Also write the common characterestic of the others.
a. Compost, Microbial fertilizer, urea, Bone-meal
b. Anthurium, Sida, Ramacham, Koovalam
c. Boer, Litchi, Rambutan, Durian
d. Phosphorus, Potassium, Nitrogen, Azetohactor
e. Rock Phosphate, Factompho- se, Muriate of Potash, Malatheon,
f. Kuthiravaly, Jyothy, Thriveni, Jaya
g. Fowl Cholera, Anthrax, Chores, Ranikatt
h. Blight disease, Root wilt, Bud rot
i. Ammonia, Urea, Compost, Factomphose.
j. APIS (njodiyan); Naaran, Melliferra.
k. Minorka, Royans, Ankona, Gramalakshmi.
Answer:
a.Urea – This is a chemical fertilizer. Others are Bio-fertilizers.
b. Anthurium – This is an ornamental plant. Others are medicinal plants.
c. Boer – It is a variety of goat. Others are fruit varieties.
d. Azetohactor – Others are essential elements.
e. Malatheon – This is a chemical pes ticide others are chemical fertilizers.
f. Kuthiravaly – This a hybrid vari ety pepper plant others are vari eties of paddy.
g. Anthrax : This is a cattle disease others are fowl diseases.
h. Blight disease, It is affect on rice plant others are diseases of coconut tree.
i. Compost – This is an organic fertiliser, others are chemical fertilisers
j. Naaran. This is prawn.Others are Honey bees used in apiculture.
k. Royans is good quality ducks, Others are fowl varieties.

Question 2.
Find out the relation between the given word pairs and on that basis fill in the blanks.
a. Rearing of honey bees : Apiculture; Rearing of silkworms : …………..
b. Malabari : Goat; Vechoor : …………..
c. Cowdung: ………….. ; Azospirillum : Microbial Fertilizer
d. Jersey: Cow; Murrah: …………..
e. Mushroom culture : Growing mushroom; Cuniculture : …………..
f. Malatheon : Chemical pesticide; Neem seed kemal: …………..
g. ………….. – Virus; Anthrax Bacteria.
h. Native variety : vechoor; Hybrid variety : …………..
i. Paddy- ………….. ; Arecanut plant : Mahali
j. Cocount caterpillar – ………….. ; Gamboosia- Larvae of mosqitoes
k. …………..: Domestic hen; Pekkins : Domestic Duck.
l. Blight disease- Paddy; Quick wilt- …………..
m. Organic fertiliser – Vermiw- ash; ………….. – Factomphose.
Answer:
a. Sericulture
b. Cow
c. Organic fertilizer
d. Buffalo
e. Rabbit farming
f. Organic pesticide
g. Foot and Mouth disease
h. Jersey cross
i. Blast disease
j. Ichneumon wasp
k. Royans is good quality ducks, Others are fowl varieties.
l. pepper.
m. chemical fertilisers

Question 3.
Arrange the following items from column B ,C with column A.

A
Agricultural sectors
B
Speciality
C
Varieties
a. Horticulture i. scientific way of rea­ring of honey 1. Ankora
b. Apiculture ii. scientific way of rearing of rabbits 2. Litchi
c. Cuniculture iii. scientific way of rearing silkworm 3. Njodiyan
iv. scientific cultiva­tion of fruits and 4. Naran

Answer:

A
Agricultural sectors
B
Speciality
C
Varieties
Horticulture scientific way of rea­ring of honey Litchi
Apiculture scientific way of rearing of rabbits Njodiyan
Cuniculture scientific way of rearing silkworm Ankora

Question 4.
Group the following statements into suitable for Polyhouse farming and Precision farming.
a. limiting the irrigation by covering soil with polyethene sheet.
b. Cover the field completely or partially by transparent polyethene sheet.
c. By regulating temperature and moisture constantly.
d. Selecting appropriate crop for agriculture only jifter understanding characters of soil, amount of elements present in soil and the presence of water.
Let's Regain Our Fields Pdf Kerala Syllabus 8th
Answer:

  • polyhouse farming – (b), (c)
  • Precision farming – (a), (d)

Question 5.
“It is essential to retain indegenious species of mango tree like Muvandan, Kilichundan even many hybrid varieties are avail-able”. Write your opinion to the farmer’s statement? Justify your answer?
Answer:

  • Farmer’s opinion is right.
  • Indigenous varieties of a locality are varieties that acquire natural immunity by adapting to the climate, the availability of food, soil texture of the place etc.
  • We can develop new high quality varieties only from indigenous varieties.

Question 6.
Find suitable term and fill the blanks;
Moovandan – Mango Tree Kasaragodu Dwaf – a. …………..
Njalippoovan – Musa Attappadi black – b. …………….
Answer:
a. Cow
b. Goat

Question 7.
Sustainable farming is an environment friendly method. Explain the reason.
Answer:
The excessive use of chemical fertilizers and pesticides may give increased profit. But this will not last long. The continuous use may spoil the natural fertility of the soil and the farmland may be changed into a barren land. By integrated cropping method, the use of outside manures, pesticides etc can be reduced. The wastes of one can be used as a manure or food for some other one. This will help to maintain the natural fertility of the soil. Moreover, the biodiversity also is conserved.

Question 8.
The practice of cultivating fruits and vegetables.
Answer:
Horticulture

Question 9.
Some statements regarding modern agricultural practices are given below. Which agricultural practice is related to this?
a. The method of farming in which nutrients are dissolved in water and are supplied on plants through dip irrigation.
b. The method plants are grown in nutrients solution
c. The method of farming by covering the rool using polythene sheets and by limiting.
Answer:
a. Polyhouse farming
b. hydroponics
c. precision farming

Question 10.
Polyhouse farming will be advantageous only in farmlands where cultivation is continuously maintained. Why?
Answer:
The cost of making a polyhouse is very high. But the yield from the crops will increase substantially.

Question 11.
Which of the following is not desirable in integrated pest control method?
a. Mechanical Pest Control
b. Excessive use of chemical pesticide
c. Friendly pests
d. Use of biopesticides
Answer:
b. Excessive use of chemical pesticide

Question 12.
What is the difference between polyhouse farming and open precision farming
Answer:
Poly house farming

  • Agricultural land is partially or completely covered by transparent polythene sheet.
  • Heat and humidity are kept constant
  • Nutrients are dissolved in water and given to plants through drip fertigation. Pest attack is compara-tively low.

Open Precision farming

  • Soil is covered by polythene sheet
  • Nature of soil, amount of elements in soil, pH of soil, presence of water etc are studied accurately using modern technology
  • Limited irrigation is needed.
  • Weed control is effective

Question 13.
What are the main characteristics of hybrid varieties?
Answer:
High yield, disease resisting capacity ability to give high yield within a short period etc. are the characteristics of hybrid varieties.

Question 14.
What are the consequences of unscientific application of chemical fertilizers?
Answer:

  • Financial loss.
  • Chemical pollution.
  • Destruction of microorganisms
  • Health issues.

Question 15.
How can grow plants without soil?
Answer:
Science has proved that cultivation is possible in the absence of soil, for example aeroponics and hydroponics. In hydroponics, plants are grow in nutrients solution. In aeroponics, plants are grow in such a way that their roots grow into air and nutrients are sprayed directly on roots.

Question 16.
What are the different varieties of buffaloes?
Answer:
Bhadawari, Jaffrabadi, Marwari, pashmina, malabari, Beetance.

Question 17.
Find out the odd one. Write the common feature of others. Rhizobium, Azetobacter, Lacto bacillus, Azospirillum
Answer:
Lactobacillus – others are nitrogen-fixing bacteria.

Question 18.
Arrange the organisms properly in the given table. Names are mentioned in the box.
Class 8 Biology Chapter 3 Kerala Syllabus
Answer:
a. Jersey, Murrah
b. Athulya, Muscovy
c. Naran, Kara
d. Grey giant, Ankora
e. Muga, Tusser
f. Mellifera, Kolan

Question 19.
What is the difference between hydroponics and aeroponics?
Answer:

  • Hydroponics is the technique by which plants are grown in nutrient solution
  • In aeroponics plants are grown in such a way that their roots are penetrating towards air and nutrients are directly sprayed to their roots.

Question 20.
What are the advantages of precision farming?
Answer:
In this method of farming the nature of soil, quality of elements in the soil, pH value of soil, presence of water etc, in the crop field are tested using modem technology and appropriate crops are selected for cultivation.

Question 21.
What are the advantage of house farming and family farming
Answer:
a. Nontoxic food
b. Maximum utilization of land/space
c. Exercise
d. Mental Pleasure
e. Collaborative work of family members

Question 22.
How do microbial fertilizers help in plant growth?
Answer:
Microbial fertilizers contains micro organisms that increase the fertility of soil. They increase the amount of growth promoting factors in soil.
Eg: Rhizobium Azetobacter raise the amount of Nitrogen in the soil.

Question 23.
Which are the foreign varieties of fowl reared in our place?
Answer:
White leghorn, Rhode Island Red, Plymouth Rock, New Hampshire

Question 24.
What are the diseases of fowls
Answer:
Ranikatt, Fowl Cholera, Salmonellosis, Diarrhoea, Chores (Bacteria) Aspergillusis (Fungus).

Kerala Syllabus 8th Standard English Solutions Unit 1 Chapter 2 The Boy Who Drew Cats

You can Download The Boy Who Drew Cats Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard English Solutions Unit 1 Chapter 2 helps you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Hindi Solutions Unit 1 Chapter 2 The Boy Who Drew Cats (Hasegawa Takejiro)

The Boy Who Drew Cats Textbook Questions and Answers

The Boy Who Drew Cats Questions And Answers Pdf Kerala Syllabus 8th Question 1.
How is the youngest child introduced in the story?
The Boy Who Drew Cats Questions And Answers Pdf Kerala Syllabus 8th
Answer
The youngest boy is introduced as a clever, but quite weak and a small boy who is not fit for hard work.

The Boy Who Drew Cats Questions And Answers Kerala Syllabus 8th Question 2.
Why did the parents take the boy to the priest?
The Boy Who Drew Cats Questions And Answers Kerala Syllabus 8th
Answer:
People of the village opined that the boy could never grow very big. So his parents thought it would be bet¬ter for him to become a priest than to become a farmer.

The Boy Who Drew Cats Activities Kerala Syllabus 8th Question 3.
‘… but perhaps you will become a great artist.’
Do you think the opinion of the priest about the boy is appropr¬iate? Why?
The Boy Who Drew Cats Activities Kerala Syllabus 8th
Answer:
Yes. Because whenever the boy was alone, he drew the pictures of cats.

The Boy Who Drew Cats Lesson Activities Kerala Syllabus 8th Question 4.
What was the advice of the priest to the boy?
The Boy Who Drew Cats Lesson Activities Kerala Syllabus 8th
Answer:
The priest advised the boy to avoid large places at night and to keep to the small.

The Boy Who Drew Cats Worksheet Answers Kerala Syllabus 8th Question 5.
Why didn’t the boy go home after he left the temple?
The Boy Who Drew Cats Worksheet Answers Kerala Syllabus 8th
Answer:
The boy did not go home because he felt sure that his father would punish him for being disobedient to the priest.

The Boy Who Drew Cats Pdf Kerala Syllabus 8th  Question 6.
When the boy entered the temple he did not see anyone. What could be the reason?
The Boy Who Drew Cats Pdf Kerala Syllabus 8th
Answer:
The reason was that a goblin rat had! frightened the priests away and had taken possession of the place.

The Boy Who Drew Cats Notes Kerala Syllabus 8th Question 7.
Why did he choose a small place to sleep?
The Boy Who Drew Cats Notes Kerala Syllabus 8th
Answer:
The boy felt afraid of the place and he resolved to look for a small place to sleep.

Kerala Syllabus 8th Standard English Solutions Unit 1 Chapter 2 Question 8.
Where did the boy find a safe place to sleep on?
Kerala Syllabus 8th Standard English Solutions Unit 1 Chapter 2
Answer:
He found a little cabinet, with a sliding door and got into it and shut himself up.

The Boy Who Drew Cats Conversation Between Boy And Priest 8th Question 9.
What was the dreadful voice that the boy heard?
The Boy Who Drew Cats Conversation Between Boy And Priest 8th
Answer:
The dreadful voice might have been the result of the fighting and screaming between the cats and the goblin rat.

The Boy Who Drew Cats Story Kerala Syllabus 8th  Question 10.
How do you think the goblin rat was killed?
The Boy Who Drew Cats Story Kerala Syllabus 8th
Answer:
The goblin rat was killed by the cats which the boy had drawn.

8th Standard English Notes Kerala Syllabus 8th Question 11.
Was the advice of the priest helpful to the boy? Why?
8th Standard English Notes Kerala Syllabus
Answer:
Yes. Because a dreadful fight was going on between the cats and the goblin rat and the boy was safe in the Cabinet.

The Boy Who Drew Cats Textbook Activities And Answers

The Boy Who Drew Cats Character Sketch Kerala Syllabus 8th Activity 1.

Did you enjoy the story, ‘The Boy who Drew Cats’? If you were the young boy, how would you tell the story?
You may begin like this.
The Boy Who Drew Cats Character Sketch Kerala Syllabus 8th
Answer:
I was the youngest child of a poor Japanese family. Because of my ill health, I was sent to a priest home to be trained as a priest. I liked to draw the pictures of j cats. Although I obeyed all the rituals and i I rule of the priest home I could not stop I drawing cats. My teacher ordered me not to draw cats anymore. But I could not stop drawing.

At last, my teacher advised me to become a painter, not a priest. I planned to quit the place. Before leaving the teacher advised me to take shelter in small places and asked me to avoid big places.

I started my journey to the next village. I found a temple. It was covered with dust and cobwebs. I did not know that it was j the place of a monstrous rat. I entered in and saw walls which were empty. I started painting cats. When I felt sleepy I slept in a small room. After sometime hearing some sound I woke up. When it was morning, I came out and saw a big rat killed. Because of the pictures of the cats the big rat was killed. After that incident, I became a famous painter in Japan.

The Boy Who Drew Cats Short Summary Kerala Syllabus 8th Activity 2.

How was the goblin rat killed?
Write the events that led to the death of the goblin rat to complete the flow-chart given below.
The Boy Who Drew Cats Short Summary Kerala Syllabus 8th
Answer:
Kerala Syllabus 8th Standard English Solutions Unit 1 Chapter 2 The Boy Who Drew Cats 14

The Boy Who Drew Cats Additional Questions & Answers

Question 1.
Read the passage from the story ‘ The Boy who Drew Catstfand answer the questions that follow.
Whenever he found himself alone, he drew cats. He drew them on the margins of the priest’s books, and on all the screens of the temple, and on the walls, and on the pillars. Several times the priest told him this was not right: but he did not stop drawing cats. He drew them because he could not really help it.

He had what is called ‘the genius of an artist’ and just for that reason he was not quite fit to be an acolyte- a good acolyte should study books. One day after he had drawn some very clever pictures of cats upon a paper screen, the old priest said to him severely: ‘ My boy, you must go away from this temple at once. You will never make a good priest, but perhaps you will become a great artist. Now, let me give you a last piece of advice, and be sure you never forget it. Avoid large places at night, keep to small!’

The boy did not know what the priest meant by saying, ‘ Avoid large places keep to small.’ He thought and thought, while he was tying up his little bundle of clothes to go away; but he could not understand those words and he was afraid to speak to the priest anymore, except to say good-bye.
1. The boy continued to draw cats because ……………………
a. He hated cats.
b. He wanted to become an artist
c. He wanted to please the priest
d. He was unable to stop drawing cats
2. Pick out the word from the passage which means ‘ one who helps the priest’,
a. Genius
b. Acolyte
c. Artist
d. Cats
3. Which were the places the boy choose to draw cats?
4. Why did the priest decide to send the boy away from the temple?
5. Pick out the expression from the passage that shows the boy wandered about the priest’s advice.
Answer:
1. He was unable to stop drawing, cats.
2. Acolyte
3. On the margins of the priest’s books, and on all screens of the temple, and on the pillar.
4. He would never make a good priest.
5. He thought and thought

Question 2.
Edit the following passage.
Whenever he found himself alone, he draws (a) cats. He drew them on the mar¬gins of the priests (b) books, and on all the screens (c) of the temple and on the walls and on the pillars.
Answer:
a. drew
b. priest’s
c. screens

Question 3.
Complete the passage given using suitable phrasal verbs from the ones given below.
The boy ………. (a) ……….. for the big temple in the next village. When he ……….. (b) ………… the village it was already dark. He……….. (c)……… the temple on a hill. He went at once to the temple and knocked. There was no sound inside. He ………….. (d)………. knocking but still nobody came. At last, he pushed gently at the door and found that it was not fastened.
Kerala Syllabus 8th Standard English Solutions Unit 1 Chapter 2 The Boy Who Drew Cats 15
Answer:
a. set out
b. got out
c. came across
d. went on

Question 4.
Given below are signboards to create awareness on Road Safety. Read them carefully and identify the verb phrases. One is done for you.
a) School Ahead, Go Slow
b) Wear Seatbelts While Driving
c) Pedestrians, Keep To Your Left

Verb Phrase
a. Go slow
b. ………….
c. ………….
Answer:
b. Wear seat belts
c. Keep to your left

Speech

Speech is an effective verbal communication made by a person addressing a group of people. Usually, a speech begins with an appropriate salutation.

Tips to remember

  • Good beginning with a salutation
  • Introduce the topic well
  • Conclude the speech well

Question 5.
You are the secretary of the Arts Club in your school. The club decides to honor the boy in the story ‘ The Boy who Drew Cats’ in your school assembly. Prepare a speech of appreciation.

(Hints: The boy always drew cats- send away from temple by the priest as he always drew cats – goes to next village – draws cats in the haunted temple – goblin rats were killed- blood on the mouth of the cats drawn by the boy – became a famous artist)
Answer:
A very warm welcome to all present here. Today I am very happy to introduce to you the famous painter Erico. He was the youngest child of a poor Japanese family. Because of his ill health, he was sent to a temple to be trained as a priest. He loved to draw cats. Although he learned whatever that was taught he couldn’t stop drawing cats. Though the old priest ordered him not to draw cats anymore he couldn’t stop drawing. At last, his teacher advised him to become a painter, not a priest. So he planned to quit the place.

Before leaving, the teacher advised him to take shelter in small places and asked him to avoid big places. He started his journey to the next temple. The temple was covered with dust and cobweb. He did not know that it was the place of a monstrous rat. When he saw the walls empty he painted cats. He felt sleepy and slept in a small cabinet. After sometime hearing some sound he woke up. In the morning when he woke up he found that the big rat had been killed by the cats he drew. It was the words of his teacher ‘Avoid large places at night, keep to small’ that saved him from the monstrous rat. After that incident, he became a famous painter in Japan.

Question 6.
Look at the following word pyramid.
Kerala Syllabus 8th Standard English Solutions Unit 1 Chapter 2 The Boy Who Drew Cats 16
Now, construct a similar word pyramid using the word ‘ book’.
Answer:
Kerala Syllabus 8th Standard English Solutions Unit 1 Chapter 2 The Boy Who Drew Cats 17

The Boy Who Drew Cats Summary in English

The main character in the story, youngest son of poor, hardworking farmer, lives in a country village of old Japan. Because he is small, weak, and bright, his parents sent him to the village priest to be trained for priesthood. The boy learns well and pleases his master in almost all ways, but he persists in one act of disobedience drawing cats whenever he can. Although warned to stop, he continued it. The disobedience of the boy causes the old priest to send him away with the advice to stop trying to become a priest but instead become an artist. The priest cautions the boy to avoid large places at night and keep to the small. Puzzled by the strange warning, the boy reluctantly leaves his temple home and walks to the next village, where there is a large temple at which he hopes to continue his religious training.

He arrives at the temple at night, only to find it deserted and covered with thick layers of dust and cobwebs. He does not know the temple has been abandoned because a bloodthirsty goblin lived there. Earlier, soldiers entered the temple at night to kill the goblin but did not survive the attempt. The unsuspecting boy sits quietly and waits for the temple priests to appear. He notices large, white screens, wonderful surfaces for drawing cats, and soon has drawing ink and brushes ready. He unhesitatingly draws cats, not stopping until he is too tired to continue.

Sleepily he remembers the old priest’s warning as he lies down, so he crawls into a small cabinet and pulls the door before he sleeps. Hours later, the boy wakes to sounds of horrible screaming and fighting. He cowers silently while the fight rages, and only ventures out after daylight streams into the room. He finds a floor wet with blood and, lying dead, a monster goblin-rat the size of a cow. Scanning the temple, he notices the wet, blood-red mouths of the cats he has drawn on the screens. Suddenly he understands the priest’s advice and realizes that his cats have destroyed the goblin in the vicious fight he has overheard. The boy later becomes a famous artist, whose pictures of cats can still be seen in Japan.

The Boy Who Drew Cats Summary in Malayalam

Kerala Syllabus 8th Standard English Solutions Unit 1 Chapter 2 The Boy Who Drew Cats 18
Kerala Syllabus 8th Standard English Solutions Unit 1 Chapter 2 The Boy Who Drew Cats 19

The Boy Who Drew Cats Glossary

Kerala Syllabus 8th Standard English Solutions Unit 1 Chapter 2 The Boy Who Drew Cats 20

Kerala Syllabus 8th Standard Maths Solutions Chapter 10 Statistics in Malayalam

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Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers in Malayalam

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Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters in Malayalam

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Kerala Syllabus 8th Standard Maths Solutions Chapter 7 Ratio in Malayalam

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Kerala Syllabus 8th Standard English Solutions Unit 3 Chapter 3 The Sower

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Kerala State Syllabus 8th Standard Hindi Solutions Unit 3 Chapter 3 The Sower (Victor Marie Hugo)

The Sower Textbook Questions And Answers

The Sower Poem Questions And Answers Kerala Syllabus 8th Standard Question 1.
What are the various activities involved in farming?
Kerala Syllabus 8th Standard English Solutions Unit 3 Chapter 3 The Sower 1
Answer:
Ploughing, Manuring, Sowing, Weeding, Irrigating, Harvesting.

The Sower Poem Activities Kerala Syllabus 8th Standard Question 2.
Where is the speaker sitting
Kerala Syllabus 8th Standard English Solutions Unit 3 Chapter 3 The Sower 2
Answer:
In a cool porchway

The Sower Question Answer Kerala Syllabus 8th Standard Question 3.
What time of the day is it?
Kerala Syllabus 8th Standard English Solutions Unit 3 Chapter 3 The Sower 3
Answer:
Evening

Let’s revisit

The Sower Poem Summary In Malayalam Kerala Syllabus 8th Standard Question 1.
‘But one sower lingers still’. Why does the sower ‘linger’? What does the use of the word ‘linger’ suggest about the sower?
Answer:
He lingers so that he can sow the seeds. It suggests that he is dedicated to his work.

The Sower Poem Questions And Answers Pdf Kerala Syllabus 8th Standard Question 2.
Why does the speaker feel a thrill on seeing the sower now? Identify the words used by the poet to establish this.
Answer:
He is impressed on seeing the silhouette of the sower dominating over the fields. Black and high his silhouette Dominate the furrows deep!

The Sower Questions And Answers Kerala Syllabus 8th Standard Question 3.
How does the sower go about performing his task?
Answer:
He marches along the field to and fro and scatters the seeds wide.

The Sower Notes Kerala Syllabus 8th Standard Question 4.
In normal circumstances pea¬rls, diamonds and rubies are referred as precious. In this poem, ‘grain’ is referred to as precious. Why does the poet consider grain precious?
Answer:
lt is these grains that grow as corn and provide us food that keeps us alive.

The Sower Poem Notes Kerala Syllabus 8th Standard Question 5.
The poet speaks of the sower as ‘old and in rags’ in the beginning of the poem. How does this opinion change towards the end of the poem? Pick out the lines from the poem.
Answer:
The sower grows in stature and the poet seems to respect him more for what he does.
‘Now his gesture to mine eyes Are august; and strange – his height Seem to touch the starry skies.’

The Sower Summary Kerala Syllabus 8th Standard Question 6.
Bring out the contrast between the poet and the sower.
Answer:
The poet is sitting idle, watching the sower whereas the sower works hard even after twilight has set in and everyone has gone home.

The Sower Textbook Activities And Answers

The Sower Poem Meaning In Malayalam Kerala Syllabus 8th Standard Question 1.
Look at the words:
cool
rule
fast
past
These are rhyming words. Now, pick out the rhyming words from the poem.
The Sower Poem Answer:
Lands — stands
still — thrill
deep — reap
plain — grain
wide — stride
kight — height
eyes — skies

The Sower Solutions Kerala Syllabus 8th Standard Question 2.
Locate the word pictures used in the poem.
Answer:
shadows shoot across the lands Marches he along the plain, To and fro and scatters wide From his hand’s precious grain

The Sower By Victor Marie Hugo Kerala Syllabus 8th Standard Question 3.
‘Seems to touch the starry skies’ is an example of alliteration from the poem. Pick out other examples from the poem
The Sower Question Answer:
shadows shoot across the lands looking on, I feel a thrill

The Sower 8th Class Kerala Syllabus  Question 4.
Write the rhyme scheme of the first stanza.
Answer:
Rhyme scheme ab -ab

The Sower Additional Questions & Answers

The Sower Class 8 Kerala Syllabus Questions 1 to 4:
Read the lines from the poem ‘ Song of the Flower’ and answer the questions that follow :
Sitting in a porchway cool,
Fades the ruddy sunlight fast
Twilight hastens on to rule
Working hours are well – nigh past
Shadows shoot across the lands;
But one sower lingers still,
Old, in rags, the patient stands,
Looking on, I feel a thrill
1. Where is the poet sitting?
2. What time of the day is it?
3. Pick out the word pictures in the poem.
4. Find out the rhyme scheme of the first stanza.
Answer:
1. In a cool porchway
2. Evening
3. Shadows shoot across the lands
4. abab

The Sower Poem Explanation Kerala Syllabus 8th Standard Question 5.
Prepare a short profile of Victor Marie Hugo using the hints given below.
Born: February 26, 1802, France
Popular as: Greatest and best known French writer
Spouse: Adele Foucher (1822)
Notable works: The Hunchback of Notredame, Sea Devils
Died: May 22, 1885
Answer:
Victor Marie Hugo:
Victor Marie Hugo was born on February 26, 1802, in France. He was one of the greatest and best known French writers. He married Adele Foucher in 1822. The Hunchback of Notredame, Sea Devils, etc are his notable works. He passed away on May 22 in 1885.

The Sower Poem Stanza Wise Explanation Kerala Syllabus 8th Standard Question 6.
There are a few errors in the given passage. They are given in bold letters. Edit this passage.
The City police has decided to took (a) stern action against drivers which (b) attempt to overtake in (c) the left side on city roads. All driver (d) who violate this rule will be punished (e).
The Sower Answer:
a. take
b. who
c. on
d. drivers
e. punished

Hss Live Guru 8th English Kerala Syllabus Question 7.
Read the following sentences given below and rewrite them after fill¬ing the blanks using appropriate relative pronouns. You may use the words given in the box.
a. The beggar …………… is hungry shouted for food.
b. The banker …………… was younger in those days was suddenly carried away by excitement.
c. The old banker ………………. money was lost was walking up and down.
d. The majority of the guests among …………….. were many journalists and intellectual men disapproved of dealth penalty.
e. The King was not ready to throw away the crown ……………….. was the sign of power.
f. Jail is the place …………….. one’s personal freedom is negated.
Kerala Syllabus 8th Standard English Solutions Unit 3 Chapter 3 The Sower 4
Answer:
a. Who
b. Who
c. Whose
d. Whom
e. Which
f. Where

Questions 8 to 12. Read the newspaper headline s given below and answer the questions that follow.
a. 20 KILLED IN ROAD ACCIDENTS
b. INDIA CLINCHES SERIES AGAINST ‘WEST INDIES’
c. FROGS MARRIED OFF IN HOPE OF RAIN
d. CYCLONE KILLS OVER TWENTY
e. M.B.A CLASSES TO BEGIN ON SEPT.20th

8. Which headline is related to the field of sports?
9. The headline which speaks of nature’s fury is ……………
10. Identify the headline that hints at superstitious belief.
11. The headline that talks of a road mishap is ……………..
12. The headline that can be categorized as educational news is ……………
The Sower Poem Brainstorming Answer:
8. INDIA CLINCHES SERIES AGAINST WEST INDIES
9. CYCLONE KILLS OVER TWENTY
10. FROGS MARRIED IN THE HOPE OF RAIN
11. 20 KILLED IN ROAD ACCIDENTS
12. M.B.A CLASSES TO BEGIN ON SEPT. 20th

Questions 13 to 17. Study the table showing the details of the sales in a book shop of a few novels by well-known authors and answer the questions that follow.
Kerala Syllabus 8th Standard English Solutions Unit 3 Chapter 3 The Sower 5
13. Which is the most recently published novel out of the given list?
14. Pick out the novel that is sold less number of copies than the others.
15. Who is the author of the novel which is sold the most?
16. Name the author whose novel stands second on the basis of the sales.
17. How many copies of the novel stands second on the basis of the sales?
Answer:
13. Blindness
14. If on a winter’s night a traveler
15. Paulo Coelho
16. Gabriel Garcia Marquez
17. 411

The Sower Summary in English

A Farmer, even though it is getting dark, is working hard in his field. All people have gone after their work. But this farmer whom the poet meets is still in the farm ploughing and preparing the soil for sowing the seeds. The poet feels happy when he watches this farmer. The farmer is making furrows and thus preparing his plot for sowing the seeds. After sowing the seeds he can happily get ready for the harvest.

The farmer walks everywhere in this farm by preparing the soil for sowing. He scatters the precious seeds with much care and enthusiasm. After some time the sunlight diminishes slowly. It is getting dark. The sight of the farmer who is like a royal and dignified person touches the skies. The poet describes the farmer as a divine person.

The Sower Summary in Malayalam

Kerala Syllabus 8th Standard English Solutions Unit 3 Chapter 3 The Sower 6

The Sower Glossary

Kerala Syllabus 8th Standard English Solutions Unit 3 Chapter 3 The Sower 7

Kerala Syllabus 8th Standard Maths Solutions Chapter 1 Equal Triangles

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Equal Triangles Text Book Questions and Answers

Textbook Page No. 11

Equal Triangles Class 8 Questions And Answers Kerala Syllabus Question 1.
In each pair of triangles below, find all pair of matching angles and write them down.
Equal Triangles Class 8 Questions And Answers Kerala Syllabus
Solution:
(i) ∠A = ∠R (The angles opposite to 5cm sides)
∠B = ∠P (The angles opposite to the sides of length 4 cm)
∠C = ∠Q (The angles opposite to the sides of length 6 cm)

(ii) ∠L = ∠Y (The angles opposite to the sides of length 10cm)
∠M = ∠Z (The angles opposite to the side of length 4 cm)
∠N = ∠X (The angles opposite to the side of length 8cm)

Kerala Syllabus 8th Standard Maths Notes Question 2.
In the triangles below AB = QR, BC = RP, CA = PQ
Kerala Syllabus 8th Standard Maths Notes
Compute ∠C of ∆ABC and all angle of ∆PQR.
Solution: C = 80° (Use the property that the sum of three angles of a triangle is 180°)
AB = QR
∴ ∠C = ∠P
∠C = 80°
∴ ∠P = 80°

BC = RP
∴ ∠A = ∠Q
∠A = 40°
∴ ∠Q = 40°

CA = PQ
∴ ∠B = ∠R
∠B = 60°
∴ ∠R = 60°
(The angle opposite to equal sides are equal)

Equal Triangles – Class 8 Solutions Kerala Syllabus Question 3.
In the triangle below.
AB = QR BC = PQ CA = RP
Equal Triangles - Class 8 Solutions Kerala Syllabus
Compute the remaining angles of both the triangles.
Solution:
AB = QR ∴ ∠C = ∠P
BC = PQ ∴ ∠A = ∠R
CA = RP ∴ ∠B= ∠Q
∠A = 60° ∴ ∠R = 6o°
∠Q = 70° ∴ ∠B = 70°
∠A = 60, ∠B = 70° then ∠C = 180 – (60° + 70°)
∴ ∠C = 50° ∴ ∠P = 50°

Equal Triangles Class 8 Notes Pdf Kerala Syllabus Question 4.
Equal Triangles Class 8 Notes Pdf Kerala Syllabus
Are the angles of ∆ABC and ∆ABDequal in the figure above? Why?
Solution:
The side AB is common to both the triangles in the figure.
The side of ∆ABC are equal to the sides of ∆ABD. So the angles of ∆ABC are equal to the sides of ∆ABC.

8th Standard Maths Guide Kerala Syllabus Question 5.
In the quadrilateral ABCD shown below, AB = AD, BC = CD
8th Standard Maths Guide Kerala Syllabus
Compute all the angles of the quadrilateral?
Solution:
AB = AD, BC = CD
AC is the common side
The sides of the triangles ABC and ADC are equal. So their angles are also equal. AB = DC
∴ ∠ACD = ∠ACB = 50° (Angles opposite to equal sides of a triangle are equal)
BC = CD
∴ ∠BAC = ∠DAC = 30° (Angles opposite to equal sides are equal in a triangle )
∴ ∠D = ∠B = 100°

Textbook Page No. 15

8th Class Maths Notes Kerala Syllabus Question 1.
In each pair of triangles below find the pairs of matching angles and write them down.
8th Class Maths Notes Kerala Syllabus
Solution:
(i) BC = PR (If two sides of a triangle and the angle made by them are equal to two sides of another triangle and the angle made by them, then the third sides of the triangle are also equal.)
∴ ∠B = ∠R
∴ ∠C = ∠P
∴ ∠A = ∠Q (The opposite angles of equal sides of two triangles are also equal)

(ii) MN = XY (If two sides of triangle and the angle made by then are equal to two sides of another triangle and the angle made by them, then the third sides of the triangle are also equal)
∴ ∠L = ∠Z
∠M = ∠Y
∠N = ∠X (If two sides of a triangle are equal, the angles opposite to these sides are also equal)

Class 8 Maths Chapter 1 Kerala Syllabus Question 2.
In the figure below, AC and BE are parallel lines.
Class 8 Maths Chapter 1 Kerala Syllabus
(i) Are the lengths of BC and DE equal. Why?
(ii) Are BC and DE parallel? Why?
Solution:
(i) Given AC and BE are parallel lines.
∴ ∠CAB = ∠EBD
When we consider the triangles ∆CAB, ∆EBD (Corresponding angles)
BC = DE (The two sides of ∆ CAB and the angle made by them are equal to the two sides of ∆ EBD and the angle made by them. So the thirif side of triangle are also equal.)

(ii) Yes, they are parallel.
∠ABC = ∠BDE (The angles opposite to the equal sides of equal triangles are equal) But they are corresponding angles. BC and DE are parallel.

Kerala Syllabus Class 8 Maths Solutions Question 3.
Is ABCD in the figure, a parallelo¬gram? Why?
Kerala Syllabus Class 8 Maths Solutions
Solution:
AC = BD
AB is the common side.
The angles between the sides AC, AB and BD, AB are equal.
∴ BC = AD
The opposite sides of quadrilateral ∆CBD are equal. The angles opposite to equal sides of triangles ∆ABC and ∆ABD are equal. So the opposite angles in quadrilateral ACBD are also equal.
∴ ACBD is a parallelogram.

Kerala Syllabus 8th Standard Maths Notes Pdf Question 4.
In the figure below, M is the midpoint of the line AB. Compute the other two angles of ∆ABC
Kerala Syllabus 8th Standard Maths Notes Pdf
Solution:
AM = BM (Given M is the mid point of AB)
CM = CM (common)
∠AMC = 90° = ∠BMC
∴ The two sides in ∆AMC and ∆BMC and the angle made by them are equal.
So the third side and other angles are equal.
∠A = 50° ∴ ∠B = 50°
∠ACM = 40° ∴ ∠BCM = 40°
∴ ∠C = 80°

Class 8 Maths Chapter 1 Notes Kerala Syllabus Question 5.
In the figure below, the lines AB and CD are parallel and M is the mid point of AB.
Class 8 Maths Chapter 1 Notes Kerala Syllabus
(i) Compute the angle of ∆AMD, ∆MBC and ∆DCM?
(ii) What is special about the quadrilateral AMCD and MBCD?
Solution:
Given AB = 12 cm and M is the mid-point of AB.
∴ AM = MB = 6 cm
In quadrilateral AMCD,
AM = CD
AB||CD ∴ AM||CD
∴ AMCD is a parallelogram.
∴ ∠AMD = ∠CDM (Alternate interior angles)
∠ADM = ∠CMD (Alternate interior angles)
∠A = ∠DCM = 40° = ∠CMB
∴ ∠MCB = 80° [180 – (60 + 40)]
(i) The angles of ∆AMD, ∆MBC and ∆DCM are 40°, 60° and 80° respectively.
(ii) Both of them are parallelograms.

Textbook Page No. 21

Class 8 Maths Scert Solutions Kerala Syllabus Question 1.
In each pair of triangles below, find matching pairs of sides and write their names.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 11
Solution:
(i) BC = PQ
AC = PR (The two angles and the side in between them in ∆ ABC are equal to the two angles and the side in between them in ∆PQR. So the third angles of the triangles ∠C and ∠R are also equal. Also BC and PQ, opposite to the 50° angle are also equal. The sides AC and PR opposite to the 70° angle are also equal.

(ii) ∠N = 70°
∠Z= 80°
MN = XZ
∠M = ∠Z(Sides opposite to equal angles are also equal)

8th Maths Notes Kerala Syllabus  Question 2.
In the figure, AP and BQ equal and parallel are lines drawn at the ends of the line AB. The point of inter section of PQ and AB is marked as M.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 12
(i) Are the sides of ∆AMP equal to the sides of ∆BMQ? Why?
(ii) What is special about the position of M on AB.
(iii) Draw a line 5.5 cm long. Using a set square, locate the midpoint of this line.
Solution:
(i) Yes, they are equal
∠P = ∠Q
∠A = ∠B (alternate angles formed by cutting the parallel lines AP and QB by PQ and AB.)
AP = QB
∴ The third angle of ∆APM and ∆BMQ and opposite sides of equal angles are equal.
(ii) AM = BM. So M is the midpoint of AB.
(iii) Draw a line segment of length 5.5 cm. Draw perpendiculars of equal lengths upward at one end of the line and downwards at the other end. Join the ends. This line divides the first one equally.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 13

Hsslive Guru Maths 8th Standard Kerala Syllabus Question 3.
In the figure, ABCDE is a pentagon with all sides of the same length and all angles of the same size. The sides AB and AE extended, meet the side CD extended at Px and Q.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 14
(i) Are the sides of ∆BPC equal to the sides of ∆EQD? Why?
(ii) Are the sides of AP and AQ of ∆ APQ equal? Why?
Solution:
(i) Yes, this are equal the sides and angles of a pentagon are equal.
∴ BC = DE
∠PBC = ∠PCB (Exterior angles of a regular pentagon)
∠QDE = ∠QED (Exterior angles of a regular pentagon)
∆QDE = ∆QED (If two angles and side of one triangle are equal are equal to two angles and corresponding side of the other triangle then their sides are equal.
BP = EQ and PC = DQ

(ii) AB = AE sides of regular pentagon.
BP = EQ
∴ AP = AQ [AB + BP = AC + EQ]

8th Standard Maths Notes State Syllabus Question 4.
In ∆ABC and ∆PQR shown below.
AB = QR BC = RP CA = PQ
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 15
(i) Are CD and PS equal? Why?
(ii) What is the relation between the areas of ∆ABC and ∆PQR?
Solution:
(i) AB = QR
BC = RP
∠A = ∠Q
∴ ∆ABC and ∆QRP are equal triangles. Given all sides of ∆ABC are equal to the sides of ∆QRP
∴ CD and PS are equal. (Opposite sides of equal angles

(ii) AB = QR and CD = PS
⇒ 1/2 AB × CD = 1/2QR × PS
∴ The areas of ∆ABC and ∆PQR are equal.

Question 5.
In the quadrilateral ABCD shown below the sides AB and CD are parallel. M is midpoint of the side BC. The lines DM and AB extended meet at N.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 16
(i) Are the areas of ∆DCM and ∆BMN equal? Why?
(ii) What is the relation between the areas of the quadrilateral ABCD and the triangle ADN.
Solution:
(i) M is the midpoint of the line BC.
∴ CM = MB; BN || DC
∴ ∠DCM = ∠NBM (Alternate angles)
∠DMC = ∠NMB (Vertically opposite angles)
∴ ∆DCM and ∆BMN are equal triangles. So their areas are equal.

(ii) The areas of ∆DCM and ∆BMN are equal and quadrilateral AB, MD common
∴ The area of the quadrilateral
ABCD and the area of ∆ADN are equal.

Question 6.
Are the two diagonals of a rectangle equal. Why?
Solution:
ABCD is a rectangle.
Consider the ∆ABD and ∆ABC
AB = AB, common AD = BC (opposite sides of rectangle); ∠A = ∠B =90°
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 17
Both sides of ∆ABD and ∆ABC and the angle formed by them are equal. So the third sides BD and AC are equal. So the diagonals of the rectangle are equal.

Textbook Page No. 26, 27

Question 1.
Some are equal isosceles triangles are drawn below, In each, one angle is given. Find the other angles.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 18
Solution:
(i) 30°, 75°, 75°
(ii) 40°, 70°, 70°
(iii) 20°, 8o°, 8o°
(iv) 100°, 40°, 40°

Question 2.
One angle of an isosceles triangle is 90°. What are the other two angles?
Solution:
The other two angles are equal. So they are 45°, 45°

Question 3.
One angle of an isosceles triangle is 6o°. What are the other two angles.
Solution:
The other two angles are equal. So they are 60°, 60°

Question 4.
In the figure below, O is the centre of the circle and A, B are points on the circle.
Compute ∠A and ∠B?
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 19
Solution:
OA = OB (radius of circle)
∆AOB is a isosceles triangle; ∴ ∠A = ∠B
∠O = 60°
∴ ∠A = ∠B = 60°

Question 5.
In the figure below, O is the centre of the circle and A, B, C are points on the circle.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 20
What are the angles of ∆ABC?
Solution:
∆AOB, ∆AOC, ∆BOC are isosceles triangles. Each triangles are with angles 120°, 30° and 30°.
∴ ∠A = ∠B = ∠C = 30° + 30° = 60°

Text Book Page No. 29

Question 1.
Draw a line of 6.5 centimetres long and draw its perpendicular bisector.
Solution:
Draw a line segment AB of length 6.5 c.m with A and B as centres draw arcs on both sides of AB with equal radii. The radius of each of these arcs must be more the half the length of AB. Let these arcs cut each other at points C and D. Join CD which cuts AB at M
Then AM = BM. Also ∠AMC = 90°
Hence, the line segment CD is the perpendicular bisector of AB as it bisects AB at M and is also perpendicular to AB.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 21

Question 2.
Draw a square, each side 3.75 centimetres long?
Solution:
Draw AB = 3. 75 cm at. A Construct ∠PAB = 90° from AP, cut AD = 3.75 cm
Taking D as centre, draw an arc of radius 3.75 cm and taking B as centre, draw one more are of radius 3.75 cm.
Let the two arcs intersect at point C. Join BC and DC.
Then ABCD is the required square.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 22

Question 3.
Draw an angle of 750 and draw its bisector?
Solution:
Draw a line segment AB of any suitable length with A as centre. Draw an arc of any size to cut AB at D. With D as centre. Draw another arcs of some size to cut the previous arc at C.
Now ∠CAD = 60°. Draw ∠EAB = 90° and bisect ∠EAC.
∴∠PAC = 150 ∠DAC + ∠CAP = 60 + 15 = 75°
∴ ∠BAP = 75°
Then bisect
∠BAP AQ to the bisectors of ∠PAB
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 23

Question 4.
Draw a circle of radius 2.25 centimetres.
Solution:
Draw a line of length 4.5 cm. Draw its perpendicular bisector it meet at point ‘O’.
‘O’ is the centre of the circle and radius = 2.25 cm. Then complete the circle.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 24

Question 5.
Draw ∆ ABC, with AB = 6 cm,
∠A = 22\(\frac{1}{2}\)°, ∠B = 67\(\frac{1}{2}\)°
Solution:
Draw the line AB in 6 cm length. Draw angle A at 45° and draw its bisector. Draw angle 135° at B and draw its bisector. Mark the point as C where bi-sectors meet.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 25

Question 6.
Draw a triangle and perpendicular bisectors of all three sides. Do all these three bisectors intersect at the same point?
Solution:
Draw a triangle ABC. By using compass mark the arcs on both sides from each ends.
Draw the same for all sides. They intersect at same point
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 26

Question 7.
Draw a triangle and the bisectors of the three angles. Do all three bisectors intersect at the same point.
Solution:
Draw a triangle PQR and by using com-pass draw the bisectors of angles. All three bisectors meet at the same point.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 27

Question 8.
Prove that if both pairs of opposite sides of a quadrilateral are equal, then it is a parallelogram.
Solution:
When the diagonal of a quadrilateral with equal opposite sides is drawn, we get two equal triangles. The angels opposite to the diagonal in the triangles are equal. That is the opposite sides and angles in the quadrilateral are equal. So the quadrilateral is a parallelogram.

Question 9.
In the figure, ABCD is parallelogram and AP = CQ
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 28
Prove that PBQD is a parallelogram
Solution:
DC = AB ………. (1)
AD = CB
QC = AP ……. (2) (as ABCD is a parallelogram)
(1) – (2) ⇒ DC – QC = AB – AP; ∴ DQ = PB
When, ∆ APD, ∆ CQB are considered.
AD = CB
AP = QC
∠A = ∠C, The two sides and angle formed by them in these triangles are equal. So the third sides PD and BQ are equal.
∴ Two pairs of opposite sides in the quadrilateral PBQD are equal. So PBQD is a parallelogram.

Question 10.
Prove that if all sides of a parallelogram are equal, them each diagonal is the perpendicular bisector of the other.
Solution:
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 29
The diagonal BD divides the parallelogram into two isosceles triangles. [The angles opposite to the equal sides in an isosceles triangles are equal.] So the diagonal DB bisect ∠D and ∠B.
Similarly the diagonal AC bisect A and ∠C. 4x +4y = 360° ⇒ x + y = 90°
The four triangles formed by intersec¬ting the diagonals are equal triangles. Each one 90 angle. So each diagonal is the perpendicular bisector of the other. In ∆AMD ∠AMD = 180 – (x – y) = 180 – 90 = 90° ⇒ BD ⊥ AC

Question 11.
In the figure below O is the centre of the circle and AB is the diameter. C is the point on the circle.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 30
(i) Compute ∠CAB
(ii) Draw another figure like this with a different number for the size of ∠COB. Calculate ∠CAB
Solution:
(i) ∠BOC = 50°
∴ ∠COA = 180° – 50°= 130° (straight angle)
OA = OC (radii)
∴ AOC is an isosceles triangle.
∴ ∠A = ∠C = \(\frac{180-130}{2}\) = 25°

(ii) ∠O = 70°
∴ ∠COA = 180 – 30 = 150°
OA = OC
∴ ∠CAB = ∠ACO =\(\frac{180-150}{2}=\frac{30}{2}\) = 15°
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 49

Question 12.
In the figure below, O is the centre of the circle and AB is a diameter. C is a point on the circle.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 31
(i) Compute ∠ACB
(ii) Draw another figure like this, changing the size of ∠COB and calculate ∠ACB.
Solution:
(i) ∠BOC = 50°
∠AOC = 130°
∆ AOC and ∆ BOC are isosceles triangles.
∠OAC = ∠OCA = \(\frac{180-130}{2}\) = 25°
∠OBC = ∠OCR = \(\frac{180-50}{2}\) = 65°
∠ACB = ∠OCA+ ∠OCB
25° + 65° = 90°

(ii) ∠AOC = 180 – 80 = 100°
∠OAC + ∠OCA = 180 – 100 = 8o°
∴ ∠OAC = ∠OCA = 80 ÷ 2 = 40°
∆ OBC
∠OBC + ∠OCB = 180 – 80 = 100°
∠OBC =∠OCB = 100 ÷ 2 = 50°
∠ACB = ∠OCA + ∠OCB
40° + 50° = 90°

Question 13.
How many different isosceles triangles be drawn with one angle 50° and any one side 7 centimetres.
Solution:
An isosceles triangle can be drawn with one angle 50° as angles either 50°, 50°, 8o° or 50°, 65°, 65°. In both the cases, 7 cm can be taken as equal sides or can be without 7 cm one as side. So there can be 4 ways of drawing diagram.

Question 14.
Draw ∆ ABC with AB = 7 cm, ∠A = 67\(\frac{1}{2}\), ∠B = 15° without using protector.
Solution:
Draw AB with length 7 cm. Extend both the sides. Draw the perpendicular from A. Draw the bisector through the left 90° angle among the 90° angles obtained. Draw an angle as 90° + 45° = 135°, Draw its bisector.
Now ∠A = 67\(\frac{1}{2}\). Draw an angle 60° in B to construct an equilateral triangle. Draw the bisector of its bisector. Then ∠B = 15°. We get ∆ ABC.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 48

Equal Triangles Additional Questions & Answers

Question 1.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 32
O is the centre of the circle in the diagram. If AB = BC,
(a) Then prove that ∠AOB = ∠BOC
(b) If OA = AB = BC, then find the values of ∠AOB and ∠BOC?
(c) Find out how many equilate¬ral triangles can be drawn in a circle with length of its side is radius.
Solution:
(a) OA = OB = OC, AB = BC
∆ OAB and ∆ OBC are equal triangles.
∴ ∠AOB and ∠BOC are equal which are opposite to the equal sides AB and BC.
(b) If OA = AB then ∆ OAB is an equilateral triangle.
If OB = BC, ∆ OBC is equilateral triangle.
∴ ∠AOB = ∠BOC = 60°
(c) Each angle at O is 60°. The angle at the centimeter is O is 360° and 6 triangles can be drawn.

Question 2.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 33
If AB = AD, BC = CD in the diagram, then prove that ∠ABC = ∠ADC
Solution:
The three sides of triangles ∆ ABC are equal. The angles opposite to the sides are also equal.
AB = AD, BC = DC, AC = AC
AC is the common side. So the angles opposite to this side ∠ADC and ∠ABC are also equal,
i e ∠ABC = ∠ADC

Question 3.
Draw a rhombus with sides and a diagonal as 5 cm.
Solution:
Draw a line of length 5 cm. Draw equilateral triangles on both ends of the line with length 5 cm and line as one side.

Question 4.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 34
In the figure, AB = DE, BC = EF and AC = DF. Can ∠BPD = ∠C? Prove it?
Solution:
The sides of ∆ DEF and ∆ ABC are equal. The angles opposite to equal sides are equal.
∠E = ∠B, ∠F = ∠C, ∠D = ∠A
But ∠C = 180 – (∠B + ∠A)
∠P = 180 – (∠B + ∠D)
∴ ∠D = ∠A
∴ ∠C = ∠P = 180 – (∠B + ∠A)
∴ ∠C = ∠P

Question 5.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 35
In the figure, AB = PQ, AC = PR, BC = QR. PQ is parallel to AB.
(a) Then show that BC is parallel to QR.
(b) Also show that PR is parallel to AC
Solution:
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 36
AB = PQ, AC = PR, BC = QR
∴ The angles of ∆ ABC and ∆ PQR are equal.
∠A = ∠P, ∠B =∠Q, ∠C = ∠R
(a) AB||PQ, ∴ ∠B = ∠PMN (corresponding angle);
∴ ∠PMN = ∠Q
∴ MN||QR
∴ BC||QR

(b) BC||QR
∠R = ∠MNP = ∠C
∴ NP||AC, PR||AC

Question 6.
Diagonals of three parallelograms with equal areas are given. Draw the parallelograms.
(i) length of diagonal 7 cm
(ii) length of diagonal 6 cm
(iii) length of diagonal 5 cm.
Solution:
(i) Draw a line of length 7 cm. Draw triangles of sides 7 cm, 6 cm, 5 cm at both the ends of the line to get a parallelogram by joining both the triangles.
(ii) Draw a line of length 6 cm and follow the above method.
(iii) Draw a line of length 5 cm and follow the above method.

Question 7.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 37
O is the centre of circle and AC, BD are diameters in the figure. Prove that AB = CD
Solution:
Consider ∆ODC and ∆OAB.
OD = OC = OA = OB (radi ∠AOB = ∠DOC; Two triangles are equal So the third sides of the triangles AB and CD are equal.)

Question 8.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 38
ABCD in the figure is a parallelogram. P, Q, R and S are the mid points of the sides of the parallelogram. The prove that PQ = RS, and QR = PS.
Solution:
Consider the triangles ∆APS and ∆CRQ
AP = CR, (half of the equal lines AB and CD)
AS = CQ (half of the lines with equal lengths AD and BC.)
∠A = ∠C (opposite angles of the parallelogram are equal)
When two sides and the angle made by them, in a triangle are equal then the third sides are also equal.
∴ QR = PS
Similarly if ∆ DSR and ∆ BQP are considered, PQ = RS is obtained.

Question 9.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 39
P is the midpoint of the sides AB and DF in the figure.
(a) Prove that BD = AF
(b) Is EF parallel to BC? Why?
(c) If D is the midpoint of BC, A is the midpoint of EF and Q is the mid point of DE then can Q be the midpoint of AC? Why?
Solution:
(a) Consider ∆APF and ∆DBP
FP = DP and AP = PB
∠APF = ∠DPB. The sides and the angle made by them in the triangles are equal. So the third sides BD and AF are equal.

(b) FP = PD. So the angles opposite to them are also equal.
∠FAP and ∠DBP are equal.
∴ FA||BD andBC|| EF.

(c) Consider BD = AF, in Question (a)
∴ BC = EF, Consider ∆ AEQ and ∆ DCQ
AE = DC, QE = DQ
∠AEQ = ∠CDQ
Two sides of a triangle and the angle made by them are equal. So the third sides are also equal, ie AQ = QC.
∴ So Q is the midpoint of AC

Question 10.
Draw a parallelogram if one of its diagonal is 8 cm length and one side is 6 cm. and the angle formed by the side and the diagonal is 40.
Solution:
Draw a diagonal of length 8 cm. Draw a line of 6 cm with 40 angle at its one end. Draw the same in its opposite direction.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 40

Question 11.
One angle of an isosceles triangle is 80. Find the other possible angles of the triangle.
Solution:
8o°, 8o°, 20°
8o°, 50°, 50°

Question 12.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 41
O is the centre of the circle in the diagram. Radiaus is 3 cm and ∠AOB = 60°. Find the perimeter of ∆ AOB?
Solution:
∆ OAB is an isosceles triangle.
∴ ∠A =∠B
∠O = 60
∴ ∠OAB is an equilateral triangle so perimeter = 3 + 3 + 3 = 9 cm.

Question 13.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 42
OM is perpendicular to AB in the diagram. Prove that M is the mid point of AB.
Solution:
OA = OB
∴ ∆ OAB is an isosceles triangle.
∴ ∠A = ∠B
When ∆ OMA and ∆ OMB are considered, OM is the common side
∠AMO = ∠BMO = 90 ∠AOM = ∠BOM
One side of the triangle and angles at the ends of sides are equal. So the other two sides are also equal.
∴ AM = MB
∴ M is the midpoint of AB.

Question 14.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 43
In the figure ∠ABC = ∠ADC and AB = AD. Prove that A BCD is an isosceles triangle?
Solution:
AB = AD
∴ ∆ ABD is an isosceles triangle.
∴ ∠ABD = ∠ADB
It is given that ∠ABC = ∠ADC
∴ ∠CBD = ∠CDB
∴ CD = CB
∴ ∆ BCD is an isosceles triangle.

Question 15.
In ∆ ABC, AB = AC = 10 cm. M is the midpoint of BC. If BC = 12 cm, Find AM? Also find the area of ∆ ABC?
Solution:
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 44

Question 16.
Show that, the triangle obtained by joining the mid points of the sides of an isosceles triangle is also an isosceles triangle.
Can we get an equilateral triangle by joining the mid points of the sides of an equilateral triangle?
Solution:
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 45
∆ ABC is an isosceles triangle. P, Q, R are the mid points of the sides of the triangle. Consider ∆ PBR and ∆ QRC.
PB = QC
BR = CR
∠B = ∠C
Two sides and the angle made by them are equal. The third sides PR and QR also equal.
∴ ∆ PQR is an isosceles triangle. Similarly the triangle obtained by joining the mid points of the sides of an equilateral triangle is an equilateral triangle.

Question 17.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 46
In the figure ∠B = ∠C = 90. If AB = CE and BE = CD, then find angles of ∆AED?
Solution:
If ∆ ABE and ∆ ECD are considered, AB = EC, BF = DC and ∠B = ∠C. Two sides of the triangle and the angle made by them are equal. The third sides AE and DE are also equal. ∆ADE is an isosceles triangle.
∆BAE = ∆DEC
∠BEA = ∠EDC (Angles opposite to the equal sides are also equal)
∠BAE + ∠BEA = 90
∴ ∠BEA + ∠+ BEA = 90
∴ ∠AED = 90°
∴ ∆AED is an isosceles triangle.
∴ ∠EAD = ∠EDA = 45°

Question 18.
Construct the following triangles by using only scale and compass.
(a) In ∆ ABC, AB = 6 cm, ∠A = 45°, ∠B = 75°
(b) In ∆ PQR, PQ = 7 cm,
∠P = 52\( \frac{1}{2}\)° , ∠ Q = 82\( \frac{1}{2}\)° 2
Solution:
(a) Draw AB = 6 cm
Draw AP making angle 45° with AB.
Draw BQ making angle 75° with AB.
Let AP and BQ intersect at C.
∴ ABC is the required triangle.
Kerala Syllabus 8th Standard Maths Solutions Part 1 Chapter 1 Equal Triangles 47

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters

You can Download Cell Clusters Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters

Cell Clusters Questions and Answers

Diversity among cells

Each part of human body is composed of different kinds of cells. These cells do not act independently. But they act in groups of similar cells. These cell clusters are called tissues.

Cell Clusters Class 8 Kerala Syllabus

Cell Clusters Class 8 Kerala Syllabus Tissues

The group of cells with common origin and perform a specific function are called tissues. The coordinated action of tissue, help to perform various physiological functions effectively.

Cell Differentiation

The single celled zygote divides continuously and forms the foetus that consists of cells differ in shape, size and contents. Foetal cells gradually attain change in structure and function. This process is called cell differentiation.

Indicators (Text Book Page No:24)

Cell Clusters Class 8 Notes Pdf Kerala Syllabus Question 1.
Formation of foetus.
Answer:
The single celled zygote formed by the fusion of sperm and ovum continuously divides and the foetus is formed.

8th Class Biology Notes Pdf Kerala Syllabus Question 2.
Significance of cell differentiation.
Answer:
Foetal cells gradually attain change in structure and function due to cell differentiation. Thus, different kinds of cells and tissues are formed. It helps to make the life processes effective.

Kerala Syllabus 8th Standard Biology Notes Stem cells

Stem cells are specialised cells that can be transformed into any kind of cells.

Stem cells can either transform into other cells through division or they can exist as such. When the cells in the tissues, get destructed, new cells emerge from stem cells.

It is expected that the research of stem cells may cause drastic change in the treatment of blood cancer, diabetes, Parkinson’s disease etc as desired cells are formed from stem cells.

Indicators (Text Book Page No:25)

Class 8 Science Notes Kerala Syllabus Question 3.
What are the peculiarities of stem cells when compared to other cells?
Answer:
Stem cells are specialized cells that can be transformed into any kind of cell. They can either transform into other cells or can retain their existence. When the cells in the tissues get destroyed new cells originate from stem cells.

Cell Clusters Class 8 Questions And Answers Pdf Kerala Syllabus Question 4.
How is the destruction of cells in tissues compensated?
Answer:
When the cells in the tissues get destroyed, new cells originate from cells. Stem cells are found in bone marrow, skin, digestive fact etc.

Cell Clusters Class 8 Notes Kerala Syllabus Question 5.
Why is stem cell research gaining importance?
Answer:
Nowadays desired cells can be produced from stem cells in research laboratories. The research of stem cells can cause a drastic change in the treatment of blood cancer, diabetes, Parkinson’s diseases etc, and in the production of artificial organs.

Animal Tissues

Animal tissues are mainly classified into 4
(a) Epithelial tissue
(b) Nervous tissue
(c) Muscular tissue
(d) Connective tissue

1. Epithelial Tissue:- Covers and protects the body.
Covers the inner lining of the alimentary canal Performs the functions like protection, absorption, production of secretions, etc.
2. Nervous Tissue:- Controls and Co-ordinates physiological Activities Helps to respond to particular stimuli
3. Muscular Tissue:- Makes the movement of the body possible.
4. Connective Tissue:-

  • Connects different tissues.
  • Provide support and strength to the body
  • Performs the functions like material transport, defence, etc.

Meristematic tissues Meristematic cells are specialized cells seen at the tip of the stem and root in plants. They divide rapidly and helps in the growth of plants.

Table (Text Book Page No:27)

Compare the figures of a meristematic cell and a mature cell. Find out the differences and complete the table.
Cell Clusters Class 8 Notes Pdf Kerala Syllabus
Answer:

Characteristics Meristematic Cells Mature plant Cells
Relative size of Nucleus (a) ………………. (b) ……………
Thickness of cell wall (c) ………………. (d) ……………
Quantity of Cytoplasm (e) ………………. (f) ……………

Basic Science Class 8 Chapter 2 Kerala Syllabus Various Plant Tissues

Parenchyma, Collenchyma, and Sclerenchyma are the main plant tissues. They differ in structure and perform different functions.

The specialised tissues in plants that conduct water and salts that are absorbed by the roots to the leaves and the food prepared in the leaves to various parts, are called vascular tissues. These are called complex tissues as they are. formed by different types of cells. Xylem and Phloem are the main vascular tissues.

Work sheet (Text Book Page No:30)

8th Class Biology Notes Pdf Kerala Syllabus

Answer:
Kerala Syllabus 8th Standard Biology Notes

  • Organs like stomach and Intestine are mainly formed by nervous tissue, epithelial tissue, connective tissue, and muscular tissue.
  • Temporary storage of food materials, digestion and the secretion of digestive juices.
  • Complete digestion of food materials. Secretion of-digestive juices, reabsorption of water, absorption of nutrients.
  • Organ performs the co-ordinated functions of various tissues. The function of an organ is not the same as that of the individual tissues which constitute the organ.
  • Various organs combine to form organ system. As a result physiological functions can be performed very effectively.
    eg: teeth, tongue, glands, oesophagus, stomach, small intestine, large intestine etc constitute digestive system. It enables the complete digestion of food.

Table (Text Book Page No:31)
Complete the following table by finding out the systems to which the organs listed in the table belong to:
Class 8 Science Notes Kerala Syllabus
Answer:

Organs Organ System
Heart, Blood vessels Circulatory System
Nose, Trachea, Lungs. Respiratory System
Kidney, Ureter, Urinary bladder Excretory System
Brain, Nerves Nervous System

An organism is composed of various organ systems. Therefore the structure of organisms is highly complicated.

Cells are the basic unit of life. Cell parts are composed of different substances.

In higher organisms diverse tissues act complementary to each other and perform various life processes.

Cell Clusters Let us assess (Text Book Page No:33)

Given below in the illustration are various tissues related to the structure of hand.
Cell Clusters Class 8 Questions And Answers Pdf Kerala Syllabus
Answer:
1. • A
• C
2. Formed from various cells
3. Only the corners of the cells are thicked

Cell Clusters Text Book Questions and Answers

Question 1.
Arrange an exhibition showing pictures and descriptions of different types of cells.

Answer:
Cell
Cell is the structural and functional unit of life.
• All organisms are formed of one or more cells.
• All vital activities of life take place inside the cells.
• Here dietary information that controls the functions of the cells are contained in the cell.
• There are three types of cells in the blood. Red blood cells, white blood cells and platelets.

Red Blood cells

Cell Clusters Class 8 Notes Kerala Syllabus

Red blood cells are the most abundant cells in the blood. They are also known as erythrocytes. Invertebrates, RBC carries oxygen to the tissues. In RBC, the cytoplasm contains an iron containing biomolecule, the haemoglobin. If in parts red colour to the blood. In human, RBCs has biconcave disc shape.

White blood cells

Basic Science Class 8 Chapter 2 Kerala Syllabus
WBC’S are also called leucocytes, they protect the body from pathogens and other antigens as a part of the immune system. They are formed from the stem cells of the bone marrow. The number of WBCs is an indicator of health. They are classified into neutrophil, eosinophil, basophil, monocyte and lymphocyte according to the diversity of cytoplasm, nucleus etc.

Platelets

Basic Science For Class 8 Chapter 2 Kerala Syllabus
They are also known as thrombocytes. They help in the coagulation of blood. They are formed from the cells called megakaryocytes.

Question 2.
Prepare a magazine specifying the importance, relevance, and scope of stem cells.
Answer:
Stem cell

Stem cell are specified cells that can transform to any kind of cell. They are also known as root cells. They transform to other cells by a long process of differentiation.

Stem cells can either transform to other cells or exist as such. When the cell in the tissues get destructed, new cells originate from stem cells. Stem cells are seen in bone marrow, skin, digestive tract etc.

Today it is possible to develop desired cells from stem cells under specific conditions of laboratories. It is expected to make tremendous change in the treatment of blood cancer, diabetes, parkinson’s diseases etc. and in the development of artificial organs through the research works of in stem cells.

Cell Clusters Additional Questions and Answers

Question 1.
Findi the odd one in each group.
Also write the common characterestic of the others.
a. Collenchyma, Sclerenchyma Xylem, Nervous tissue.
b. Blood, Muscle, ligament, Bone
c. Basophil, Platelet, Neutrophil, Lymphocytes
d. Collenchyma, Aerenchyma, Parenchyma, Chlorenchyma
e. Intercalary meristem, Primary meristem, Apical meristem, Lateral meristem
f. Xylem, Collenchyma, Scler-enchyma, Parenchyma.
g. Man, Pigeon, Duck, Amoeba
h. Large intestine, Digestive tract, Heart, Stomach
i. Bony tissue, Muscular tissue, Cartilaginous tissue, Blood
j. Parenchyma, Collenchyma, Epithelial tissue, Sclerenchyma.
Answer:
a.Nervous tissue; Others are plant tissues.
b. Muscle, Others are connective tissues.
c. Platelets, Others are white blood cells.
d. Collenchyma, Others are the different forms of parenchyma.
e. Primary meristem : Others are divide and new tissues. Primary meristem divide and from apical meristem and lateral meristem.
f. Xylem : It is conduction tissue and others are basic tissues.
g. Amoeba – Cell level organism and
others are organ system level organisms.
h. Heart : It is a circulatory organ. Others are digestive organ.
i. Muscular tissue : The others are connective tissues
j. Epithelial tissue : The others are plant tissues.

Question 2.
Find out the relation between the given word pairs and on that basis fill in the blanks.
a. Water and salts : xylem :: ………… : Phloem
b. Photosynthesis : Parenchyma : Support and Strength : ………..
c. Between bones and muscles : tendons ; Between bones and bone: ………..
d. Blood : Connective tissue : : Skin: ……….. ;
e. Control and Co-ordination : Nervous tissue : : Body Move ment:: ………..
f. Fibrous tissue : Connectother tissue; ……….. : Transport of materials
g. Storage of food : ……….. ; Flexibility to plant parts : Col-lenchyma
h. White blood cells : ……….. ; Redblood cell; O2 transportation
i. Xylem – ………..; Phloem: Transports synthsised food
j. Intercalary meristem : Inter node lengthening; Lateral meristem: ………..
k. Aerenchyma : Air cavities ; Chlorenchyma : ………..
l. Cell :Tissue; system : ………..
Answer:
a. Food
b. Sclerenchyma
c. Ligaments
d. Epithelial Tissue
e. Muscular Tissue
f. Blood
g. Parenchyma
h. Immunity
i. Transports water and minerals
j. Thickening of stem
k. Chlorophyll
l. Organ

Question 3.
Complete the flowchart which shows levels of organisation of human being.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 10
Answer:
a. cell organelle
b. Organ

Question 4.
Observe the following figure and answer the given questions.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 20
a. Find out correct one from the bracket shows the plant tissue given in the figure.
A. parenchyma
B. sclerenchyma
C. xylem
D. collenchyma
b. Write down the characteristic features of the plant cell in the given figure?
c. Give scientific explanation for the features of plant part are given below.
i. Desterity of shaft of colacasia.
ii. Hardness of shell
Answer:
a. B. sclerenchyma
b. composed of cells that are uniformly thick all over the cell wall.
c. i. presence of collenchyma cells, ii. presence of sclerenchyma cells.

Question 5.
Find out wrong items from the following statements, and also correct their underlined words.
a. Sclerenchyma can provides strength and support to plant parts.
b. Fibrous tissue can enables to respond by identifying the changes inside and outside the body.
c. Production of secretion is one of the function of Epithelial tissue.
d. muscular tissue enables the movement of the body.
Answer:

  • b, d are wrong statements.
  • b. Nervous tissue can enables to respond identifying the changes inside and outside the body.
  • d. Muscular tissue enables the movement of the body.

Question 6.
What is the difference between blood and blood circulatory system?
Answer:
Blood is a tissue in the blood circulatory system. But blood vessels, heart, nerves, etc. are the other factors of the blood circulatory system. The combined work of all these factors form the blood circulatory system.

Question 7.
All parts of the plant do not grow. Why?
Answer:
In plants, only the meristematic cells have the ability to divide and grow. Meristematic cells are mainly found in the root apex and stem apex. So the plant growth is concentrated at the root apex and stem apex.

Question 8.
What are tissues? Give examples.
Answer:
A group of similar cells is called tissue
eg : Parenchyma

Question 9.
“Body is composed of millions of cells. Are they perform the same function? How do they act independently?”
This is question raised in the class by the teacher for a discussion. What will be your answer to this question?
Answer:
Cells act in groups of similar kind. Group of similar cells that are originated from a single cell and perform a specific function are called tissues. They perform different functions.
Eg: muscular tissue, Nervous tissue, Epithelial tissue.

Question 10.
Our body is developed from a single-celled zygote. How do this much organs and organ system sare formed in our body?
Answer:
Zygote divides continuously and forms the foetus that consists of cells differ in shape, size, and content. Foetal cells gradually attain diversity in structure and function. Thus different organs and organ systems are formed.

Question 11.

Stem cells for Cancer treatment
Tvm:- Scientists say that the pr­oduction of tissues from stem cells may cause drastic change in the treatment of Cancer…

Did you read the news report? What are stem cells? How do they become useful in the treatment of Cancer?
Answer:
Stem cells are specialized cells that can be transformed to any kind of cell. Stemcells modify to other cells through prolonged differentiation. When cells destroyed, new cells originates from stem cells. In research laboratories desired cells are formed from stem cells. New cells, instead of cancer cells, can be produced from stem cells.

Question 12.
Complete the given flow chart
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 11
Answer:
a. Nervous Tissue
b. Connective Tissue
c. • Covers and protects the body
• The inner lining of alimentary ‘ canal
• Perform the function of protection, absorption, production of secretion, etc.
d. helps to respond by recognizing the changes within and out of the body
e. Makes the movement of body possible.

Question 13.
Categorise the given statements under suitable headings.
1. Provide shape to body.
2. Defence
3. Covers and protects internal organs.
4. Conduction of respiratory gases.
Answer:

Blood Bone, Cartilage (Connective Tissue)
2. Defence 1. Provide shape to body.
4. Conduction of respiratory gases. 3. Covers and protects internal organs.

Question 14.
Identify the figures given below.
Answer:
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 12
a. Nervous tissue
b. Muscular tissue

Question 15.
What are the differences between the process of growth in plants and animals?
Answer:
Plant growth

  • Plants grow throughout their life.
  • In plants, growth is mainly restricted at the tips of stem and root(meristem)

Animal growth

  • Animals grow upto a particular period
  • In animals, growth is not confined to any specific regions.

Question 16.
Why plant growth is confined to specific parts?
Answer:
In plants, specific cells called meristematic cells are concentrated the tips of roots and stems. They have the ability to divide rapidly.

Question 17.
Complete the table properly

Characteristics Meristematic Cells Mature Cells
Relative size of Nucleus (a) ………………. (b) ……………
Thickness of cell wall (c) ………………. (d) ……………
Quantity of Cytoplasm (e) ………………. (f) ……………

Answer:
a. Comparatively large nucleus
b. Small nucleus
c. Thin cell wall
d. Thick cell wall
e. More
f. Comparatively less

Question 18.
How do the water and salts absorbed by the roots reach the leaves?
Answer:
In plants, the water and salts absorbed by the roots are transported to the leaves through specialised tissues called vascular tissues.

Question 19.
Vascular tissues are called complex tissues. Why?
Answer:
Vascular tissues are formed by the union of different types of cells. Hence they are called complex tissues.

Question 20.
Picture of section of stem is given. Label the parts.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 13
Answer:
a. Collenchyma
b. Parenchyma
c. Sclerenchyma
d. Phloem
e. Xylem

Question 21.
Pair the given cells with their specific character

Parenchyma, Collenchyma,  Sclerenchyma.

a. formed of cells whose cell wall is uniformly thickened in all parts.
b. seen in tender parts of plant.
c. formed of cells whose cell wall has thickenings in the comers only
Answer:
parenchyma – b
Collenchyma –
c sclerenchyma – a

Question 22.
Write examples for vascular tissues?
Answer:
Xylem and Phloem

Question 23.
Arrange the given statements in the right column.
1. Formed of inter related cells seen as ducts.
2. Carries water and salts absorbed by the roots to the leaves.
3. Carries the food prepared in the leaves to various plant parts.
4. Provide support and strength to plant
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 15
Answer:
Xylem 2, 4 Phloem -1, 3

Question 24.
Complete the worksheet based on the hints given.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 16
Hints
A. Carries the food prepared in the leaves to various plant parts.
B. Carries water and salts absorbed by the roots to the leaves.
C. Seen in the tender parts of the leaves.
D. Thickenings seen only in the corners of the cell wall
E. Cells have uniform thickenings in the cell wall.
Answer:
A – Phloem
B – Xylem
C – Parenchyma
D – Collenchyma
E – Sclerenchyma

Question 25.
Which are the tissues that constitute the organs like stomach and intestine?
Answer:
Nervous tissue, Epithelial tissue, Connective tissue, Muscular tissue

Question 26.
Identify the organ system to which the given organs are associated?
a. Heart, Blood Vessels
b. Nose, trachea, lungs
c. Kidney, Ureter, Urinary Bladder.
d. Brain, Nerves
Answer:
a – Circulatory system
b – Respiratory system
c – Excretory system
d – Nervous system

Question 27.
Which of the statements is not related to tissues?
a. Different types of cells are seen.
b. Similar kind of cells are seen.
c. Performs specific function.
d. Formed from different cells
Answer:
d. Formed from different cells

Question 28.
Which indicator helps to identify collenchyma when tissues are observed through a microscope.
a. No thickenings in the cell wall.
b. all parts of the cell wall is thick,
c. No nucleus in the cell
d. Thickenings are seen only in the comers of the cell.
Answer:
d. Thickenings are seen only in the comers of the cell.

Question 29.
Complete the flow chart showing the levels of organization in organisms
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 17
Answer:
a. tissue
b. Organ system
c. Organism
d. Species

Question 30.
What is meant by voluntary muscles and involuntary muscles? Give examples:
Answer:
Tissues that work according to our wish are called voluntary muscles.
eg: muscles of limbs.
Muscles which works independently and are out of our control are called involunatary muscles.
eg : Muscles of the alimentary canal, muscles of the eyelids, etc.

Question 31.
Xylem vessels are thicker than phloem vessels. Why?
Answer:
The cell wall of the xylem vessels is comparatively thicker than that, of the phloem vessels. Lignin is used to make the cell wall strong. So the xylem vessels are thicker.

Kerala Syllabus 8th Standard Hindi Solutions Unit 3 Chapter 2 बात उस मंगलवार की

You can Download बात उस मंगलवार की Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Hindi Solutions Unit 3 Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Hindi Solutions Unit 3 Chapter 2 बात उस मंगलवार की (डायरी)

बात उस मंगलवार की पाठ्यपुस्तक के प्रश्न और उत्तर

Bath Us Mangalvar Ki Kerala Syllabus 8th प्रश्ना 1.
मेहनत की कमाई का भोजन स्वदिष्ठ क्यों हो जाता है?
8th Standard Hindi Notes State Syllabus
उत्तर:
श्रम के कारण भूख लगती है। भूख मिटाने के लिए जब खाना खाते हैं, वह अधिक स्वादिष्ठ होता है। बेकार में बैठकर खानेवाले को इसी प्रकार की अनुभूति नहीं होती।

बात उस मंगलवार की Notes Kerala Syllabus 8th प्रश्ना 2.
मरीज़ गैर ज़रूरी इंजेक्शन चाहते हैं। क्यों?
Kerala Syllabus 8th Standard Hindi Textbook
उत्तर:
बीमारी और इलाज़ के संबंध में बहुत गलतफहमियाँ हैं। आम जनता चाहती है कि इंजेक्शन से बीमारी जल्दी से दूर होती है। यह विचार चिकित्सा के संबंध में उनकी अज्ञता के कारण है।

Bath Use Mangalwar Ki Kerala Syllabus 8th प्रश्ना 3.
“यहीं जंगल के बीच ये सभी मेरे क्लीनिक हैं।” -इससे आपने क्या समझा?
Class 8 Hindi Notes Kerala Syllabus
उत्तर:
यहाँ डॉक्टर का मनोभाव प्रकट होता है। यह डॉक्टर चिकित्सा को व्यापार नहीं . मानती है। वे इसे सेवाकार्य मानती है। इसलिए शहर में उनकी अपनी क्लीनिक नहीं है। जंगल के अनपढ़, अशिक्षित, गरीब ही उनके मरीज़ हैं। वे उनके लिए काम – करती हैं।

बात उस मंगलवार की Textbook Activities

Kerala Syllabus 8th Standard Notes Hindi प्रश्ना 1.
डॉ रमणी अटकुरी की चरित्रगत विशेषताएँ लिखें।
Hss Live Guru 8th Hindi Kerala Syllabus
उत्तर:
रमणी अटकुरी एक ईमानदार डॉक्टर हैं। वे डॉक्टरी को व्यापार मानती नहीं। उनके अनुसार डॉक्टर को समाज की सेवा करनी चाहिए। इस आदर्श को वे अपने जीवन में निभाती है। वह एक आदर्श डॉक्टर हैं।

Hsslive Guru 8th Class Hindi Kerala Syllabus प्रश्ना 2.
आपके दृष्टिकोण में एक डॉक्टर के गुण क्या-क्या हैं? टिप्पणी लिखें।
8th Standard Hindi Guide Kerala Syllabus
उत्तर:
मेरी राय में डॉक्टर को ईमानदार होना चाहिए। अपने पेशे को धन कमाने का उपाय न मानना चाहिए। मरीज़ों के पास जाकर उनकी सेवा-सुश्रूषा करनी चाहिए। मरीज़ों के साथ सहानुभूति पूर्ण व्यवहार अपनाना चाहिए। अपने उत्तरदायित्व को निभाते हुए समाज के लोगों को स्वस्थ रखने के कामों में भाग लेना चाहिए।

Bath Us Mangalvar Ki Question Answer Kerala Syllabus 8th प्रश्ना 3.
संवाद लिखें।
Kerala Syllabus 8th Standard Hindi Solutions Unit 3 Chapter 2 बात उस मंगलवार की 13
8th Std Hindi Notes Kerala Syllabus
उत्तर:
Kerala Syllabus 8th Standard Hindi Notes

Hss Live Guru 8 Hindi Kerala Syllabus प्रश्ना 4.
संवाद लिखें।
Kerala Syllabus 8th Standard Hindi Solutions Unit 3 Chapter 2 बात उस मंगलवार की 15
उत्तर:
Hindi Class 8 Kerala Syllabus
8std Hindi Notes Kerala Syllabus

बात उस मंगलवार की Summary in Malayalam and Translation

8 Std Hindi Textbook Kerala Syllabus
Bath Us Mangalvar Ki Kerala Syllabus 8th

बात उस मंगलवार की शब्दार्थ Word meanings

बात उस मंगलवार की Notes Kerala Syllabus 8th

Kerala Syllabus 8th Standard Maths Solutions Chapter 3 Polygons

You can Download Polygons Questions and Answers, Activity, Notes, Kerala Syllabus 8th Standard Maths Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Maths Solutions Chapter 3 Polygons

Polygons Text Book Questions and Answers

Textbook Page No. 49

Class 8 Maths Polygon Kerala Syllabus Question 1.
Find the sum of angles of a polygon with 52 sides?
Solution:
Sum of angles = (52 – 2) 180
= 50 × 180
= 9000

Hsslive Guru 8th Class Maths Kerala Syllabus Question 2.
The sum of angles of a polygon is 8100°. Find the number of its sides?
Solution:
Let sides = n
∴ Sum of angles = 8100
(n – 2) 180 = 8100
n – 2 = \(\frac{8100}{180}\)
= 45
n = 47, Number sides of the polygon = 47

Hss Live Maths 8th Standard Kerala Syllabus Question 3.
Can 1600 be the sum of angles of a polygon. Can 900° be the sum?
Solution:
1600 is not the multiple of 180°. So it cannot be the sum of the angles. 900° is the multiple of 180. So 900° can be the sum of angles of a polygon.

Polygons Chapter For Class 8 Kerala Syllabus  Question 4.
All the angles of a polygon with 20 sides are equal. Find the measure of each angle?
Solution:
Sum of angles of a polygon with 20 sides. = (20 – 2) × 180
= 18 × 180
Each angle = \(\frac{18 \times 180}{20}\) = 162°

Hss Live Guru 8th Maths Kerala Syllabus Question 5.
The sum of angles in a polygon is 1980. Find the sum of angles of the polygon with one side more. Find the sum with one side less?
Solution:
When one side of a polygon increases, the sum of angle increases by 180°.
∴ Sum of angles = 1980 + 180 = 2160°
When the number of sides decreases by 1, the sum of angles also decreases by 180°.
∴ Sum of angles = 1980 – 180 = 1800°

Textbook Page No. 51

Polygon Chapter Class 8 Kerala Syllabus Question 1.
Two angles of a triangle are 40°, 60° each. Find the measure of the exterior angles?
Solution:
Third angle = 180 – (40 + 60) = 80°
Exterior angles = 180 – 40, 180 – 60,
180 – 80, 140,
ie, 140°, 120° and 100°

Class 8 Polygon Chapter Kerala Syllabus Question 2.
Find all the angles in the figure.
Class 8 Maths Polygon Kerala Syllabus
Solution:
∠ABC = 180 – 105 = 75°
∠C = 180 – (35 + 75)
= 180° – 110° = 70°
Exterior angle at A = 180° – 35° = 145°
Exterior angle at B = 180° – 75° = 105°
Exterior angle at C = 180° – 70°
= 110°
Hsslive Guru 8th Class Maths Kerala Syllabus

Hsslive Maths Class 8 Kerala Syllabus Question 3.
Find all the exterior angles of the quadrilateral in the figure.
Hss Live Maths 8th Standard Kerala Syllabus
Solution:
4th angle = 360 – (130 + 70 + 60) = 100°
Exterior angles = 120°, 110°, 50°, 8o°

8th Class Maths Notes Kerala Syllabus Question 4.
Find all the angles of the given diagrams
Polygons Chapter For Class 8 Kerala Syllabus
Solution:
(i)
Hss Live Guru 8th Maths Kerala Syllabus
Angles of the triangle are 70°, 35°, 75°
Exterior angles are 35°, 110°, 105°

(ii)
Polygon Chapter Class 8 Kerala Syllabus
Angles of the quadrilateral are
∠C = 85°, ∠B = 85°, ∠A = 100°, ∠D = 90°
Exterior angles are 95°, 80°, 90°, 95°
All angles at D are 90°
Angles at C are 85°, 95°, 85°, 95°

(iii)
Class 8 Polygon Chapter Kerala Syllabus
Angles of the quadrilateral are :
∠ A = 65°, ∠B = 75s°, ∠C = 100°, ∠D = 120°
Exterior angles are 115°, 105°, 80°, 60°
Angles at C are 100°, 80°, 100°, 80°

Hss Live Guru 8 Maths Kerala Syllabus  Question 5.
Prove that in a triangle, the exterior angle at a vertex is the sum of interior angles at the other two vertices
Solution:
Hsslive Maths Class 8 Kerala Syllabus
∠DBC + ∠ABC = 180° (linear pair)
∠A + ∠C + ∠ABC = 180° (Angles of a triangle)
∴ ∠A + ∠C = ∠DBC
So in a triangle the exterior angle at a vertex is the sum of interior angles at the other two vertices.

Textbook Page No. 54

Hsslive Guru 8th Maths Kerala Syllabus Question 1.
All the angles of an eighteen sided polygon are equal. Find each exterior and interior angles?
Solution:
The angles of the polygon are equal. So the exterior angles are also equal. Sum of the exterior angles = 360°
Measurement of an interior angle = \(\frac{360}{18}\) = 20°

Hsslive Guru 8 Maths Kerala Syllabus Question 2.
The sides PQ and RS of the quadrilateral PQRS are parallel. Find all the exterior and interior angles of the quadrilateral.
Solution:
8th Class Maths Notes Kerala Syllabus
PQ and RS are parallel to each other ∠P = 500, The exterior angle at S = 500
∴ ∠S = 130°
∠Q = 360 – (50 + 130 + 110)
= 360 – 290 = 70°
The angles of the quadilateral are ∠P = 50°, ∠Q = 70°, ∠R = 110°, and ∠S = 130°
The exterior angles are 130°, 110°, 70°, 50°

Hss Live Guru Class 8 Maths Kerala Syllabus Question 3.
Hss Live Guru 8 Maths Kerala Syllabus
Draw a quadrilateral and mark the exterior angles at two vertices. Is there any relationship between the sum of these angles and the sum of the interior angles at the other two vertices?
Solution:
Hsslive Guru 8th Maths Kerala Syllabus
Sum of the angles of the quadrilateral = 360°
Sum of the exterior angles = 360°
Let the sum of two exterior angles = x
The sum of two interior angles at the same vertex = 360 – x. So the sum of the other two interior angles = x. So sum of two exterior angles at two vertices is equal to the sum of two interior angles at the other two vertices.

Kerala Syllabus 8th Standard Maths Notes Question 4.
An exterior angle in a polygon with all angles are equal is twice of an interior angle.
(i) Find the measure of each angle in it?
(ii) Find the number of sides?
Solution:
The exterior angles are equal as all the angles are equal.
Let the interior angle is x, then the exterior angle is 2x.
x + 2x = 3x = 180
∴ x = 60
(i) Interior angles are 6o° each and exterior angles are 120° each.
(ii) The polygon has 3 sides. It is a triangle.

Hsslive Guru Maths 8th Standard Kerala Syllabus Question 5.
The sum of exterior angles in a polygon is twice the sum of the interior angles.
(i) Find how many sides the polygon has?
(ii) Find the number of sides, if the sum of the exterior angles is half of the sum of the interior angles.
(iii) Find the number of sides if the sum of the exterior angles is equal to the sum of the interior angles?
Solution:
The sum of exterior angles in a polygon is 360°.
(i) The sum of the interior angles in a triangle is 180°. It is a triangle and has 3 sides.
(ii) If the sum of the interior angles is 720°, the polygon has six sides.
(iii) If the sum of the interior angles is 360°, the polygon is a quadrilateral it has 4 sides.

Textbook Page No. 58

Hss Live Guru Maths 8th Kerala Syllabus Question 1.
Draw a hexagon with all sides equal but angles different?
Solution:
The sum of angles of a hexagon is 720. Make 720, the sum of 6 different angles. Draw a line with a definite length and make an angle at its end. Then draw the next line and angle. Draw 6 lines continuously and you get a hexagon.

Std 8 Maths Kerala Syllabus Kerala Syllabus Question 2.
Draw a hexagon with all the angles are equal and sides are different.
Solution:
Draw a line. Draw another line by taking an angle 1200 at its one end. Then draw different lines by taking angles 1200. By joining we get a hexagon.

Hsslive 8th Class Maths Kerala Syllabus Question 3.
Find the measurements of each angle in an equal polygon with 15 sides. Find each exterior angles?
Solution:
The sum of angles of a polygon with
15 sides = (15 – 2) × 180
= 13 × 180 = 2340
One angle = \(\frac{2340}{15}\) = 156°
Exterior angle = 180 – 156 = 24°

Kerala Syllabus 8th Standard Maths Notes Pdf Question 4.
One angle of a regular polygon is 1680. Find the number of its sides?
Solution:
Exterior angle = 180 – 168 = 12°
Sum of exterior angles = 360°
∴ Number of sides = \(\frac{360}{12}\) = 30

Hsslive Class 8 Maths Kerala Syllabus Question 5.
Can you draw a regular polygon with each exterior angle 6°. Can you draw it if the exterior angle is 7? .
Solution:
Sum of exterior angles = 360°
One of the exterior angle = 6°
Number of sides = \(\frac{360}{6}\) = 60, yes we can draw; If one exterior angle is 7°
Number of sides = \(\frac{360}{7}\) = 51.42
Not a whole number. The polygon cannot be drawn.

Question 6.
A regular pentagon and regular hexagon are jointly drawn in the figure. Find PQR?
Hsslive Guru 8 Maths Kerala Syllabus
Solution:
Sum of angles of a regular pentagon = (5 – 2) 180°
= 3 × 180 = 540°
One angle of a regular pentagon
= \(\frac{540}{5}\) = 108°
Sum of angles of a regular hexagon = (6 – 2) 180°
= 4 × 180 = 720°
One of its angle = \(\frac{720}{6}\) = 120°
∴ ∠PQR = 360 – (108 + 120)
= 360 – 228 = 132°

Question 7.
A square, a regular pentagon and a regular hexagon are jointly drawn in the figure Find ∠BAC
Hss Live Guru Class 8 Maths Kerala Syllabus
Solution:
One angle of a regular pentagon = 108°
One angle of a regular hexagon = 120°
One angle of a square = 90°
∴ ∠BAC = 360 – (108 + 20 + 90)
360 – 318 = 42°.

Question 8.
In the figure below, ABCDEF is a regular hexagon. Prove that ∆ BDF is an equilateral triangle?
Kerala Syllabus 8th Standard Maths Notes
Solution:
Sum of angles of a regular hexagon = (6 – 2) 180 = 720°
= 4 × 180 = 720°
One angle = \(\frac{720}{6}\) = 120°
consider ∆ EFO
∠E = 120°
∠EFD = ∠EDF = 30°(Angles opposite to equal sides of an isosceles triangle are equal)
Similarly ∠AFB = 30°
∴∠DFB = 120 – (30 + 30) = 60°
∴ ∠FBD = 60°,
∠FDB = 60°
∴ ∆ FDB is an equilateral triangle.

Question 9.
In the figure below ABCDEF is a regular hexagon. Prove that ACDF is rectangle.
Hsslive Guru Maths 8th Standard Kerala Syllabus
Solution:
One angle of a regular hexagon = 120°
∠EFD = ∠EDF = 30°; ∠F = 120°
∴ ∠AFD = 90°
Similarly all the angles of ACDF is 90°
∴ ACDF is a rectangle.

Polygons Additional Questions and Answers

Question 1.
How many triangles can be for made if all the diagonals from a vertex of a 10 sided polygon are drawn?
Solution:
7 diagonals 8 triangles.

Question 2.
The sum of angles of a polygon is x°. Find the sum of angles of the polygon with one side is more. Find the number of sides if one side is less?
Solution:
(x + 180°), (x – 180°)

Question 3.
Can it possible that the external angle of a polygon be 13?
Solution:
\(\frac{360}{13}\) is a fraction. So it is not possible.

Question 4.
The external angles of a triangle from the three vertices are (2x + 30°), (3x – 10°), 100°. Find the value of x?
Solution:
2x + 30 + 3x – 10 + 100 = 360
5x + 120 = 360
5x = 360 – 120 = 240
∴ x = \(\frac{240}{5}\) = 48

Question 5.
Draw a circle and construct a regular pentagon with all the vertices in it?
Solution:
Draw circle and construct the pentagon.

Question 6.
Find the sum of angles of a heptagon? Find an angle if all the angles are equal?
Solution:
Sum of angles of a heptagon = (7 – 2) × 180°
= 5 × 180° = 900°
All the angles are equal. So one angle
= \(\frac{900}{7}\) = 128.571………..

Question 7.
Find the sum of the angles of the polygon with the given sides?
(a) 12
(b) 15
(c) 20
(d) 24
Solution:
(a) Sum of angles = (12 – 2 ) × 180
1800°
(b) Sum of angles = (15 – 2) × 180
= 2340°
(c) Sum of angles = (20 – 2) × 180
= 3240°
(d) Sum of angles = (24 – 2) × 180
= 3960°

Question 8.
The angles of a hexagon are (x – 10)°, x°, (x + 10)°, (x + 20)°, (x + 30)°, and (x + 40)°. Find the value of x?
Solution:
Sum of the angles of a hexagon = (6 – 2) × 180° = 720°
ie, (x – 10) + x + (x + 10) + (x + 20) + (x + 30) + (x + 40) = 720
6x + 90 = 720
ie, 6x = 630; x = 630 ÷ 6
∴ x = 105°

Question 9.
The sum of the angles of a polygon with 12 sides is 1800°. Find the sum of the angles of a polygon with 13 sides?
Solution:
When one side is increased, the angle is increased by 180°.
∴ Sum of the angles of a 13 sides Polygon = 1800° + 180 = 1980°

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 10 Force

You can Download Force Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 10 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 10 Force

When an object is pushed or pulled, a force is being applied on it. The unit of force is newton. It is indicated by the letter N.

Force is that which changes or tends to change the shape, size, volume, state of rest or state of motion of a body.

There are various types of forces such as muscular force, magnetic force, electric force, gravitational force, frictional force etc.

Contact force and non contact force

The force applied by the contact between objects is contact force. The force applied on an object without contact on it is non contact force.
Ex:pushing a trolley- contact force, Magnet attracting a nail-non contact force

Kerala Syllabus 8th Standard Physics Notes Frictional force:

When a ball is rolling on the floor after some time it stops. This is because of the frictional force between the ball and floor. On the surface of the objects which are on contact with each other there are many mounts and pits get interlocked.

When we try to move a body by applying a force parallel to their surfaces an opposing force is experienced.
When a body moves or tends to move on the surface of another body, a force is experienced parallel to the surface which opposes the relative motion between them. This is friction.

Rolling friction and sliding friction

When a body rolls over the surface of another body, the force of friction which originates is rolling friction. When a body slides on the surface of another body the force of friction which originates is sliding friction.

Rolling friction is less than sliding friction. Striking a match stick on the side of a match box, wear and tear of machines, ability to hold objects, walking etc are advantages of the friction and treading of tyres of vehicles, wearing out of tyres, fuel loss are disadvantages of friction.

The materials like oil, grease which are used to reduce friction are the lubricants. Graphite is a solid lubricant.

Force Class 8 Kerala Syllabus Thrust and Pressure

The total normal force experienced by a ‘ surface is thrust. Thrust per unit area is pressure.
Pressure = \(\frac{Thrust}{area}\)

Unit of pressure is N/m2. This is known as pascal. The force acting on a larger area exerts a smaller pressure, while a force acting on a smaller area exerts a large pressure.

Hsslive Guru 8th Class Physics Kerala Syllabus Liquid Pressure

The pressure is experienced in liquids also. As the height of liquid column increases the pressure exerted by it increases.

The pressure exerted by a liquid column increases with increase in height. The thrust acting per unit area by a liquid is liquid pressure.

Liquids exert force on all sides of the container in which they are taken.

If the height of the liquid column is Ti’, density of liquid ‘d’ and acceleration due to gravity ‘g’, then liquid pressure is P = hdg

Class 8 Physics Kerala Syllabus Atmospheric pressure

Atmospheric air also can exert pressure. An envelope of air surrounds the earth. This is earth’s atmosphere. The density of atmospheric air near the surface of earth is greater and it decreases when we go up. The weight of air column over the unit area of earth’s surface is atmospheric pressure. One atmospheric pressure is the weight of mercury column of 0.76 m height and unit area (1m²). This is standard atmospheric pressure .

The unit of atmospheric pressure is* bar. The instrument used to measure atmospheric pressure is Barometre

Force Textbook Questions and Answers

Kerala Syllabus 8th Standard Physics Notes Pdf Question 1.
Classify the following situations into contact and noncontact forces.
a. Applying break in a bicycle.
b. A mango falling from a mango tree.
c. The earth revolving around the sun.
d. The speed of a hall rolling on ground is reduced.
Answer:
Contact force
a. Applying break in a bicycle.
d. The speed of a ball rolling on ground is reduced
Non contact force
b. A mango falling from a mango tree.
c. The earth revolving around the sun.

Class 8 Physics Notes Kerala Syllabus Question 2.
State reason
a. We are able to walk on the ground without slipping.
b. It is easy to cut vegetables using a sharp knife.
c. The number of tyres is more for goods vehicles.
d. The moving parts of machines experience wear and tear.
Answer:
a. Because of the friction between ground and feet
b. When the area decreases pressure increases. As the area of the edge of the knife is veiy less, it is easy to cut
c. When the area of surface increases pressure decreases, so when more wheels are used it won’t depressed in soil.
d. Because of the friction between the area of contacts.

Hss Live Guru 8th Physics Kerala Syllabus Question 3.
Match the following

A B C
Atmospheric pressure pascal coconut oil
Lubricant attration barometer
Coconut falling down bar thermometer
Magnet friction noncontact
gravitation repulsion

Answer:

Atmospheric pressure bar barometre
Lubricant friction Coconut oil
Coconut falling down Gravitational force Non contact
Magnet attraction repulsion

Hsslive Guru Physics 8th Standard Kerala Syllabus Question 4.
Bubbles rising from the bottom of the water filled in a bottle are depicted in the figure. Which is the correct figure? Justify your answer.
Kerala Syllabus 8th Standard Physics Notes

Answer:
Fig.c. There will be more pressure in the lower side of the liquid. The pressure decreases when it comes upward.

Hss Live Guru 8 Physics Kerala Syllabus Question 5.
A toy car of about 50 g placed on a polished table with threads attached to it carrying two pans passed through pulleys is depicted.
Force Class 8 Kerala Syllabus

(a) What do you observe if too g each is placed on both pans?
(b) What do you observe if too g is placed on pan A and 200 g in pan B
(c) Justify your answers.
Answer:
a. The car will be stationary because the force excreted from both sides are equal.
b. The pan of weight 200 will pull with more force and it moves that direction.
c. In the first phase same force is excerted. But in second phase force is exerted to one direction.

Force Additional Questions and Answers

Force And Pressure Class 8 Worksheets With Answers Pdf Kerala Syllabus Question 1.
Tabulate the following as contact force and non contact force.
1. pushing a trolley
2. falling coconut from coconut tree
3. magnet attract iron nail
4. Drowing water from the well
5. pushing the car
6. earth is moving around the sun
Answer:
contact force – 1,4,5
non contact force – 2,3,6

8th Standard Physics Notes Kerala Syllabus Question 2.
Ramu tried to push a round stone and rectangle stone of same weight. Which is easy to push.
Answer:
It is easy to push round stone because rolling friction is less than sliding friction.
3. Which are the following occation friction is beneficial and non beneficial
1. striking a match stick on the side of a match box
2. wear and tear of machines.
3. ability to hold objects
4. wearing out of tyres
5. walking 6. loss of fuel
Answer:
Beneficial : 1, 3, 5
non beneficial : 2, 4, 6

8th Std Physics Notes Kerala Syllabus Question 4.
Why does the design of ship and aeroplane are streamlining?
Answer:
To reduce friction while moving through air and water.

Kerala Syllabus 8th Standard Physics Question 5.
Write one solid lubricant and liquid lubricant
Answer:
Solid-graphite
Liquid-oil

Kerala Syllabus 8th Standard Science Notes Question 6.
Fill up suitably.
Force – newton
Pressure – ………..
Answer:
Pressure – pascal

8th Standard Physics Notes Pdf Kerala Syllabus Question 7.
Write the reason
1. Constructing a knife with sharp edge
2. Constructing the building with wide basement
Answer:
1. When the area of contact decreases the pressure increases
2. When area increases the chance to depress the base in soil is very less.

Class 8 Basic Science Solutions Kerala Syllabus Question 8.
Given figures, one can filled with water and other can filled with kerosine.
Hsslive Guru 8th Class Physics Kerala Syllabus
Which balloon is filled with water
Answer:
fig 2. When density of liquid increases the pressure increases, water is denser than kerosene.

Physics Class 8 Kerala Syllabus Question 9.
What are the factors effecting liquid pressure? Write the mathematical equation.
Answer:
The height of the liquid column is ‘h’, density of liquid ‘d’ and acceleration due to gravity ‘g’ are the factors. The equation is liquid pressure P = hdg

Basic Science Class 8 Kerala Syllabus Question 10.
As mountaineers climb higher altitude there is a possibility of nasal bleeding. Why?
Answer:
Going higher altitude the atmospheric pressure decreases. So blood veins are broken and blood is oozing out.

Question 11.
What is standard atmospheric pressure
Answer:
Atmospheric pressure at the sea level is equal to the pressure exerted by 0.76m of mercury column. This is standard atmospheric pressure.

Question 12.
When a hole is formed at the bottom of the water bottle, Anu closed the lid of the bottle tightly. The flow of water stops. Explain
Answer:
When the lid is closed tightly, the effect of atmospheric pressure is not experienced on the surface of the water in the bottle. Hence water cannot flow out from the bottle.

Question 13.
Mercury remains at a height of 76cm in a barometer. Raju made a hole at the top end of the tube. What will you observe? How can it be explained?
Answer:
Mercury will flow downwards when a hole is made at the top part of the tube. The atmospheric pressure exerted on the mercury in the vessels supports the mercury level in the tube. When a hole is made at the top, air enters the tube and the pressure is distributed throughout uniformly and hence the mercury level comes down.

Question 14.
A sharp nail pierces more easily through wood than a blunt nail. What is the reason?
Answer:
The surface area of the sharp tip of a nail is very small. Hence the pressure that it can exert on the wood is very great. That is why it pierces easily. For a blunt nail, the surface area of the tip is greater. Hence it can exert only a small amount of pressure.

Question 15.
What is meant by limiting friction.
Answer:
Limiting friction is the force of static friction acting just before the body starts moving.