Students can Download Maths Chapter 6 Construction of Quadrilaterals Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 8th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
You can Download Water Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 16 help you to revise complete Syllabus and score more marks in your examinations.
Water is a precious natural resource. Water is essential not only for sustaining life but also for agriculture, industry, energy production and transportation. Water is formed by the combination of hydrogen and oxygen. Hydrogen and oxygen are produced by decomposition of water. Decomposition of water is done by electrolysis of water.
Basic Science Class 8 Chapter 16 Characteristics of water
Water is a substance found in nature in all the three states of matter namely solid, liquid and gas. The temperature at which a liquid boils at normal atmospheric pressure is its boiling point. The boiling point of water is 100°C. Once water starts boiling, the temperature will not change because all the heat supplied is utilised for the change of state. Water has the ability to hold more heat than other liquids. The ability to hold heat by a liquid is its heat capacity. Because of its high heat capacity water is utilised in radiators to regulate heat, to cool hot objects, to control the temperature of earth etc.
Water becomes ice in 0°C. The temperature of a liquid at which it freezes to solid at normal atmospheric pressure is called its freezing point. When water is converted to ice its volume increases and the density decreases. The unbalanced attractive force on the surface of a liquid causes the liquid, surface behaves like a stretched membrane. This is surface tension. The surface tension decreases the surface area of a liquid. The liquid drops remain like a ball to decrease the surface area. Adding soap to water is a way to decrease surface tension of water.
Class 8 Chemistry Notes Kerala Syllabus The components of water
When electricity is passed through water it decomposes to hydrogen and oxygen. For this, we can make water voltmeter shown in fig.
Iron nail and plastic bottle can be used for this. Adding a few drops of acid into water and battery are also used. When electricity is passed through it oxygen and hydrogen are filled in the test tubes. This process is called electrolysis. The apparatus used for the electrolysis of water is called Hoffmann Water Voltameter.
Kerala Syllabus 8th Standard Physics Notes Reaction of water with metals
Coldwater reacts with sodium, potassium etc. to liberate hydrogen. Magnesium reacts with hot water and iron reacts with steam to liberate hydrogen gas.
Hss Live Guru 8 Chemistry Kerala Syllabus Water the Universal solvent
Several substances dissolve in water. Since water can dissolve various substances and widely used for preparing solutions it is a universal solvent.
8th Standard Chemistry Textbook Soft water and hard water
The water in which soap does not lather easily is called hard water. Water in which soap gives lather readily is called soft water. The hardness of water containing calcium or magnesium bicarbonate is removed during boiling. This type of hardness is called temporary hardness. But the hardness of water containing the chlorides and sulphates of calcium and magnesium is not removed by boiling. Such hardness is called permanent hardness. Pure water has neither the properties of acid nor those of alkali, hence it is a neutral solvent.
Std 8 Chemistry Notes Kerala Syllabus Water pollution
The following activities cause the water pollution
Kerala Syllabus 8th Standard Basic Science Notes Questions 1.
When water is heated at its boiling point or melting point, its temperature does not change.
a. What is meant by boiling point and melting point?
b. What are the boiling and freezing points of water?
c. Why is there no change in temperature?
Answer:
a. The temperature at which a liquid boils at normal atmospheric pressure is its boiling point. The temperature of a liquid at which it freezes to solid at normal atmospheric pressure is called its freezing point.
b. 100°C, 0°C.
c. Once water starts boiling, the temperature will not change because all the heat supplied is utilized for the change of state. So the temperature does not vary.
8th Class Chemistry Notes Kerala Syllabus Question 2.
A definite quantity of water and coconut oil are heated in separate test tubes using the same amount of heat.
a. In which case does the temperature increase slowly?
b. What is the reason for this?
c. Write any one practical application of this property.
Answer:
a. water
b. The high heat capacity of water
c. 1. To regulates the heat of radiator in vehicles
2. To regulate the heat of earth
8th Standard Water Lesson Kerala Syllabus Question 3.
100 ml. each of coconut oil and water are taken in two beakers and kept in the freezer.
a. What difference can be observed in their volumes during freezing?
b. What do you infer from the observation?
c. When water is frozen in glass bottles, it is advised not to fill the bottles completely. Explain the reason.
Answer:
a. The volume of the beaker containing water is greater.
b. Due to the anomalous expansion of water.
c. The volume of water in the bottle filled with water is increased and the bottle is broken.
Hss Live Guru 8 Physics Kerala Syllabus Question 4.
Soap decrease the surface tension of water.
a. What is surface tension?
b. How does the decrease in surface tension benefit washing of clothes?
Answer:
a. The unbalanced attractive force on the surface of a liquid causes the liquid surface behaves like a stretched membrane. This is surface tension
b. When we add soap to water it reduces the surface tension of water and water can enter into the fine threads of the fabric.
Basic Science For Class 8 Chapter 16 Question 5.
Surface tension tends to minimise the surface area of a liquid. Suggest an experiment to prove this. (Follow the pattern: Required materials, procedure, expected observation).
Answer:
’Materials required: metallic loop, thread, soap water, pin Method of experiment: Tie a thread to a metallic loop, immerse it in soap water and create a soap film as shown in fig. a, b, c.
Prick a portion of film using a pin. The shape of remaining portion of soap film will be as such the surface area is reduced.
Basic Science Class 8 Ch 16 Kerala Syllabus Question 6.
Providing excess food for fish in an aquarium is a threat to its survival. Justify.
Answer:
When we add more food materials into water, the dissolved oxygen is utilized to decompose the food materials. So the amount of oxygen is reduced and the water is polluted.
Hsslive Guru 8th Class Chemistry Question 7.
Some substances when dissolved in water cause hardness of water.
a. Which of the following substances cause hardness of water? Sodium chloride, Calcium bicarbonate, Calcium carbonate, Calcium sulfate, Magnesium sulfate, Calcium chloride, Magnesium carbonate
a. The hardness due to which of the above salts cannot be removed by boiling?
Answer:
a.calcium bicarbonate, calcium sulfate, calcium chloride.
b. The hardness of water containing the chlorides and sulphates of calcium and magnesium is not removed by boiling.
Kerala Syllabus 8th Standard Chemistry Notes Question 1.
a. What are the gases liberated in the water voltameter arranged for decomposition of water?
b. Which gas is liberated at the negative pole of the battery?
Answer:
a. Hydrogen and oxygen
b. Hydrogen
c. Ratio of hydrogen and oxygen – 2: 1
8th Class Biology Notes Pdf Question 2.
Heat equal quantity of water and coconut oil in two vessels.
a. The temperature of which liquid will be raised rapidly?
b. Which has more heat capacity?
Answer:
a.Coconut oil
b.water
Hss Live Guru 8th Chemistry Kerala Syllabus Question 3.
Take water in a vessel and put a needle on the surface of water carefully.
a. Will the needle be immersed in water. Justify your answer.
b. Why do the water drops assume spherical shape?
Answer:
a. No. Because of the surface tension of water.
b. The surface tension reduces the surface area.
Hsslive Guru 8th Basic Science Question 4.
Name two metals which react with cold water.
Answer:
Potassium, sodium
Hsslive Guru Chemistry 8 Kerala Syllabus Question 5.
Which is the metal reacts with hot water? Which is the gas produced? Write the chemical equation.
Answer:
Magnesium; Hydrogen.
Mg + H2O → MgO + H2
Hss Live Guru 8th Basic Science Question 6.
Why does water is known as universal solvent?
Answer:
Since water can dissolve various substances and widely used for preparing solutions it is a universal solvent.
8th Physics Guide Kerala Syllabus Question 7.
What are hard water and soft water?
Answer:
The water in which soap does not lather easily is called hard water. Water in which soap gives lather readily is called soft water.
Class 8 Physics Notes Kerala Syllabus Question 8.
Give the reason for temporary hardness of water.
Answer:
The hardness of water is due to the presence of calcium or magnesium bicarbonate. This type of hardness is called temporary hardness.
8th Standard Chemistry Textbook Kerala Syllabus Question 9.
Hardness of water is not removed even after it is boiled. What may be the reason for this?
Answer:
This is due to chloride and sulphate of magnesium and calcium are dissolved in water.
Class 8 Basic Science Chapter 16 Question 10.
Why is air continuously introduced into the water in an aquarium?
Answer:
Aquatic plants and animals make use of oxygen dissolved in water. Water gets polluted as the amount of oxygen in it decreases. Air is introduced into water to make up the quantity of oxygen.
Question 11.
Write two remedies to prevent water pollution.
Answer:
Question 12.
Write two methods to reduce the surface tension of water
Answer:
You can Download Equal Triangles Questions and Answers, Activity, Notes, Kerala Syllabus 8th Standard Maths Solutions Chapter 8 help you to revise complete Syllabus and score more marks in your examinations.
Textbook Page No 149
Area Of Quadrilateral Class 8 Scert Chapter 8 Question 1.
Draw a parallelogram of sides 5 centimetres, 6 centimetres and area 25 square centimetres.
Solution:
Class 8 Mathematics Area Of Quadrilaterals Kerala Scert Solutions Question 2.
Draw a parallelogram of area 25 square centimetres and perimeter 24 centimetres.
Solution:
Area of the parallelogram to be 25 cm2. The product of one side and distance to the opposite side to be 25 cm2.
One side and distance to the opposite side can be 5 cm each.
The other side is = 12 – 5 = 7 cm.
Draw a square with side cm and draw an arc of radius 7 cm with A as centre such that it intersects the side CD. Mark the point E and join AE. Draw an arc of radius 7 cm with B as centre and draw another arc of radius 5 cm with E as centre, Two arcs meet at the point
P. JoinBF and CF.
Area Of Quadrilaterals Class 8 Kerala Syllabus Question 3.
In the figure, the two bottom corners of a parallelogram are joined to a point on the top side.
Solution:
Area of dark triangle = \(\frac{1}{2} b h\) cm2
Area of parallelogram = bh
= 2 × area of triangle 10 cm2
Class 8 Mathematics Construction Of Quadrilaterals Kerala Scert Solutions Question 4.
The picture below shows the parallelogram formed by the intersection of two pairs of parallel lines?
What is the area of this parallelogram? And the perimeter?
Solution:
One side of the parallelogram = 4 cm
Distance to the opposite side = 3 cm
Area of the parallelogram = 4 × 3 = 12 sq.cm
BC × DE = 12
BC × 2 = 12
BC = 6, AD = 6
Perimeter = 4 + 6 + 4 + 6 = 20 cm.
Area Of Quadrilateral Questions For Class 8 Kerala Syllabus Question 5.
Compute the area of the parallelogram below:
Solution:
Textbook Page No 153
Hss Live Guru 8 Maths Kerala Syllabus Chapter 8 Question 6.
Draw a square of area 4½ square centimetres.
Solution:
Draw a circle of radius 3 cm. Draw the diameter AC and construct the perpendicular bisector of AC which meets the circle at B and D. Join AC, AD, CD and BD to get the square ABCD.
Area Of Quadrilateral Formula Class 8 Kerala Syllabus Question 7.
Draw a non-square rhombus of area 9 square centimetres.
Solution:
Similarly we can draw the other two rhombus
AC = 9 cm,
OB = OD = 1 cm
AC = 3 cm, OD = OB = 3 cm
8th Standard Maths Notes Kerala Syllabus Chapter 8 Question 8.
The area of the dark triangle in the figure is 5 square centimetres. What is the area of the parallelogram?
Solution:
iii. Perimeter = 4 × 15 = 60 m
iv. Rhombus is a parallelogram. So,
Area of parallelogram = one side × distance to the opposite side
15 × h = 216
h = \(\frac{216}{15}=14.4 \mathrm{cm}\)
8th Standard Maths Notes PdfKerala Syllabus Chapter 8 Question 9.
A 68 centimetre long rope is used to make a rhombus on the ground. The distance between a pair of opposite corners is 16 metres.
i. What is the distance between the other two corners?
ii. What is the area of the ground bounded by the rope?
Solution:
Length of one side of a parallelogram
Diagonals are perpendicular bisector
Kerala Syllabus 8th Standard Maths Notes Pdf Chapter 8 Question 10.
In the figure, the midpoints of the diagonals of a rhombus are joined to form a small quadrilateral:
i. Prove that this quadrilateral is a rhombus.
ii.The area of the small rhombus is 3 square centimetres. What is the area of the large rhombus?
Solution:
The diagonals of the rhombus intersect at O and they bisect each other at right angles.
OA = OC
\(\frac{\mathrm{OA}}{2}=\frac{\mathrm{OC}}{2}\)
OQ = OS
Similarly, since OB = OD
ie., OR = OP
The diagonals of PQRS bisect each other. (Since OQ = OS and OR = OP)
Since the diagonal AC is perpendicular to BD, the diagonals of PQRS are also mutually perpendicular bisectors.
therefore PQRS is a rhombus
ii. Area of quadrilateral PQRS = 3 sq. cm.
8th Area Of Quadrilateral Formula Class 8 Kerala Syllabus Question 11.
What is the area of the largest rhombus that can be drawn inside a rectangle of sides 6 centimetres and 4 centimetres?
Solution:
If we join the midpoints of the opposite sides of rectangle, we get the longest diagonal of the rhombus.
Textbook Page No 156
Hsslive 8th Maths Kerala Syllabus Chapter 8 Question 12.
Draw a rectangle of sides 7 centimetres and 4 centimetres. Draw isosceles trapeziums of the same area, with the following specifications.
i. Lengths of parallel sides 9 centimetres, 5 centimetres.
ii. Lengths of non-parallel sides 5 centimetres.
Solution:
i. Draw a rectangle of length 7 cm and breadth 4 cm.
Extent AB either side by 1 cm and mark E and F. Also mark the points G and H on CD at a distance 1 cm from C and D respectively. Draw the rhombus EFGH.
ii. Draw the rectangle ABCD.The arc of radius 5cm drawn with A as centre intersects DC at P. The arc of 5cm drawn with C as the centre intersect extented AB at Q.
Std 8 Maths Kerala Syllabus Kerala Syllabus Chapter 8 Question 13.
Calculate the area of the isosceles trapezium drawn below:
Solution:
8th Std Maths Notes Kerala Syllabus Chapter 8 Question 14.
The parallel sides of an isosceles trapezium are 8 centimetres and 4 centimetres long; and non – parallel sides are 5 centimetre long. What is its area?
Solution:
Divide the isosceles trapezium into a rectangle and two triangles.
Textbook Page No 158
Area Of Quadrilateral Class 8 Kerala Syllabus Chapter 8 Question 15.
The lengths of the parallel sides of a trapezium are 30 centimetres, 10 centimetres and the distance between them is 20 centimetres. What is its area?
Solution:
Question 16.
Compute the area of the trapezium shown below:
Solution:
Consider the ∆ PQS, it is a right angled triangle
Textbook Page No 159
Question 17.
Compute the area of the hexagon below.
Solution:
Total area = 108 + 108 + 168 = 384 cm2
Question 18.
This is a picture drawn in the lesson Construction of Quadrilaterals.
What is the area of the large trapezium made up of the four smaller ones?
Solution:
Question 19.
What is the area of the quadrilateral shown below?
Solution:
Question 20.
Prove that for any quadrilateral with diagonals perpendicular, the area is half the product of the diagonals.
Solution:
Question 21.
Compute the area of the quadrilateral shown below:
Solution:
Question 22.
Compute the area of the parallelogram shown below:
Solution:
The diagonal divides the parallelogram into two equal triangles.
Area of the parallelogram = 2 × area of triangle = 2 × 96 = 192 cm2
Question 23.
The three blue lines in the picture below are parallel:
Prove that the areas of the quadrilaterals ABCD and PQRS are in the ratio of the lengths of the diagonals AC and PR.
i. How should the diagonals be related for the quadrilaterals to have equal area?
ii. Draw two quadrilaterals, neither parallelograms nor trapeziums, of area 15 square centimetres
Solution:
i. For the quadrilaterals to have equal area, the diagonals AC and PR should be equal.
ii. Draw three parallel lines, distance between them is 5 cm each. Also draw two quadrilaterals with diagonals PQ and RS 6 cm each.
Additional Questions And Answers
Question 1.
Which of the following has greater area; a parallelogram with one side 12 cm and distance between parallel sides is 6 cm or a square having diagonals 12 cm each?
Solution:
Question 2.
Area of a trapezium is 128 sq. cm and the distance between its parallel sides is 8 cm. Length of one parallel side is 28 cm. Find the length of the other parallel side.
Solution:
Length of the other Parallel side = 4 cm
Question 3.
Find the area of a rhombus of one side 6 cm and one diagonal 6 cm.
Solution:
Question 4.
It was decided to lay tiles in the room of Hari’s house with tiles in the shape of trapezium. The parallel sides of a tile is 40 cm and 20 cm and the distance between them is 10 cm. If 200 tiles are needed to lay tiles in the hall, find the area of the hall in square metres.
Solution:
Number of tiles required = 200
Area of the hall = 200 × 300
= 60000 cm2
= 6 m2
Question 5.
In the pictures given below, which is has more area?
Solution:
Rectangle has the maximum area among the parallelograms with same sides.
Question 6.
The ratio of the two adjacent sides of a parallelogram is 3 : 2. Distance between longer sides is 10 cm. If area is 900 cm2, find the sides of the parallelogram?
Solution:
Let the sides be 3x and 2x
Question 7.
If one side of a parallelogram is ‘a’ and height of that side is h, prove that area = ah.
Solution:
In the figure, ABCD is a parallelogram. AB = a
The perpendicular distance from D to AB = h
By drawing the diagonal BD, we can divide the parallelogram into two equal triangles. ∆ ABD and ∆ BCD are equal triangles. ∴ their areas are equal, ie; area of the parallelogram is two times of the area of ∆ ABD.
Question 8.
PQRS is a rhombus. If the diagonals are 8 cm and 9 cm each, compute the area.
Solution:
Let the diagonals of the rhombus are d1 and d2
Question 9.
The perimeter of a rhombus is 40 cm. If the length of one diagonal is 16 cm. What is the length of the other diagonal. Find the area?
Solution:
Perimeter = 40 cm
One side = 10 cm .
∆ POQ is a right angled triangle Since d1 = 16 cm, OP = 8 cm
∴ 82 + OQ2 = 102
Question 10.
Draw a square of area 12.5 cm2 and write the geometric principles of the construction.
Solution:
Draw a line of length 5 cm and its perpendicular bisector.
Draw a circle with the midpoint of this line as centre and the distance to one end as radius. This line will be the diameter of the circle. Mark the point of intersection of the perpendicular bisector as the circle and complete the square.
Question 11.
Compute the area of the trapezium shown below.
Solution:
Question 12.
Compute the area of the quadrilateral ABCD
Solution:
Question 13.
In the figure AB || CD. AD = BC = 13 cm
Distance between the parallel side is 12 cm. If CD = 20 cm. Find the area of ABCD.
Solution:
To compute AB
Question 14.
Draw a square and mark the mid-points. How is the area of the first square with the second square.
Solution:
Draw the square ABCD.
The perpendicular bisector of the side AB bisects CD also at right angles.
Similarly draw the perpendicular bisectors of AD and BC. Which intersect the sides. Join the points of intersection to complete the square.
We see, 8 triangles in the figure. The square PQRS is formed by joining 4 among these. The area of the square PQRS is half of the area of the square ABCD.
Question 15.
What is the maximum area of parallelogram of sides 8 cm and 5 cm? What is the speciality of the parallelogram of maximum area?
Solution:
The area will be maximum for a rect¬angle and the maximum area is 40 cm2.
Question 16.
See the quadrilateral Ravi drew. In it AC = 12 cm and perpendiculars to AC are 6 cm and 12 cm.
a. Find the area of the quadrilateral.
b. Raju has to draw a parallelogram of this area. If its base is 12 cm what should be its height?
Solution:
b. If the height of the parallelogram is h its area is bh = 12 × h
This is 54 sq. cm
Question 17.
The area of a rhombus is 112 sq. cm and one of its diagonal is 16 cm long. Find the length of the other diagonal.
Solution:
Area of the rhombus \(=\frac{1}{2} \mathrm{d}_{1} \times \mathrm{d}_{2}\)
= 112 sq. cm
Question 18.
Draw a parallelogram of sides 6 cm, 4 cm and area 18 cm2.
Solution:
Draw a line AB of length 6 cm. Draw a circle with centre A and radius 4 cm
In the circle (as shown in the picture) mark a point D and join AD. Draw an arc of radius 4 cm with centre B as centre and radius 4 cm and draw another arc with D as centre and radius 6 cm. Two arcs meet at C. Join CD and BC.
ABCD will be a parallelogram.
Question 19.
In the figure ABCD, AB parallel to CD and the distance between them is 8 cm. ?
AB = 12 cm, CD = 10 cm. Compute the area of quadrilateral (trapezium) ABCD?
Solution:
Question 20.
Compute the area of the quadrilateral ABCD in the figure.
Solution:
You can Download सुख-दुख Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.
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?
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?
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?
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।
Sukh Dukh Kavita Ka Saransh Kerala Syllabus 8th प्रश्ना 1.
कविता से शब्द युग्मों का चयन करके लिखें।
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।
दाईं ओर:
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:
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:
1. आनन,
5. गगन,
6. जीवन,
9. अविरत
नीचे की ओर:
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।
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 शब्दार्थ Word meanings
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.
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
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
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:
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)
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
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
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.
Students can Download Basic Science Chapter 20 Static Electricity Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 8th Standard Basic Science Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
You can Download Let’s Regain Our Fields Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.
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
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.
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
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.
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.
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.
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 |
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.
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.
Answer:
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:
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
Open Precision farming
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:
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.
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:
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).
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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.
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
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
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.
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
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.
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.
(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?
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
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.
(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.
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.
(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.
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.
(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
(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.
(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°
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.
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?
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.
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.
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.
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
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.
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.
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
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.
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
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:
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.
(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°
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.
(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.
Question 1.
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.
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.
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.
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:
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.
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.
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.
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.
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.
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.
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.
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:
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:
∆ 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.
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.
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.
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 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:-
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.
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)
Answer:
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:
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.
Answer:
1. • A
• C
2. Formed from various cells
3. Only the corners of the cells are thicked
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
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
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
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.
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.
Answer:
a. cell organelle
b. Organ
Question 4.
Observe the following figure and answer the given questions.
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:
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
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:
a. Nervous tissue
b. Muscular tissue
Question 15.
What are the differences between the process of growth in plants and animals?
Answer:
Plant growth
Animal growth
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.
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
Answer:
Xylem 2, 4 Phloem -1, 3
Question 24.
Complete the worksheet based on the hints given.
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
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.
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.
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?
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?
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?
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।
Kerala Syllabus 8th Standard Notes Hindi प्रश्ना 1.
डॉ रमणी अटकुरी की चरित्रगत विशेषताएँ लिखें।
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।
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।
उत्तर:
Hss Live Guru 8 Hindi Kerala Syllabus प्रश्ना 4.
संवाद लिखें।
उत्तर:
बात उस मंगलवार की शब्दार्थ Word meanings
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.
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.
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 Maths Class 8 Kerala Syllabus Question 3.
Find all the exterior angles of the quadrilateral in the figure.
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
Solution:
(i)
Angles of the triangle are 70°, 35°, 75°
Exterior angles are 35°, 110°, 105°
(ii)
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)
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:
∠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:
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.
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:
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?
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
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?
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.
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.
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°