Kerala Syllabus 9th Standard Social Science Solutions Chapter 6 India, the Land of Synthesis in Malayalam

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Kerala State Syllabus 9th Standard Social Science Solutions Chapter 6 India, the Land of Synthesis in Malayalam Medium

India, the Land of Synthesis Textual Questions and Answers in Malayalam

India The Land Of Synthesis

Kerala Syllabus 9th Standard Social Science Solutions Chapter 6 India, the Land of Synthesis in Malayalam 2
Kerala Syllabus 9th Standard Social Science Solutions Chapter 6 India, the Land of Synthesis in Malayalam 3

Kerala Syllabus 9th Standard Social Science Solutions Chapter 6 India, the Land of Synthesis in Malayalam 4
Kerala Syllabus 9th Standard Social Science Solutions Chapter 6 India, the Land of Synthesis in Malayalam 5

Synthesis In Malayalam
Kerala Syllabus 9th Standard Social Science Solutions Chapter 6 India, the Land of Synthesis in Malayalam 7
Kerala Syllabus 9th Standard Social Science Solutions Chapter 6 India, the Land of Synthesis in Malayalam 8

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India The Land Of Synthesis Notes Pdf
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Kerala Syllabus 9th Standard Social Science Solutions Chapter 6 India, the Land of Synthesis in Malayalam 12
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Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 4 By the Hands of the Nature

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Kerala State Syllabus 9th Standard Social Science Solutions Part 2 Chapter 4 By the Hands of the Nature

By the Hands of the Nature Textual Questions and Answers

By The Hands Of The Nature Kerala Syllabus 9th Question 1.
By The Hands Of The Nature Kerala Syllabus 9th
Observe the diagram. Haven’t you seen how rainwater carries away the loose rock particles from elevated regions and deposits elsewhere? What all changes take place on the surface of the earth as a result of both the process mentioned above?
Answer:
Carries loose material away – erosion
Deposits in low lying regions – deposition

HSSLive.Guru

By The Hands Of The Nature Notes Pdf Kerala Syllabus 9th Question 2.
What are the different process of weathering?
Answer:
Weathering is the process where rock is dissolved, worn away or broken down into smaller and smalleds pieces. There are three different weathering processes such as mechanical, chemical and organic/biological.

Kerala Syllabus 9th Standard Geography Notes Question 3.
Varied features are seen in the river course at every stage. Observe the diagram and answer the following questions by analyzing the features of these three courses.
i) In which stage is the intensity of erosion more?
ii) Which process results in the landforms in the lower course?
iii) In which stage of the river is the amount of sediments more?
Answer:
i) Upper course
ii) Active depositional process
iii) Middle course

Hss Live Guru Class 9 Social Science Kerala Syllabus Question 4.
What could be the reason behind the round shape and polished surface of pebbles?
Answer:
The rock particles are carried by the river rub against the rocks along the bed and both the sides of the river. This results in the wearing down of rocks. Such erosion is known an abrasion or corrosion. Through these processes, the river can polish even the hard rocks along its course.

Kerala Syllabus 9th Standard Social Science Notes Question 5.
Why are gullies formed along steep slopes?
Answer:
Gullies are formed through intense erosion resulting from an increase in the velocity of water flow.

Hss Live Guru 9th Social Science Kerala Syllabus Question 6.
By The Hands Of The Nature Notes Pdf Kerala Syllabus 9th
Kerala Syllabus 9th Standard Geography Notes
Observe the pictures to understand how oxbow lakes take birth from meanders.
Answer:
Meanders are usually formed in the middle and lower courses of wide rivers. The transformation happening to the meanders through further erosion and deposition: Meanders may further curve through continuous erosion and deposition. Finally, the river takes a straight course. The curves may get detached from the main river to form isolated water bodies. Such water bodies are called oxbow lakes.

9th Standard Social Science Notes Pdf Kerala Syllabus Question 7.
Prepare notes by discussing the agricultural importance of flood plains.
Answer:
The deposition of alluvium along both the flooded banks may cause the formation of plains called flood plains. Flood plains contain rich and fertile soil suitable for cultivation. Agricultural improvement is possible in flood plains. This has historically led to the development of various civilizations along the flood plains. Flood plains are along the banks of rivers. Thus the area is rich in water availability. The physiography of flood plains is suitable for both agricultural activities and settlement.

Hsslive Guru 9th Social Science Kerala Syllabus Question 8.
Complete the table based on what you have learnt about the landform created by rivers.
Hss Live Guru Class 9 Social Science Kerala Syllabus
Answer:

Landforms Course of formation Erosional/ Depositional
Waterfall Upper course Erosional
Meanders Middle and lower course Erosion and deposition

Hsslive Guru 9th Geography Kerala Syllabus Question 9.
Why is water is called universal solvent?
Answer:
Water is called as universal solvent because most of the mineral present in the rocks gets dissolved as water pass through them.

9th Standard Social Science Notes Kerala Syllabus Question 10.
Identify the landform created by the merging of stalacticles and stalagmites.
Answer:
Limestone pillars

9th Standard Social Science Notes Pdf Malayalam Medium Question 11.
Name the districts in which the tourism-oriented beaches in Kerala can be found? )
Answer:
Thiruvananthapuram, Alappuzha, Kannur, Kozhikode, Kasaragod, and Kollam

Hsslive Guru Class 9 Social Science Kerala Syllabus Question 12.
Which is the major geomorphic agent creating landforms in deserts?
Answer:
Rock pedestals, Deflation hollows, Oasis, Sand dunes, Yasdangs and Desert Pavements.

Kerala Syllabus 9th Standard Social Science Notes Pdf Question 13.
What could be the reason for the increased erosion at the bottom of the rocks as shown in the figure?
Kerala Syllabus 9th Standard Social Science Notes
Answer:
Mushroom rocks are created by wind blowing material and eroding the rocks to form a mushroom rock. It also can be formed by glacial erosion. This can happen when a glacier retreats and leaves behind rocks causing the rock on the bottom to erode faster leaving behind a mushroom rock. One day, erosion will weaken the base of these rocks and the top part will come tumbling down.

By the Hands of the Nature Model Questions and Answers

Hsslive Guru Social Science 9th Kerala Syllabus Question 14.
What are glaciers?
Answer;
Thick masses of ice slowly move downhill in snow-clad region are called glaciers.

Kerala Syllabus 9th Standard Social Science Guide Question 15.
Define geomorphology
Answer:
Geomorphology is the branch of geography which deal with the study of origin and evolution of landforms.

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Question 16.
What are geomorphic processes?
Answer:
The processes that help in the formation of landforms are called geomorphic processes.

Question 17.
Distinguish between erosion and deposition.
Answer;
The transfer of rock particles framed by chemical, physical or biological weathering processes from one place to another by external agencies such as running water, wing, glaciers, sea waves, etc. is called erosion. These materials will be deposited in low lying regions and this process is called deposition.

Question 18.
Identify the external agencies that help in erosion?
Answer:
Running water, wing, glaciers, sea waves.

Question 19.
Place of origin of river is called
Answer:
Source

Question 20.
The place at which river discharges into the sea or water bodies called ……………
Answer:
River mouth.

Question 21.
Complete the flow chart
Hss Live Guru 9th Social Science Kerala Syllabus
Answer:
9th Standard Social Science Notes Pdf Kerala Syllabus

Question 22.
Match the following

A B
Upper course Oxbow lakes
Deltas
Valleys

Answer:

A B
Upper course Oxbow lakes
Middle course Deltas
Lower course Valleys

Question 23.
Point out the features of upper course
Answer:

  • Place of origin of the river
  • River flows through steep slopes
  • Intense rate of erosion
  • Less amount of sediments
  • No deposition
  • Landforms such as valley, waterfall, etc. are seen

Question 24.
What are the features of middle course?
Answer:

  • Flows through the foothills
  • Both erosional and depositional processes are active
  • More sediments are carried down
  • Features such as meanders and oxbow lakes are seen

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Question 25.
Mention important peculiarities of lower course.
Answer:

  • Flows through the plains
  • Active depositional process
  • The quantity of water, as well as sediments, is high
  • Depositional landforms such as flood plains are deltas are seen

Question 26.
Point out the factors affecting the intensity of river erosion.
Answer:

  • Velocities of water flow
  • Slope of the terrain
  • Rock structure

Question 27.
Explain the process of abrasion or corrosion.
Answer;
The rock particles like gravel, sand, pebbles, etc. carried by the river rub against the rocks along the bed and the sides of the river. This results in the wearing down of rocks. Such erosion is known as abrasion or corrosion. Through these processes, the river can polish even hard rocks along its course.

Question 28.
River bed erosion is more prevalent in ………..
a) Upper course of the river
b) Middle course of the river
c) Lower course of the river
Answer:
a) Upper course of the river

Question 29.
Examine the process that results in waterfalls.
Answer:
Waterfalls are generally formed at the upper course of rivers as a result of erosion. Soft rocks are easily eroded in the valleys where soft and hard rocks are found intermingled. This results in the formation of waterfalls.

Question 30.
What do you mean by flood plains?
Answer:
Rivers overflowing their banks during rainy seasons. Flood water may cover extensive areas on both sides of the river. The deposition of alluvium along both the flooded banks may cause the formation of plains. Such plains are called flood plains.

Question 31.
How is deltas formed?
Answer:
Velocity of the river decreases when it nears the river mouth. Most rivers branch out to distributaries at this stage where the volume of both water and sediments is high. The sediments brought by the river are deposited between these distributaries forming almost triangular-shaped landforms called deltas. These features are called as they resemble the Greek alphabet A (Delta).

Question 32.
Name the largest delta in the world.
Answer;
The Sundarbans in West Bengal

Question 33.
Why are the erosional and depositional landforms of underground water mainly confined to limestone region?
Answer;
Rocks like limestone easily dissolve in rainwater which percolates to form underground water. Hence the erosional and depositional landforms of underground water are mainly confined to limestone regions.

Question 34.
Categorize the following based on their formation.
1. Limestone caves
2. Stalactites
3. Stalagmites
4. Limestone Pillars
Answer:
Erosion:
Limestone caves
Depositions:

  • Stalactites
  • Stalagmites
  • Limestone Pillars

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Question 35.
Give an example of limestone caves.
Answer;
The Borra caves near Visakhapatnam in Seemandhra

Question 36.
What are coastlines? Give different types of coastlines.
Answer:
Coastal landforms are created by the erosional and depositional processes carried out by the waves. There are two types of coastlines namely.

  • Rocky coasts
  • Non-rocky coasts

Question 37.
Define sea cliff.
Answer:
The steep hillocks facing the sea are called sea cliffs.

Question 38.
Give an example of sea cliff in Kerala.
Answer:
Varkala beach

Question 39.
Landforms created by wind are mostly seen in
a) Valleys
b) Beaches
c) Deserts
d) Farmlands
Answer:
c) Deserts

Question 40.
What is the process of deflation?
Answer:
The strong whirlwind carry away the dry desert sands from one place to another. This process of wind erosion is called deflation.

Question 41.
What do you mean by barchans?
Answer;
The sand dunes formed in the deserts are as a result of the deposition by wind. The sand dunes commonly formed in crescent shapes are called barchans.

Let Us Assess

Question 42.
Describe the characteristic of different stages in a river’s course.
Answer;
The course of a river can generally be divided into three stages based on the difference in slope from its source to mouth.

  • Upper course
  • Middle course
  • Lower course

Upper course is that part of the river where it rapidly flows down along steep slopes from the place of origin. The intensity of erosion is severe in this course. Middle course is that stage of the river where it flows through gently sloping foothills. As the velocity of the flow decreases, the intensity of erosion declines and deposition begins.
Lower course is the stage where the river flows through the plains. The rate of deposition will be higher due to the slow pace of the river and the increase in the amount of sediments during this stage.

HSSLive.Guru

Question 43
Compare the ‘v-shaped’ valleys with ‘u shaped’ valleys based on processes of formation.
Answer:
Deepening of gullies occurs through intense erosion resulting from an increase in the velocity of water flow. The valleys take a distinct shape as a result of the intensity of erosion along the river bed. Such valleys are called V-shaped valleys.

The rate of erosion along the river bed decreases as the river leaves the upper course. However lateral erosion dominates, the river flowing through comparatively gentle slopes takes deviations when the sediments or rock forms create obstruction to the flow. Such bending course of a river is leading to u shaped valleys.

Question 44.
List with example the agricultural and environmental significance of deltas and flood plains.
Answer:
The deposition of alluvium along both the flooded banks may cause the formation of plains called flood plains. Flood plains contain rich and fertile soil suitable for cultivation. Agricultural improvement is possible in flood plains. This has historically led to the development of various civilizations along the flood plains. Flood plains are along the banks of rivers. Thus the area is rich in water availability. The physiography of flood plains is suitable for both agricultural activities and settlement.

HSSLive.Guru

Question 45.
Illustrate the formation of mushroom rocks with the help of a diagram
Answer:
As a result of the continued erosion caused by sand and other rock particles carried by strong winds, rocks in deserts get worn down. This process of wind erosion is called abrasion. Figure given below shows here is of a rock formed in this manner. Such rocks seen in deserts resembling mushrooms are called mushroom rocks.
Hsslive Guru 9th Social Science Kerala Syllabus

Question 46.
Explain the formation of any two erosional landforms created by glaciers.
Answer:
Glacial landforms are generally confined to the high mountain ranges and the poles. The movement of a glacier along the mountain slope is depicted in the given picture.
Hsslive Guru 9th Geography Kerala Syllabus
Observe the changes happening to the valley at different stages. Different types of valleys as shown in the pictures are formed by glacial erosion. Armchair like valleys so formed are called cirques.
9th Standard Social Science Notes Kerala Syllabus
9th Standard Social Science Notes Pdf Malayalam Medium

Question 47.
Prepare a table showing the erosional and depositional landforms created by any three external forces.
Hsslive Guru Class 9 Social Science Kerala Syllabus
Answer:
Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 4 By the Hands of the Nature 13

Question 48.
Identify the landforms shown in the pictures and explain how they are formed
Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 4 By the Hands of the Nature 14
Answer:
A — Oxbow lakes:
Meanders are usually formed in the middle and lower courses of wide rivers. The transformation happening to the meanders through further erosion and deposition. Meanders may further curve through continuous erosion and deposition. Finally, the river takes a straight course. The curves may get detached from the main river to form isolated water bodies. Such water bodies are called oxbow lakes.
B — Mushroom Rocks:
As a result of the continued erosion caused by sand and other rock particles carried by strong winds, rocks in deserts get worn down. This process of wind erosion is called abrasion. Figure given below shows here is of a rock formed in this manner. Such rocks seen in deserts resembling mushrooms are called mushroom rocks.

Kerala Syllabus 9th Standard Chemistry Solutions Chapter 7 The World of Carbon in Malayalam

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Kerala Syllabus 9th Standard Chemistry Solutions Chapter 4 Periodic Table

You can Download Periodic Table Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Chemistry Solutions Part 1 Chapter 4 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Chemistry Solutions Chapter 4 Periodic Table

Periodic Table Textual Questions and Answers

Earlier Attempts for Classification of Elements

Kerala Syllabus 9th Standard Chemistry Notes Chapter 4 Question 1.
Explain the earlier attempt of classification by Lavoiser?
Answer:
Antoine Lavoisier classified the known elements into metals and nonmetals. But he was not able to duly classify metalloids.

9th Class Chemistry Periodic Table Kerala Syllabus Question 2.
Explain Newland’s law of octaves?
Answer:
Newlands arranged elements in the increasing order of atomic mass. He noticed that every eighth element has properties similar to those of the first elements. But this peculiarity could be noticed in elements upto calcium only.

Octaves of Newlands
Kerala Syllabus 9th Standard Chemistry Notes Chapter 4

9th Class Chemistry 4th Chapter Kerala Syllabus Question 3.
Define Mendeleev’s periodic law?
Answer:
In 1869 Mendeleev arranged the known 63 elements in horizontal and vertical columns and gave shape to the periodic table. He found that the chemical and physical properties of elements repeat at a regular intervals when they were arranged in the increasing order of atomic masses. Based on this Mendeleev proposed the periodic law of elements. The law states that physical and chemical properties of elements are periodic function of their atomic masses.

Chemistry Notes For Class 9 Periodic Table Kerala Syllabus Question 4.
What is meant by groups and periods in the periodic table?
Answer:
The vertical columns in the periodic table are known as groups and the horizontal rows are called periods.

Kerala Syllabus 9th Standard Chemistry Notes Question 5.
Evaluate the Mendeleev’s periodic table and find the following.
a) Total number of periods
b) Total number of Groups
c) Are the same elements showing similar properties arranged in the same group or same period?
Answer:
a) 6
b) 8
c) Same group

Class 9 Chemistry Notes Kerala Syllabus Question 6.
Advantages of Mendeleev’s periodic table?
Answer:
1. For the first time elements were comprehensively classified in such a way that elements of similar properties were placed in the same group. This has made the study of chemistry easy.
2. When the classification was made in such a way that the elements of similar properties came in the same group. It was noticed that certain their proper group. The reason for this was wrongly determined atomic masses and consequently, those wrong atomic masses were corrected.
Eg. The atomic mass of beryllium was known to be 14. Mendeleev reassessed it as a and assigned beryllium a proper place.
3. Columns were left vacant for elements which were not known at the time and their properties were predicted also. This gives an impetus to experiments in chemistry.

Ex Mendeleev give names Eka aluminum and Eka silicon to those elements which were to come below aluminum and silicon respectively in the periodic table and predicted their properties. Later when these elements gallium and germanium were discovered the prediction of Mendeleev turned out to be true.

Hsslive Guru 9th Chemistry Kerala Syllabus Question 7.
Explain the limitation of Mendeleev’s Periodic table?
Answer:
1. Elements with large difference in properties were included in the same group.
eg. Hard metal like copper [Cu], Silver [Ag] were included along with soft metals like sodium [Na] potassium [K],
2. No proper position could be given to element hy¬drogen. Non-metallic hydrogen was placed along with metals like sodium [Na] and potassium [K]
3. The increasing order of atomic mass was not strictly followed throughout.
eg. Co and Ni, Te and I
4. As isotopes are atoms of same element having different atomic masses, they should have been given different position while arranging them in the order of atomic mass. But this was not done.

Counting Atomic Calculator is a free online tool that displays the atomic mass for the given chemical formula.

Modern Period Table

Periodic Table 9th Class Pdf Kerala Syllabus Question 8.
State and explain modern periodic table and mod-ern periodic law?
Answer:
In 1913 Mosely through his x-ray diffraction experiments proved that the properties of elements depended on the atomic number not on the atomic mass.

According to this the periodic law of Mendeleev and the periodic table were modified consequently the modern periodic table was prepared by arranging elements in the increasing order of atomic number. The modern periodic law states that the physical and chemical properties of elements are periodic function of their atomic number.
9th Class Chemistry Periodic Table Kerala Syllabus

Periodic Table Chapter Class 9 Kerala Syllabus Question 9.
How many periods in the modern periodic table?
Answer:
7

Labour India Class 9 Chemistry Kerala Syllabus Question 10.
Which is the shortest period?
Answer:
I period

9th Class Chemistry Chapter 4 Kerala Syllabus Question 11.
Number of elements in the third period?
Answer:
8

Class 9 Chemistry Periodic Table Kerala Syllabus Question 12.
Total number of groups?
Answer:
18

Periodic Table In 9th Class Kerala Syllabus Question 13.
Explain representative elements?
Answer:
Elements of group 1 and 2 also those in groups of 13 – 18 are called representative elements it belongs to metals, nonmetals, and metalloids.

Periodic Table Chemistry Class 9 Kerala Syllabus Question 14.
Do in representative elements do they include metalloids [eg. Si, Ge, As, Sb…) exhibiting the characteristics of metals and non-metals?
Answer:
Yes

Periodic Table Notes Pdf Class 9 Kerala Syllabus Question 15.
As there elements existing in solid, liquid and gaseous state find examples?
Answer:

  • In solid-state- sodium, aluminum, carbon
  • In liquid state – Bromine
  • In gaseous state – oxygen, neon, argon

9th Class Chemistry Chapter 4 Notes Kerala Syllabus Question 16.
Write the electronic configuration of elements with atomic number 1-10
Answer:

Element Atomic number Electronic configuration
Hydrogen 1 1
Helium 2 2
Lithium 3 2, 1
Beryllium 4 2, 2
Boron 5 2, 3
Carbon 6 2, 4
Nitrogen 7 2, 5
Oxygen 8 2, 6
Fluorine 9 2, 7
Neon 10 2, 8

The atom of the elements of these group show the periodically in electron filling they contain 1 -8 electron in their outermost shell. The elements of these groups are called representative elements.

Noble Gases

Kerala Syllabus 9th Standard Notes Chemistry Question 17.
List the elements in group 18
Answer:
Helium, Neon, Argon, Krypton, Xenon, Radon

9 Class Chemistry Chapter 4 Kerala Syllabus Question 18.
Now try to write their electronic configuration
Answer:
2He – 2
10Ne – 2, 8
18Ar – 2, 8, 8
36Kr – 2, 8, 18, 8
54Xe – 2, 8, 18, 18, 8
86Rn – 2, 8, 18, 32, 18, 8

Question 19.
How many electrons are there in the outermost shell of each element?
Answer:
8

Question 20.
The elements do not normally take part in chemical reactions. Find the reason?
Answer:
They have a stable configuration in the outermost shell.

Transition Elements

Question 21.
Which group of elements belong to transition elements?
Answer:
Elements of group 3-12 in the periodic table are transition elements.

Question 22.
Find out whether elements familiar to you are present in these groups?
Answer:
Copper, silver, gold, iron

Question 23.
Aren’t transition elements metals?
Answer:
Yes

Question 24.
What are the characteristics of transition elements?
1. They from coloured compound,
2. They show similarity in properties as well as in a period.
3. In compounds, they exhibit different oxidation state
eg. Fe2+ and Fe3+

Lanthanides and Actinoids

Question 25.
Which element is next to lanthanum with atomic number 57 of group 6 in the periodic table?
Answer:
Cerium with atomic No. 58

Question 26.
Find out the position allotted to the elements with atomic number 58-71?
Answer;
Separate position at the bottom of the periodic table.

Question 27.
Is the same way aren’t the elements with atomic number 90 to 103 of period 7 give separate positions at the bottom of the periodic table?
Answer:
Yes. These elements are called inner transition elements.

Question 28.
What is meant by inner transition elements?
Answer:
Inner transition elements from Cerium [Ce] to Lutecium [Lu] of period 6 are called lanthanides. Inner transition elements from Thorium (Th] to Lewrencium [Lr] of period 7 are called actinoids. Lanthanoids are also called rare earth. Actinoids are man-made artificial elements (except thorium and uranium).

Periodic trends in the periodic table

Question 29.
Electronic configuration of group I elements of the periodic table are given
Answer:

Element Atomic number Electron configuration Group Period
H 1 1 1 1
Li 3 2, 1 1 2
Na 11 2, 8, 1 1 3
K 19 2, 8, 8, 1 1 4
Rb 37 2, 8, 18, 8, 1 1 5
Cs 55 2, 8, 18, 18, 8, 1 1 6
Fr 87 2, 8, 18, 32, 18, 8, 1 1 7

Question 30.
What is the peculiarity seen in the electronic configuration of the outer most shell of these elements?
Answer:
All these elements we can see one electron in the outermost shell.
Hence elements of group I exhibit similarity in chemical properties.

Question 31.
Which are the electrons shows the chemical properties of elements?
Answer:
Outermost electrons.

Question 32.
Is there any relationship between the group number and the number of electrons present in the outermost shell? What is it?
Answer:
Same, group number equal tot he number of election in the outermost shell for the elements in groups 1 and 2.

Question 33.
Observe figure the electronic configuration of the second-period elements of the group from 13-18 given below.
9th Class Chemistry 4th Chapter Kerala Syllabus

(i) Won’t we get the group number of these elements by adding 10 to the number of elements by adding 10 to the number of electrons in the outermost shell?
Answer:
Yes

(ii) Analyze table 3.1 and find whether there is any relation between the number of shells in an atom and the number of periods?
Answer:
Number of shells in an atom and the period number is same.

Size of an Atom in Group

Question 34.
Are you familiar with the Bohr model of an atom? See the Bohr model of atoms of certain elements, in group I.
Chemistry Notes For Class 9 Periodic Table Kerala Syllabus
Kerala Syllabus 9th Standard Chemistry Notes

(i) Which among them is the biggest?
Answer:
Potassium (K)

(ii) Which one is the smallest?
Answer:
Hydrogen [H]

(iii) What happens to the size of an atom when we move down the group?
Answer:
Increases

(iv) What is the reason for this?
Answer:
Number of shells increases.
As we move from top to bottom of a group in the periodic table the size of the atom increases as there is an increase in the number of shells.

Atomic Size in Period

See (Fig 3.3) the representation of Bohr model of elements with atomic numbers 3 to 9 in the second period of the periodic table.
Class 9 Chemistry Notes Kerala Syllabus

Question 35.
Is there are in the number of shells with the increase in atomic number?
Answer:
No.

Question 36.
What happens to the nuclear charge with increase in atomic number?
Answer:
On moving from left to right in a period, as nuclear charge increases, the force of attraction on the outer-most electrons increases and consequently the size of atom decreases.

Ionisation Energy

Question 37.
You have understood how sodium chloride is formed by combining sodium and chlorine atoms. The Bohr model of sodium and chlorine are given below
Answer:
Hsslive Guru 9th Chemistry Kerala Syllabus

(i) Which among these atoms lose electrons?
Answer:
Sodium atom

(ii) Which one gains electrons
Answer:
Chlorine atom

Question 38.
How the ions are formed?
Answer:
Atom becomes charged when there is transfer of electrons [Lose or gain electrons] they are called ions.

Question 39.
Define ionization energy?
Answer:
The amount of energy required to liberate the most loosely bound electrons from the outermost shell of an isolated gaseous atom of an element is called ionization energy.

Question 40.
What are the factors affecting the ionization energy? Nuclear charge
Answer:
Size of the atom

Question 41.
When the size of an atom increases, does the attraction of the nucleus on the outermost electron increase or decrease?
Answer:
Decrease

Question 42.
Then what is the change in ionization energy?
Answer:
As the size of atom increases ionization energy decreases.

Question 43.
Can you find out how ionization energy changes as we move from top to bottom in a group?
Answer:
Ionization energy decreases.

Question 44.
What is the general trend in the variation of ionization energy on moving across a period from left to right?
Answer:
Ionization energy increases

Question 45.
Find how ionization energy changes with increase in nuclear charge?
Answer:
On moving from left to right in a period, as nuclear charge increases, the size of the atom decrease hence ionization energy increase.

Question 46.
Define electronegativity?
Answer:
In the case of two atoms joined by a covalent bond, electronegativity is the ability of each atom to attract the bonded electrons.

Question 47.
How size of an atom influence the electronegativity?
Answer:
As the size of an atom increases the distance between the nucleus and the outermost electron increases, hence the electronegativity decreases. As we move in the same period form left to right size of atom decrease hence electronegativity increases.

Question 48.
What is the basis for the chemical properties of metals and non-metals?
Answer:
Metals are the elements which give away the electrons and those that accept electrons are generally non-metals. Metals are electropositive elements because they lose electrons to form positive ions. Non-metals are called electronegative elements because they gain electrons in chemical reactions to form negative ions.

Question 49.
What is relationship between metallic character and the size of an atom?
Answer:
As the size of the atom increases metallic character also increases.

Question 50.
How do the metallic character and nonmetallic character vary while moving from left to right in a period? Arriving at a conclusion by assessing the size of atom?
Answer:
In the periodic table, while moving from to top to bottom in groups metallic character generally in-creases while non-metallic character decreases.
In a period as we move form left to right metallic character generally decreases while non-metallic character increases.

Question 51.
Don’t you think that there is a relationship between ionization energy and metallic -non-metallic character? Is the element with highest ionization energy metallic or non-metallic?
Answer;
Non-metallic

Question 52.
Then what about those having the low ionization energy?
Answer:
Metals

Question 53.
Isn’t there a relationship between electronegativity and metallic, non-metallic character? Explain the relationship?
Answer:
Non-metals are more electronegative.

Metalloids

Question 54.
Explain metalloids?
Answer:
Elements exhibiting the properties of both metal as well as nonmetal are called metalloids, eg. Silicon [Si], germanium [Ge] Arsenic [As], Antimony [Sb] and Tellurium [Te] belongs to this category.

Question 55.
You must understand certain periodic trends in the periodic table? Based on these (✓)the correct option given below in table 3.7.
Periodic Table 9th Class Pdf Kerala Syllabus
Answer:

Trends In a group from  top to bottom In period from  left to right
Size of atom ✓ Increases/  decreases Increases/  decreases ✓
Metallic character ✓ Increases/ decreases Increases/ decreases ✓
Non-metallic character Increases/ decreases ✓ ✓ Increases/ decreases
Ionization energy Increases/ decreases ✓ ✓ Increases decreases
Electronegativity Increases/ decreases ✓ ✓ Increases/ decreases

Let Us Assess

Question 1.
The table given below lists the contributions and names of scientists who made earlier attempts in the classification of elements. Fill in the blanks.
Periodic Table Chapter Class 9 Kerala Syllabus
Answer:

Contribution/Findings Name of Scientist
Triads Dobereiner
Law of octaves Newlands
Classification of elements into metals and non-metals Antonie Lavoisier
Modern periodic law Henry Moseley

Question 2.
Complete the table
Answer:

Element Atomic number Electronic  configuration Group  number Period  number
Lithium 3 2,1 1 2
Oxygen 8 2J3 16 2
Argon 18 2,8,8 18 3
Calcium 20 2,8,8,2 2 4

Question 3.
Symbols of certain elements are given. Write their electronic configuration and find the period and group in which they are included.
Labour India Class 9 Chemistry Kerala Syllabus
Answer:
a) \(_{6}^{12} C\)
Electronic configuration 2, 4
Period – 2
Group – 14
b) \(_{12}^{24} \mathrm{Mg}\)
Electronic configuration 2,8,2
Period-3
Group – 2
c) \(_{17}^{35} \mathrm{Cl}\)
Electronic configuration 2,8,3
Period – 3
Group – 17
d) \(_{13}^{27} \mathrm{Al}\)
Electronic configuration 2,8,3
Period – 3
Group – 13
e) \(\begin{array}{l}{20} \\ {10}\end{array} \mathrm{Ne}\)
Electronic configuration 2,8
Period – 2
Group -18

Question 4.
There are three shells in the atom of element ‘X’, 6 electrons are present in its outermost shell.
a) Write the electronic configuration of the element.
b) What is its atomic number?
c) In which period does this element belong?
d) In which group is this element included?
e) Write the name and symbol of this element.
f) To which family of element does is this element belong to?
g) Draw and illustrate the Bohr atom model of this element.
Answer:
a) 2, 8, 6
b) 16
c) 3
d) 16
e) Sulphur, ‘S’
f) Oxygen family
9th Class Chemistry Chapter 4 Kerala Syllabus

Question 5.
Electronic configurations of elements P, Q, R, and S are given below. (These are not actual symbols).
P – 2, 2
Q -2, 8, 2
R – 2, 8, 5
S – 2, 8
a) Which among these elements are included in the same period?
b) Which are those included in the same group?
c) Which among them is a noble gas?
d) To which group and period does the element R belong?
Answer;
a) P and S, Q and R – belongs to same period
b) P and Q, belongs to same group
c) S
d) R belongs to 3rd period and 15th group

Question 6.
An incomplete form of the periodic table is given below. Write answers to the questions connecting the position of elements in it.
Class 9 Chemistry Periodic Table Kerala Syllabus
a) Which is the element with the biggest atom in group 1?
b) Which is the element having very lowest ionization energy in group 1?
c) Which element has the smallest atom in period 2?
d) Which among them are transition elements?
e) Which of the elements L and M has the lowest electronegativity?
f) Among B and I which has higher metallic character?
g) Which among these are included in the halogen family?
h) Which is the element that resembles E the most in its properties?
Answer:
a) D
b) D
c) M
d) G, H
e) L
f) B
g) M, N
h) F

Periodic Table Model Questions and Answers

Question 1.
Symbols of certain elements are given write down the electronic configuration and find the period and group in which they are included.
Periodic Table In 9th Class Kerala Syllabus
Answer:
a) \(\begin{array}{l}{23} \\ {11}\end{array} \mathrm{Na}\)
Electron configuration 2, 8, 1
Period – 3
Group – 1
b) \(_{17}^{35} \mathrm{Cl}\)
Electron configuration -2, 8, 7
Period – 3
Group-17
c) \(_{9}^{19} \mathrm{F}\)
Electron configuration -2, 7
Period – 3
Group – 17

Question 2.
A, B, C, D are four elements. The electronic configuration is given below and find the answers in the following (Hint. The symbols are not real)
A – 2, 2
B – 2, 8, 5
C – 2, 7
D – 2, 8, 2
a) Find the elements belongs to same period?
b) Find the elements belongs to same group.
c) ‘C’ belongs to which period and group?
Answer:
a) B, D, and A, C because the number of shells are same.
b) A, D because the number of electrons in the outermost shell is same.
c) ‘C’ belongs to second period and 17th group

Question 3.
Table given below lists the contributions and names of scientists who made earlier attempts in the classification of elements. Make them in the correct order.

Contribution/Findings Name of scientist
Octet Rule John Dalton
Triads New Lands
Modern periodic table Lavoisier
Classify into metals Henry Moseley
Non-metals
Atomic theory Dobereiner

Answer:

Contribution/Findings Name of Scientist
Triads Dobereiner
Law of octaves Newlands
Classification of elements into metals and non-metals Antonie Lavoisier
Modern periodic law Henry Moseley

Question 4.
An incomplete form of the periodic table is given below.
write answers in the question connecting the position of elements in it.
Periodic Table Chemistry Class 9 Kerala Syllabus
1. Which element has the largest atomic size in group I?
2. Write the transition elements?
3. Which element has the lowest ionization energy in the 2nd period?
4. Which element belongs to Noble gas elements?
5. Compare L, M which element has the lowest electronegativity?
6. Write the element belongs to Halogen family?
Answer:
1. D
2. F, G, H
3. C
4. N
5. L
6. M

Kerala Syllabus 9th Standard Social Science Solutions Chapter 8 Population, Migration, Settlements in Malayalam

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Kerala Syllabus 9th Standard Maths Solutions Chapter 12 Proportion in Malayalam

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Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 5 Ocean and Man

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Ocean and Man Textual Questions and Answers

Ocean And Man Class 9 Kerala Syllabus Question 1.
Identify the location of each ocean from the world map. List the straits, bays and the seas of each ocean with the help of an atlas.
Answer:
Ocean And Man Class 9 Kerala Syllabus

Kerala Syllabus 9th Standard Social Science Notes Question 2.
The following table contains the names of some major islands and peninsulas in the world. With the help of an atlas find out the names of the oceans to which they below.
Answer:
Kerala Syllabus 9th Standard Social Science Notes

  1. Indian Ocean
  2. Indian Ocean
  3. Pacific Ocean
  4. Atlantic Ocean
  5. Atlantic Ocean

Kerala Syllabus 9th Standard Social Science Notes Pdf Question 3.
Complete the table using fig. 5.7 in textbook
Kerala Syllabus 9th Standard Social Science Notes Pdf
9th Standard Social Science Notes Pdf Kerala Syllabus
Answer:
9th Standard Social Science Notes Kerala Syllabus
9th Standard Social Science Notes Pdf Kerala Syllabus Question 4.
The equatorial regions; record a high amount of salinity as compared to the polar regions why?
Answer:
The temperature is high in equatorial regions compared to polar regions. Density increases as temperature rises. High density is associated with high salinity. There is possibility of high evaporation in equatorial regions. That is why equatorial regions record high amount of salinity.

9th Standard Social Science Notes Kerala Syllabus  Question 5.
Why is salinity is less at river mouths?
Answer:
Salinity is less at river mouths because of huge amount of freshwater added from hundred of rivers.

9th Std Social Science Notes Kerala Syllabus Question 6.
Which are the warm and cold currents of the Atlantic Ocean? Identify the continents near which they flow.
9th Std Social Science Notes Kerala Syllabus
Answer:

Warm currents Continents near which they flow
1. North Atlantic current 1. Europe
2. Gulf Stream 2. North America
3. Florida current 3. North America
4. North equatorial currents 4. South America & Africa
5. Equatorial counter currents 5. South America & Africa
6. South Equatorial currents 6. South America & Africa
7. Brazilian current 7. South America
Cold current Continent near which they flow
1. Labrador current 1. North America
2. West wind drift 2. South America
3. Benguela current 3. Africa
4. Canaries current 4. Europe of Africa

Kerala Syllabus 9th Standard Social Science Notes English Medium Question 7.
Complete the following table based on the currents of the Indian Ocean.
Kerala Syllabus 9th Standard Social Science Notes English Medium
Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 5 Ocean and Man 8
Answer:

Currents Warm/Cold Direction
1. South equatorial current 1. Warm 1. From East to West
5. South West Monsoon current 5. Warm 2. From West to East
6. Agulhas current 6. Warm 3. From North to West
7. West Australian current 7. Cold 4. From South to North
8. West wind Drift 8. Cold 5. From West to East

Kerala Syllabus 9th Standard Social Science Notes Malayalam Medium Question 8.
You have learnt the uses of Oceans. Conduct a seminar on the topic ‘influence of Oceans in human life’.
Answer:
“Influence of Oceans in Human Life”
Oceans play significant role in the life of human beings. We cannot neglect oceans because oceans are useful in

  • Influencing our climate.
  • Providing mineral deposits.
  • Helping power generation.
  • Providing source of food.

We shall now explain the influence of Ocean in detail.
Climate:
Oceans have a decisive role in controlling the climate along the coastal regions. The sea breeze during the day and the land breeze in the night regulate the temperature over the coasts. Oceans play a part in the formation of weather phenomena like rain, wind, and cyclones. Generally, the coastal regions have moderate climate, whereas severe summer and winter prevail in regions away from the sea.
Oceans as a source of food:
Fish is an important item of food. Fishing is major
activity in Japan, Peru, China, Norway, and the United States of America. Marine organisms are the source of many medicines. They are used for the production of antibiotics, steroids, and vitamins.

Ocean and Man Model Questions and Answers

9th Standard Social Science Notes Pdf In English Kerala Syllabus Question 9.
Nearly ………… % of earth’s surface area is covered with water.
Answer:
71%

9th Class Social Science Textbook Kerala Syllabus Question 10.
Name the Oceans
Answer:
a) The Pacific Ocean
b) The Atlantic Ocean
c) The Indian Ocean
d) The Artie Ocean
e) The Antarctic Ocean

9th Standard Social Science Textbook Kerala Syllabus Question 11.
Match the following

A B
Pacific ocean Southern ocean
Atlantic ocean Warton trench
Indian ocean Puerto Rico
Antarctic ocean Challenger Deep

Answer:

A B
Pacific ocean Challenger Deep
Atlantic ocean Puerto Rico
Indian ocean Warton trench
Antarctic ocean Southern ocean

Kerala Syllabus 9th Standard Social Science English Medium Question 12.
Distinguish between bay and strait.
Answer:
The portion of the sea surrounded by land on three sides is called a bay. On the other hand, the narrow stretch of sea between two landmasses is known as strait.

Question 13.
Complete the table

Islands Peninsula
Sri Lanka
Maldives
Green land
Sumatra

Answer:

Islands Peninsula
Sri Lanka Indian Peninsula
Maldives Arabian Peninsula
Green land Alaska Peninsula
Sumatra Labrador Peninsula

Question 14.
Define Peninsula
Answer:
The landmasses surrounded by sea on three sides are called peninsula.

Question 15.
Identify the important features of seawater.
Answer:

  • Temperature
  • Salinity
  • Density

Question 16.
Where do you find the highest Ocean temperature?
Answer:
Between 10° latitudes on either side of the equator

Question 17.
As we more away from the equator, temperature considerably.
a) Increases
b) Decreases
c) First increases and then decreases
d) Remains constant
Answer:
a) Increases

Question 18.
What is the reason for the variation in temperature over different latitudinal zones?
Answer:
Variation in the amount of insolation received on the earth is the major reason, The Ocean currents and winds also influence the temperature of seawater.

Question 19.
Define Salinity?
Answer:
The concentration of salt content in seawater is known as salinity.

Question 20.
The average amount of saltiness of seawater is
a) 2.5%
b) 3.5%
c) 4.5%
d) 5.5%
Answer:
b) 3.5%

Question 21.
Name the major contents of seawater
Answer:

  • Sodium chloride
  • Magnesium chloride
  • Magnesium Sulphate

Question 22.
Point out the conditions leading to variations in salinity.
Answer:

  • Salinity will be more in landlocked seas.
  • Salinity increases in areas of high evaporation.
  • Salinity decreases in areas where snow meltwater reaches in large quantity.
  • Salinity decreases at river mouths.
  • Heavy rainfall leads to reduction in salinity.

Question 23.
What causes movements in seawater.
Answer:
The density of seawater is not uniform everywhere. This is due to the variations in salinity and temperature of sea water. Density decreases as temperature increases, and it increases as salinity increases. You have understood that the temperature, salinity and the density of seawater are not uniform everywhere. These variations lead to movements in seawater.

Question 24.
Which are the movements of seawater.
Answer:

  • Waves
  • Tides
  • Ocean currents

Question 25.
What are sea waves?
Answer:
The up and down motion of the water along the surface of the sea is called sea waves.

Question 26.
Mark the missing portion.
Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 5 Ocean and Man 9
Answer:

  1. WaveLength
  2. Crest
  3. Wave height
  4. Trough

Question 27.
Name the following:
(i) The summit of the wave
(ii) Bottom part of the wave
(iii) Distance between 2 adjacent crests
(iv) Vertical distance between the crest and the trough
Answer:
(i) Wave crest
(ii) Wave trough
(iii) Wavelength
(iv) Wave height

Question 28.
………. is the reason for waves.
Answer:
The friction exerted by winds on the ocean surface

Question 29.
As the speed of the wind increases, the strength of the wave
a) Decreases
b) Increases
c) Remains constant
Answer:
b) Increases

Question 30.
Name two damages caused by strong waves
Answer:

  • Cyclones
  • Shelving of shores

Question 31.
Some measures are taken to prevent damage and to protect the lives of people living in the coastal areas. Identify the measures.
Answer:

  • Depositing boulders along the seashore.
  • Construction of interlocking concrete structures (Pulimuttu)
  • Planting of mangroves.

Question 32.
Name the sea waves generated by earthquakes and volcanos
Answer:
Seismic sea waves or tsunami waves.

Question 33.
Prepare a note on mud bank
Answer:
Mudbank is a phenomenon that develops in the Arabian Sea during the onset or the end of the monsoon season. Planktons grow luxuriantly in the turbulent muddy water along the seashore during the monsoon rains. Schools of fish such as shrimp, sardine, and mackerel arrive to feed on the planktons and the mud, giving fishermen a good catch. This phenomenon is known as mud bank.

Question 34.
What do you mean by tides?
Answer:
Tides are he periodic rise and fall of water level in the ocean.

Question 35.
Distinguish between high tide and low tide
Answer:
The rise in the level of ocean water is the high tide and the lowering of the water level is known as the low tide.

Question 36.
What are the reasons for tides?
Answer:
Tides are formed as a result of the gravitational pull exerted by the moon and the sun along with the centrifugal force due to the earth’s rotation.

Question 37.
‘ Illustrate with the help of a figure the occurrence of high tide and low tide.
Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 5 Ocean and Man 10
Answer:
The water level on the part of the earth facing the moon rises. The rise in water level due to the gravitational pull exerted by the moon leads to high tide. You might have noticed that the water level at the opposite side also has risen. The centrifugal force due to the earth’s rotation is the reason for the rise in water level here. It can be seen that the water level goes down at places located 90° away from the places of tidal influence. This is due to the draining of water towards the tidal regions. The phenomenon of fall of water level is known as low tide.

Question 38.
Point out the important effects of tides.
Answer:
High tides and low tides have many effects. Let’s have a look at them.

  • The debris dumped along the seashores and ports are washed off to the deep sea.
  • The formation of deltas is disrupted due to strong tides.
  • Brackish water can be collected in salt pans during high tides.
  • The fishermen make use of the tides for going and returning from the sea in catamarans.
  • Tidal energy can be used for power generation.
  • Ships can be brought to shallow harbors during high tides.

Question 39.
Prepare a note on ocean currents
Answer:
It is a type of seawater movement. Ocean currents are the continuous flow of seawater from one direction to another. They can be classified as warm currents and cold currents. Warm currents are the currents that flow from the tropical or subtropical regions towards the polar or subpolar regions. The water that flows in will be warmer than the water at the destination.

Similarly, cold currents are the currents that flow in from the polar or the subpolar regions towards the tropical or subtropical regions.In this case, the water that flow in will be colder than the water at the destination. The temperature and salinity of seawater varies from ocean to ocean. This difference leads to density difference in seawater. The difference in density is one of the factors that cause ocean currents.

Question 40.
Complete the chart.
Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 5 Ocean and Man 11
Answer:

  1. Warm currents
  2. Cold currents

Question 41.
What are the effects of ocean currents?
Answer;

  • Influence the climate of coastal regions.
  • Fog develops in the regions where warm and cold currents meet.
  • The regions where the warm and cold currents meet provide favorable conditions for the growth of fish.

Question 42.
Which are the mineral deposits in ocean.
Answer;

  • Iron ore
  • Coal
  • Petroleum
  • Natural gas

Question 43
The oil field started in 1974 in Mumbai is known as
Answer:
Mumbai High

Question 44.
Name some countries where fishing is an important activity.
Answer:

  • Japan
  • Peru
  • China
  • Norway
  • USA

Question 45.
What are the medicinal user of marine organisms?
Answer:
Marine organisms help in production of :-

  • antibiotics
  • steroids
  • vitamines

Question 46.
Prepare a flow chart showing method of purifying seawater.
Answer:
Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 5 Ocean and Man 12

Let Us Assess

Question 47.
Which among the following statements is not related to the Indian Ocean?
a. The southern part of this ocean extends up to the Antarctic Ocean.
b. The average depth is more than that of the Atlan tic Ocean.
c. The Puerto Rico trench is situated in this ocean.
d. It ranks third in area.
Answer:
B & C are not related

Question 48.
which among the following places record the least salinity? why?
1. Land-locked sea,
2. Areas of heavy rainfall
3. Areas of high evaporation
Answer;
Landlocked sea. Because salinity increases in landlocked seas.

Question 49.
Is there any relation between the intensity of waves and the wavelength? substantiate
Answer:
The Intensity of waves is defined as the power delivered per unit area. The unit of intensity will be W.m2. The wave energy comes from simple harmonic motion of its particles. The total energy will equal the maximum kinetic energy. As intensity of waves increases the wavelength also increases.

Question 50.
High tide occurs twice a day. Explain this statement.
Answer:
The rise in water level due to the gravitational pull exerted by the moon leads to high tide. Tides can occur as two high waters and two low water each day. The tides are occurred by the gravitational. The moon’s gravity pulls the ocean surface closer to it and the moon makes two trips around the earth each day. Hence there in high tide twice a day.

Question 51.
Explain spring tides and neap tides with the help of diagrams.
Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 5 Ocean and Man 13
The sun, moon, and earth come in a straight line on full moon and new moon days. The tidal force will be intense due to the combined influence of sun and moon. As a result, the tides formed on these days will be stronger. These are known as spring tides. The moon and the sun will be at an angular distance of 90° from the earth after seven days from the full moon and new moon days. As the sun and the moon attract the earth from an angular distance of 90° the tides casued are weak. Such weak tides are known as neap tides. Note the positions of the earth, moon, and sun in the given diagram.

Question 52.
Oceans play an important role in human life and environment. Justify.
Answer:
Influence of Oceans in human life
Oceans play significant role in the life of human beings. We cannot neglect oceans because oceans are useful in

  • Influencing our climate.
  • Providing mineral deposits.
  • helping power generation.
  • Providing source of food.

We shall now explain the influence of ocean in detail.

Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles

You can Download Similar Triangles Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles

Kerala Syllabus 9th Standard Maths Similar Triangles Text Book Questions and Answers

Textbook Page No. 100

Similar Triangles Class 9 Kerala Syllabus Question 1.
The perpendicular from the square corner of a right triangle cuts the opposite side into two parts of 2 and 3 centimetres length.
Similar Triangles Class 9 Kerala Syllabus
i. Prove that the two small right triangles cut by the perpendicular have the same angles.
ii. Taking the length of the perpendicular as h, prove that \(\frac{h}{2} = \frac{3}{h}\)
iii. Calculate the perpendicular sides of the large triangle.
iv. Prove that if the perpendicular from the square corner of a right triangle divides the opposite side into parts of lengths a and b and if the length of the perpendicular is h, then h² = ab.
Answer:
i. ∠ADB = 90°
Let ∠BAD = x°
∠DAC = ( 90 – x)°
Similar Triangles Class 9
∴ ∠B = 180 – (x + 90) = 90 – x
and ∠C = 180 – (90 – x + 90) = x
Angles of ΔABC = x°, (90 – x)°, 90°
Angles of ΔACD = x°, (90 – x)°, 90°
The two small triangles have the same angles.

ii. Triangle with the same angles, sides opposite equal angles are scaled by the factor.
2 : h = h : 3 ⇒ \(\frac{2}{h} = \frac{h}{3}\)
i.e \(\frac{h}{2} = \frac{3}{h}\)

iii. \(\frac{h}{2} = \frac{3}{h}\)
h² = 6 ⇒, h = \(\sqrt 6\)
In ΔABD
AB² = BD² + AD² = 2² + (\(\sqrt 6\))²
= 4 + 6 = 10
AB = \(\sqrt 10\)
In ΔADC
AC² = DC² + AD² = 3² + (\(\sqrt 6\))²
= 9 + 6 = 15
AC = \(\sqrt 15\)
The perpendicular sides are \(\sqrt 10\) and \(\sqrt 15\)

iv. Triangle with the same angles, sides opposite equal angles are scaled by the same factor.
BD : AD = AD : DC
Similar Triangles Class 9 Kerala Syllabus Chapter 7
\(\frac{h}{a} = \frac{b}{h}\)
h² = ab

Similar Triangles Class 9 Question 2.
At two ends of a horizontal line, angles of equal size are drawn and two points on the slanted lines are
Similar Triangles Class 9 Chapter 7
i. Prove that the parts of the horizontal line and parts of the slanted line are in the same ratio.
ii. Prove that the two slanted lines at the ends of the horizontal line are also in the same ratio.
iii. Explain how a line of length 6 cm can be divided in the ratio 3 : 4
Answer:
Kerala Syllabus 9th Standard Maths Chapter 7
i. ∠A = ∠B
∠AMC = ∠BMD
∴ ∠C = ∠D
(The triangles AMC and DMB are similar).
∴ \(\frac{MC}{MD} = \frac{AC}{BD} = \frac{AM}{MB}\)
\(\frac{AM}{MB} = \frac{MC}{MD}\)

ii. \(\frac{MC}{MD} = \frac{AC}{BD} = \frac{AM}{MB}\)
(From the similar triangles AMC and DMB)

iii. Draw a line segment of length 6 cm. At one end of this line AB, 3cm long and at the other end draw CD, 4cm long in the opposite direction.
Chapter 7 Similar Triangles Kerala Syllabus
∠CAB = ∠ACD
Draw BD to cut A at O.
Since AB : CD = OA : OC
∴ OA : OC = 3 : 4

Scert Class 9 Maths Similar Triangles  Question 3.
The mid point of the bottom side of a square is joined to the ends of the top side and extended by the same length. The ends of these lines are joined and perpendiculars are drawn from these points to the bottom side of the square extended.
9th Class Maths Notes Kerala Syllabus Chapter 7
i. Prove that the quadrilateral obtained thus is also a square.
ii. Explain how we can draw a square with two corners on a semicircle and the other two corners on its diameter as in the figure
Kerala Syllabus Class 9 Maths Solutions Chapter 7
Answer:
Kerala Syllabus 9th Standard Maths Notes Chapter 7
i. ΔDCM and ΔSRM are similar.
The angles of the ΔMBC and ΔMAD are equal.
ΔPMS, ΔQMR are also similar triangles.
∠D = ∠S; ∠C = ∠R; ∠A = ∠B; ∠B = ∠Q
Let the ratio of the equal sides are be k.
\(\frac{QR}{BC}\) = k ⇒ \(\frac{QR}{2a}\) = k
∴ QR = 2ak
Similarly, PS = PQ = SR = 2ka
∴ PQRS is a square.

ii.
Kerala Syllabus 9th Standard Maths Guide In Malayalam
Draw a line of AB, 8cm long. Find midpoint O on the line. Mark P and Q such that OP = OQ = 1cm. Draw the square PQRS whose sides are 2cm long.

Draw a semicircle with O as centre and OA as radius. Extend OS and OR to meet the semicircle at G and F. From G and F draw GD and FE perpendicular AB. Draw GF. Quadrilateral DEFG is the required square.

Kerala Syllabus 9th Standard Maths Chapter 7 Question 4.
The picture shows a square drawn sharing one corner with a right triangle and the other three corners on the sides of this triangle.
9th Class Maths Kerala Syllabus Chapter 7
i. Calculate the length of a side of the square.
ii. What is the length of a side of the square drawn like this within a triangle of sides 3, 4 and 5 centmetres?
Answer:
i.
Class 9 Maths Notes Kerala Syllabus Chapter 7
ΔABC and ΔAPQ are similar.
\(\frac{x + 2}{2} = \frac{x + 1}{x}\) = x(x + 2) = 2(x +1)
x² + 2x = 2x + 2
x² = 2; x = \(\sqrt 2\)
Side of square = \(\sqrt 2\)cm.

ii. AB = 4, BC = 3, AC = 5
ΔAPQ and ΔABC are similar triangles
∴ \(\frac{AP}{PB} = \frac{PQ}{BC} ⇒ \frac{4 – x}{4} = \frac{x}{3}\)
3(4 – x) = 4x
12 – 3 = 4x
12 – 7x; x = \(\frac {12}{7}\)
Hss Live Class 9 Maths Kerala Syllabus Chapter 7
The length of the side of the square = \(1\frac {5}{7}\)

Chapter 7 Similar Triangles Kerala Syllabus Question 5.
Two poles of heights 3 m and 2 m are erected upright on the ground and ropes are stretched from the top of each to the foot of the other.
Kerala Syllabus 9th Standard Maths Solutions Chapter 7
i. At what height above the ground do the ropes cross each other?
ii. Taking the heights of the poles as a and b and height above the ground of the point where the ropes cross each other as h, And the relation between a, b and h.
iii. Prove that this height would be the same, whatever be the distance between the poles.
Answer:
i. Consider ΔABC and ΔAFE. They are similar.
\(\frac{b}{h} = \frac{x + y}{x}\) ………..(1)
9th Standard Maths Notes Kerala Syllabus Chapter 7
Consider ΔADB and ΔFEB. They are also similar
\( \frac{a}{h} = \frac{x + y}{y}\) ………..(1)
\( \frac{h}{a} = \frac{y}{x + y}, \frac{h}{b} = \frac{x}{x + y}\)
(1) + (2) ⇒ \( \frac{h}{a} + \frac{h}{b} = \frac{y}{x + y} + \frac{x}{x + y}\)
h\((\frac{1}{a} + \frac{1}{b}) = \frac{x + y}{x + y}\)
h\((\frac{1}{a} + \frac{1}{b})\) = 1
\(\frac{1}{a} + \frac{1}{b} = \frac{1}{h}\) ……..(2)
a = 3 and b = 2
\(\frac{1}{h} = \frac{1}{3} + \frac{1}{2} + \frac{5}{6}\)cm
h = \(\frac{6}{5}\) = 1.2

ii. The heights of the poles are a and b and the height above the ground of the point where the ropes cross each other is h, then the relation between a, b and h is \(\frac{h}{a} + \frac{h}{b}\) = 1 or \(\frac{a + b}{ab} = \frac{1}{h}\) from equation (3).

iii. Only change its height according to the height of the poles not the distance.

Textbook Page No. 107

9th Class Maths Notes Kerala Syllabus Question 1.
Draw a triangle of angles the same as those of the triangle shown and sides scaled by \(1\frac{1}{4}\)
Class 9 Maths Chapter 7 Kerala Syllabus Chapter 7
Answer:
One side is 6 cm. Its \(1\frac{1}{4}\) part is 7.5 cm
Other sides are 4 × \(1\frac{1}{4}\) = 5 cm
8 × \(1\frac{1}{4}\) = 10 cm
Draw one side in 10 cm. Complete the triangle with given measures.
Chapter 7 Maths Class 9 Kerala Syllabus Chapter 7

Kerala Syllabus Class 9 Maths Solutions Question 2.
See this picture of a quadrilateral.
Std 9 Maths Kerala Syllabus Kerala Syllabus Chapter 7
i. Draw a quadrilateral with angles the same as those of this one and sides scaled by \(1\frac{1}{2}\)
ii. Draw a quadrilateral with angles different from those of this and sides scaled by \(1\frac{1}{2}\).
Answer:
i. Draw a quadrilateral by increasing all the sides including the diagonal by \(1\frac{1}{2}\)
4 × \(1\frac{1}{2}\) = 4 × 1.5 = 6
2 × \(1\frac{1}{2}\) = 2 × 1.5 = 3
3 × \(1\frac{1}{2}\) = 3 × 1.5 = 4.5
5 × \(1\frac{1}{2}\) = 5 × 1.5 = 7.5
6 × \(1\frac{1}{2}\) = 6 × 1.5 = 9
Kerala Syllabus 9th Standard Maths Notes Malayalam Medium Chapter 7

ii. Draw a quadrilateral by increasing all the sides increased by \(1\frac{1}{2}\) except the diagonal.
Kerala Syllabus 9th Standard Maths Guide Pdf Chapter 7

Textbook Page No. 111

Kerala Syllabus 9th Standard Maths Notes Question 1.
The picture shows two circles with the same centre and two triangles formed by joining the centre to the points of intersection of the circles with two radii of the larger circle. Prove that these triangles are similar
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 21
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 22
AP = AQ (radius)
∠APQ = ∠AQP
(Base angles of an isosceles triangle)
∠A = ∠A (Common angle)
AC = AB (radius)
∴ ∠ACB = ∠ABC
∴ \(\frac{AP}{AC} = \frac{AQ}{AB}\)
That is two sides of these triangles are scaled by the same factor.
Since two sides of these triangles are scaled by the same factor and the angle between them the same ΔAPQ and ΔACB are similar. So the triangles are similar

Kerala Syllabus 9th Standard Maths Guide In Malayalam Question 2.
The lines joining the circumcentre of a triangle to the vertices are extended to meet another circle with the same centre and these points are joined to make another triangle.
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 23
i. Prove that the two triangles are similar.
ii. Prove that the scale factor of the sides of the triangle is the scale factor of the radii of the circles.
Answer:
∴ ΔAOB and ΔPOQ are similar triangles
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 24
∴ \(\frac{AB}{PQ} = \frac{OB}{OQ}\) ……..(1)
ΔBOC, ∠QOR are similar
∴ \(\frac{OB}{OQ} = \frac{BC}{QR} = \frac{OC}{OR}\) ……..(2)
Similarly
\(\frac{OC}{OR} = \frac{AC}{PR}\) ……..(3)
(1), (2), (3) consider
∴ \(\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}\)
ΔABC and ΔPQR are similar triangles.

ii. From equation (1), (2) and (3)
The scale factor of the sides of the triangle is the scale factor of the radii of the circles.

9th Class Maths Kerala Syllabus Question 3.
A point inside a quadrilateral is joined to its vertices and the lines are extended by the same scale factor. Their ends are joined to make another quadrilateral.
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 25
i. Prove that the sides of the two quadrilateral are scaled by the same factor.
ii. Prove that the angles of the two quadrilaterals are the same.
Answer:
i.
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 26
ΔAOB, ΔPQO are similar
\(\frac{OA}{OP} = \frac{AB}{PQ} = \frac{OB}{OQ}\) …….(1)
ΔOBC, ΔOQR are similar
\(\frac{OB}{OQ} = \frac{BC}{QR} = \frac{OC}{OR}\) …….(2)
ΔOCD, ΔORS are similar
\(\frac{OC}{OR} = \frac{CD}{RS} = \frac{OD}{OS}\) …….(3)
ΔODA, ΔOSP are similar
\(\frac{OD}{OS} = \frac{AD}{PS} = \frac{OA}{OP}\) …….(4)
Consider (1), (2), (3), (4)
\(\frac{AB}{PQ} = \frac{BC}{QR} = \frac{DC}{SR} = \frac{AD}{PS}\)
Since the sides of quadrilateral EFGH and quadrilateral ABCD are scaled by the same factor.
∴ The angles of the quadrilateral are also same (Angles of similar triangles).

ii. Since the sides of quadrilateral PQRS and quadrilateral ABCD are scaled by the same factor, the angles of the large quadrilateral will be the same as the angles of the small quadrilateral.

Kerala Syllabus 9th Standard Maths Similar Triangles Exam Oriented Text Book Questions and Answers

Class 9 Maths Notes Kerala Syllabus Question 1.
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 27
ΔABC, ΔXYZ are similar triangles in the picture. Then
\(\frac{AB}{XY} = \frac{BC}{….} = \frac{….}{XZ}\)
Answer:
\(\frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}\)

Hss Live Class 9 Maths Kerala Syllabus Question 2.
In ΔXYZ PQ is parallel to XY.
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 28
a. \(\frac{YP}{PZ} = \frac{XQ}{….}\)
b. \(\frac{x}{a} = \frac{…..}{b}\)
Answer:
a. \(\frac{YP}{PZ} = \frac{XQ}{QZ}\)
b. \(\frac{x}{a} = \frac{y}{b}\)

Kerala Syllabus 9th Standard Maths Solutions Question 3.
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 29
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 30
Fill in the blanks according to the picture.
∠K = ∠A; ∠B = …….
∠C = ……; KL = ……cm
KM = ….. cm
Answer:
∠B = ∠L
∠C = ∠M
KL = 8cm
KM = 10cm

9th Standard Maths Notes Kerala Syllabus Question 4.
In the diagram AP and BQ are perpendicular to AB. AP = 4cm, BQ = 2cm Then show that AC : CB = 2 : 1
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 31
Answer:
∠PAC = 90° ∠CBQ = 90°
∠ACP = ∠BCQ (opposite angles)
∴ ΔPAC ≅ ΔBCQ
\(\frac{PA}{BQ} = \frac{AC}{CB} = \frac{PC}{CQ}\)
\(\frac{AC}{CB} = \frac{PA}{BQ} = \frac{AC}{CB} = \frac{2}{1}\)
AC : CB = 2 : 1

Class 9 Maths Chapter 7 Kerala Syllabus Question 5.
In ΔABC and ΔPQR we have \(\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP}\). In triangle ABC, the altitude through P meets BC at D, and in triangle PQR. The altitude through P meets QR at ‘S’. Prove that \(\frac{AB}{PQ} = \frac{AD}{PS}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 32
Answer:
ΔABC, ΔPQR are similar.
∠A = ∠P; ∠B = ∠Q;
∠C = ∠R;
Consider ΔADB ΔPSQ
∠B = ∠Q, ∠ADB = ∠PSQ = 90°
(AD ⊥ BC & PS ⊥ QR)
[each equal to 90°]
∴ ΔADB, ΔPSQ are similar.
[Since the corresponding sides of similar triangles are proportional]
∴ \(\frac{AD}{PS} = \frac{DB}{SQ} = \frac{AB}{PQ} ⇒ \frac{AB}{PQ} = \frac{AD}{PS}\)

Chapter 7 Maths Class 9 Kerala Syllabus Question 6.
Prove that if the area of two similar triangles are proportional to the squares on their corresponding sides.
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 33
Answer:
ΔABC ~ ΔPQR.
AM, PN are perpendicular, \(\frac{AM}{PN} = \frac{BC}{QR}\)
Area of ΔABC = \(\frac{1}{2}\) BC × AM
Area of ΔPQR = \(\frac{1}{2}\) QR × PN
\(\frac{Area of ΔABC}{Area of ΔPQR}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 34
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 35

Std 9 Maths Kerala Syllabus Kerala Syllabus Question 7.
In the given figure, PC ⊥ QR and QD ⊥ PR. Prove that ΔPCR and ΔQDR are similar.
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 36
Answer:
In ΔPCR, ∠PCR = 90°, and in ΔQDR = 90°
∠R is common to both the triangles
ΔPCR, ΔQDR. ∠PCR = ∠QDR = 90°
∠PRC = ∠QRD (common)
(Two triangles having two pairs of corresponding angles equal. The triangles are similar.)

Kerala Syllabus 9th Standard Maths Notes Malayalam Medium Question 8.
In ΔABC, P is a point on BC. Where D, E, F are the mid-points on BP, AP and CP respectively, then prove that ΔABC ~ ΔDEF.
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 37
Answer:
Consider ΔABP, ΔDPE.
ΔABP ~ ΔDPE (∠A is ∠BAP = ∠DEP and ∠ABP = ∠EDP)
∴ \(\frac{AB}{ED} = \frac{AP}{EP}\) …..(1)
similarly ΔAPC ≅ ΔEPF.
∴ \(\frac{AC}{EF} = \frac{AP}{EP}\) …..(2)
(1) = (2) \(\frac{AB}{ED} = \frac{AC}{EF} = \frac{AP}{EP}\)
ΔABC ~ ΔDEF

Kerala Syllabus 9th Standard Maths Guide Pdf Question 9.
In figure AD || BC, AB || DE prove that ΔABC – ΔEDA .
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 38
Answer:
∠DAE = ∠C
(AD || BC corresponding angles)
∠BAC = ∠DEA
(AB || DE corresponding angles)
ΔABC ~ ΔEDA

Question 10.
ΔABC is a right angled triangle. ∠B = 90° a perpendicular line from B to AC intersect AC at D. Prove that ΔABC, ΔADB and ΔBCD are similar to each other.
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 7 Similar Triangles 39
ΔABC = ΔABD
[∠ABC = ∠ADB = 90°, ∠ACB = ∠ABD = 90 – X]
∠A (common)
similarly ΔABD ~ ΔBDC
In ΔABC, consider
∠A = X°
∠B = 90°
∠A + C = 90°, ∠C = 90 – X
In ΔABD, ∠ADB = 90°
Therefore ∠A = X°, ∠ABC = 90°
∠A + ∠ABD = 90°
∠ABD = 90 – ∠A = 90 – X;
In ΔBCD, ∠BDC = 90°
∠C = (90 – X); ∠C + ∠DBC = 90°;
∠DBC = 90 – (90 – X) = X°
Hence angles of three triangles are 90°, X° and 90° – X°.
Hence three triangles are similar to each other.

Kerala Syllabus 9th Standard Chemistry Solutions Chapter 2 Chemical Bonding

You can Download Chemical Bonding Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Chemistry Solutions Part 1 Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Chemistry Solutions Chapter 2 Chemical Bonding

Chemical Bonding Textual Questions and Answers

Kerala Syllabus 9th Standard Chemistry Notes Chapter 2 Question 1.
What peculiarity do you see in the electronic configuration of noble elements except Helium? Except Helium all other elements have 8 electrons in the outermost shell, hence shall be considered to be chemically stable.

The arrangement of eight electrons in the outermost shell of atom is called octet electron configuration. In Helium atom there is only one shell. The maximum number of electrons in the first shell is 2. Hence two-electron pattern system of Helium also stable.

Kerala Syllabus 9th Standard Chemistry Chapter 2 Question 2.
The electronic configuration of some elements are given below.

Element Atomic mass Electronic configuration
Magnesium 12 2, 8, 2
Oxygen 8 2, 6
Sodium 11 2, 8, 1
Chlorine 17 2, 8, 7

Is the number of electrons in the outermost shell of these elements the same as that of the elements in Table (2.1).
Answer:
No

(i) You are familiar with the compounds of these elements. Write the names of some compounds?
Answer:
Magnesium chloride, Sodium oxide, Sodium chloride.

(ii) How are atoms in these compounds held together?
Answer:
Strong attractive force

(iii) What is meant by Chemical Bonding?
Answer:
The attractive force that holds the atoms together in the formation of a molecule is called chemical bonding.

Ionic Bonding

Class 9 Chemistry Chapter 2 Notes Kerala Syllabus Question 3.
In the formation of sodium chloride which atoms are combing.
Answer:
Sodium, chlorine

Chemical Bonding Notes Class 9 Kerala Syllabus Question 4.
How many electrons are there in the outermost shell of sodium atom?
Answer:
1

Chemical Bonding Questions And Answers Class 9 Kerala Syllabus Question 5.
How many electrons are there in outermost shell of chlorine?
Answer:
7

Chemical Bonding Questions And Answers Pdf Class 9 Kerala Syllabus Question 6.
How do chlorine and sodium attain stability?
Answer:
Sodium donates one electron to chlorine to become sodium ion [Na+] and chlorine become chloride ion [Cl]

Chemical Bonding Class 9 Notes Pdf Kerala Syllabus Question 7.
Analyze the electron transfer in each atom during the formation of sodium chloride.
Answer:
Kerala Syllabus 9th Standard Chemistry Notes Chapter 2

Chemical Bonding Class 9 Pdf Kerala Syllabus Question 8.
Draw the electron dot diagram of the transference of electron of sodium atom and chlorine atom. The diagram represents only electrons in the outermost shell because they are the only electrons participating- in chemical bonding.
Answer:
Kerala Syllabus 9th Standard Chemistry Chapter 2

Questions On Chemical Bonding Class 9 Kerala Syllabus Question 9.
Complete Table 2.3 by examining the arrangement of electrons before and after the chemical reaction during the formation of sodium chloride.
Class 9 Chemistry Chapter 2 Notes Kerala Syllabus
a) Which atom donates electron? How many electrons?
b) Which atom accepts electron? How many electrons?
Answer:
a) Sodium, one electron
b) Chlorine, one electron

Kerala Syllabus 9th Standard Chemistry Notes English Medium Question 10.
Electron transfer during the formation of sodium chloride can be written in the form of an equation
Na → Na++1e
Cl + 1e → Cl
Answer:
During the formation of sodium chloride sodium atom donates electron and gets converted to sodium ion (Na+) chlorine accepts an electron to form chloride ion (Cl ). Through this sodium and chlorine atoms complete an octet in their outermost shell to attain stability.

The oppositely charged ions thus formed are held together by electrostatic force of attraction. This attractive force is called Ionic Bond. Sodium chloride contains ionic bond.

Kerala Syllabus 9th Standard Chemistry Chapter 1 Question 11.
Define Ionic Bond?
Answer:
Ionic bond is a chemical bond formed by electron transfer in an ionic bond, the ions are held together by the electrostatic force of attraction between the oppositely charged ions.

Kerala Syllabus 9th Standard Chemistry Solutions Question 12.
Explain the formation of magnesium oxide from magnesium and oxygen?
Analyze the electron dot diagram and complete the table.
Answer:
Chemical Bonding Notes Class 9 Kerala Syllabus
To attain stability magnesium donates 2 electrons to become magnesium ion (Mg2+) and – oxygen become [O2 ] ion. This type of bonding is ionic bonding.
Chemical Bonding Questions And Answers Class 9 Kerala Syllabus

Chemical Bonding Notes Class 9 Pdf Kerala Syllabus Question 13.
How the ionic bond formation of sodium oxide is represented?
[Hint: Atomic No. of sodium 11, oxygen 8]
Answer:
Chemical Bonding Questions And Answers Pdf Class 9 Kerala Syllabus

Question 14.
Draw the electron dot diagram of following compounds. [Hint: Atomic No. Na=11, F=9, Mg=12]
Answer:
1. Sodium Flouride [NaF]
Chemical Bonding Class 9 Notes Pdf Kerala Syllabus

Question 15.
Define ionic compounds?
Answer:
Compounds formed by ionic bonding are called ionic compound.

Covalent Bonding

Fluorine [F2], Chlorine [Cl2] oxygen [O2] Nitrogen [N2] etc. are diatomic molecules. Let us examine the formation of these molecules.
The Bohr atom model of fluorine is given in figure.
Chemical Bonding Class 9 Pdf Kerala Syllabus

Question 16.
Write the atomic number of fluorine?
Answer:
9

Question 17.
The electronic configuration of Fluorine
Answer:
2, -7 ‘

Question 18.
How many electrons are required for one fluorine atom to attain the octet?
Answer:
1

Question 19.
Is there a possibility of transferring electrons from one fluorine atom to another fluorine atom?
Answer:
No.

Question 20.
How can the two fluorine atoms attain an octet arrangement?
Answer:
By sharing of electrons

Question 21.
The manner in which the two fluorine atoms in a fluorine molecule undergo chemical bonding is illustrated in fig. 2.6.
Answer:
Questions On Chemical Bonding Class 9 Kerala Syllabus

Question 22.
What happens during the formation of fluorine molecule electron transfer or electron sharing?
Answer:
Electron sharing

Question 23.
How many pairs of electrons are shared?
Answer:
One pair

Question 24.
How covalent bonds are formed?
Answer:
The chemical bond formed as a result of the sharing of electrons between the combining atoms is called a covalent bond.

Question 25.
How single bonds are formed?
Answer:
Single bonds are formed by sharing one pair of electrons. It is represented by a small line between the symbols of the combining element, eg. Fluorine molecules can be represented as F – F. The atomic number of chlorine is 17.

Question 26.
Write down the electronic configuration?
Answer:
2, 8, 7

Question 27.
Draw the electron dot diagram of the formation of chlorine molecule by combining two chlorine atoms?
Answer:
Kerala Syllabus 9th Standard Chemistry Notes English Medium
Here one pair of electrons – sharing hence single bond is formed.

Question 28.
Examine the diagram illustrating the chemical bonding in the molecule of oxygen and nitrogen.
Answer:
Kerala Syllabus 9th Standard Chemistry Chapter 1
In oxygen molecules, two pairs of electrons are shared hence this type of covalent bond is double bond.
In nitrogen molecule, three pairs of electrons are shared hence this type of covalent bond is triple bond.

Question 29.
Complete table 2.5 given below related to covalent bonding.
Kerala Syllabus 9th Standard Chemistry Solutions
Answer:

Element molecule Shared electron pairs Chemical bond
F2 One pair Single bond
Cl2 One pair Single bond
O2 Two pair Double bond
N2 Three pairs Triple bond

Question 30.
Draw the chemical bond formation of hydrogen chloride [HCI]
Chemical Bonding Notes Class 9 Pdf Kerala Syllabus
a) How many electron pairs are shared?
b) Represent chemical bond by using symbols?
Answer:
a) One pair of electrons
b) H – Cl

Question 31.
Examples of some covalent compounds are given draw the chemical bonds of the compound by using electron dot diagram.
a)CH4
b) HF
c)H2O
Answer:
a)
Kerala Syllabus 9th Standard Chemistry Solutions Chapter 2 Chemical Bonding 15
Kerala Syllabus 9th Standard Chemistry Solutions Chapter 2 Chemical Bonding 16

Electronegativity

Question 32.
Define electronegativity?
Answer:
In a covalent bond the relative ability of each atom to attract the bonded pair of electrons towards itself is called electronegativity.

Question 33.
Who proposed the electronegativity scale?
Answer:
Linus Pauling

Question 34.
Some compounds and their nature are shown in table (2.6) complete the table by finding out the electron negativity difference between the constituent elements.
Kerala Syllabus 9th Standard Chemistry Solutions Chapter 2 Chemical Bonding 17
Answer:

Compounds Electronegativity difference of the elements Character of compound
Carbon Tetra Chloride [CCl4] 3.44 – 2.55 = 0.89 Covalent bond
Sodium chloride [NaCl] 3.16 – 0.98 = 2.23 Ionic bond
Methane [CH4] 2.55 – 2.20 = 0.35 Covalent bond
Magnesium chloride [MgCl2] 3.16 – 1.31 = 1.85 Ionic bond
Sodium oxide[Na2O] 3.44 – 0.93 = 2.51 Ionic bond

Generally, the electronegativity difference of the component elements in a compound is 1.7 or more it shows ionic character. If it is less than 1.7 it shows covalent character.

Polar Nature

Question 35.
Consider the case of hydrogen molecule [HCl]
a) What is the electronegativity of hydrogen?
b) What is the electronegativity of chlorine?
c) The atomic nucleus of which of these elements has a greater tendency to attract the shared pair of electrons?
d) The chlorine atom with a higher electronegativity attracts the shared pair of electrons towards its nucleus. As a result, the chlorine atom in hydrogen chloride develops a partial negative charge g: (delta negative) and hydrogen atom develops a partial positive charge δ+ (delta positive) it can be represented below
Answer:
a) 2.2
b) 3.16
c) Chlorine
d) \(\begin{array}{l}{\delta^{+} \quad \quad \delta^{-}} \\ {H-C^{\prime}}\end{array}\) Compounds having partial electron charge separation with the molecule are called polar compound. HF, HBr, H20 are example of polar compounds.

Question 36.
Explain the properties of Ionic compounds and covalent compounds.
Answer:

Properties Ionic compound Covalent compounds
State Solid Found in the three states solids, liquids and gases
Solubility in water soluble in water Insoluble in water. But soluble in organic solvent like kerosene, benzene etc.
Electrical Conductivity conduct electri­city in fused or solution state Do not conduct Electricity
Melting point Boiling point High Generally Low

Valency:

Question 37.
What is meant by valency?
Answer:
Valency is the combining capacity of the atoms of an element. It can be treated as the number of electrons lost gained or shared by an atom during chemical combination.

Question 38.
In the formation of sodium chloride- sodium donates one electron, chlorine accepts one electron write the valencies of each element?
Answer:
1

Question 39.
In the formation of magnesium oxide- How many electrons are donated by magnesium?
Answer:
2

Question 40.
How many electrons are accepted by oxygen?
Answer:
2

Question 41.
How is valency and electron transfer related in this case?
Answer:
Same

Question 42.
In the formation of hydrogen chloride, how many electron pairs are shared?
Answer:
One pair

Question 43.
What will be the valency of each atom?
Answer:
1

Question 44.
Complete the table given below analyze the change in the electronic arrangement of elements during the formation of each compound. Find how they are related to valency.
Kerala Syllabus 9th Standard Chemistry Solutions Chapter 2 Chemical Bonding 18
Answer:

Compound Component elements Atomic number Electron configur­ation No. of elect­rons donated accepted/ shared Valency
NaCl Na 11 2, 8, 1 1 1
Cl 17 2, 8 ,7 1 1
MgO Mg 12 2, 8, 2 2 2
0 8 2, 6 2 2
HF H 1 1 1 1
F 9 2, 7 1 1
CCl4 C 6 2, 4 4 4
Cl 17 2, 8, 7 1 1

From Valency to Chemical Formula

The chemical formula of some compounds are given
Sodium chloride – NaCl
Magnesium chloride – MgCl2
Aluminium chloride – AICl3
Carbon tetrachloride – CCl4

Question 45.
The symbol of some elements and there valencies are given. Write the chemical formula of the compounds formed by them.

Element Valency
Cl 1
Li 1
0 2
Zn 2
Ca 2

Answer:
Kerala Syllabus 9th Standard Chemistry Solutions Chapter 2 Chemical Bonding 19

Question 46.
Why does the number of chlorine atoms differ in these compounds? Try to find out by analyzing the valency of the elements Na, Mg, Al, Cl and C. Analyse Table 2.9
Answer:

Examine the above table and identify how to write the chemical formula from valency. Compare your findings with the following.

  • First write the element with lower electronegativity.
  • Exchange the valency of each element and write as suffix.
  • Divide the suffix with the common factor.
  • If the suffix is 1, it need not be written.

Let Us Assess

Question 1.
Complete the table given below and answer the following questions (symbols used are not true)

Element Atomic number Electronic configuration
P 9 2, 7
Q 17 …………………………..
R 10 …………………………..
S 12 ……………………………

a) Which element in the table is the most stable one? Justify your answer.
b) Which element donates electrons in chemical reaction?
c) Write the chemical formula of the compound formed by combining element S with P.
Answer:

Element Atomic number Electronic configuration
P 9 2, 7
Q 17 2, 8, 7
R 10 2, 8
S 12 2, 8, 2

a) R – it contains an octet configuration in the outermost shell.
b) S
c) The valency of S = 2 and P is 1, hence the chemical formula
Kerala Syllabus 9th Standard Chemistry Solutions Chapter 2 Chemical Bonding 20
[The element in which the electron lose can be written first]

Question 2.
Electronegativity values of some elements are given. Using these values, find whether the following compounds are ionic or covalent.
(Electronegativity of Ca = 1, O = 3.5, C = 2.5, S = 2.58, H = 2.2, F = 3.98)
i) Sulphur dioxide (S02)
ii) Water (H2O)
iii) Calcium fluoride (CaF2)
iv) Carbon dioxide (CO2)
Answer:
i) SO2
Electronegativity difference = 3.5 – 2.58 = 0.92
If the electronegativity difference is less than 1.7, it shows covalent character.
ii) Water (H2O)
Electronegativity difference = 3.5 – 2.2 = 1.3
If the electronegativity difference is less than 1.7 it forms covalent compounds.
iii) CaF2
Electronegativity difference = 3.98 -1.0 = 2.98
If the electronegativity difference is greater than 1.7 if forms ionic compounds.
iv) CO2
Electronegativity difference = 3.5 – 2. 5 = 1.0
If the electronegativity difference is less than 1.7 it shows covalent compounds.

Question 3.
Some elements and their valencies are given

Element Valency
Ba 2
Cl 1
Zn 2
O 2

a) Write the chemical formula of barium chloride
b) Write the chemical formula of zinc oxide
c) The chemical formula of calcium oxide is CaO. What is the valency of calcium?
Answer:
Kerala Syllabus 9th Standard Chemistry Solutions Chapter 2 Chemical Bonding 21
c) The valency of calcium is 2.

Question 4.
Examine the following chemical equations and answer the questions.
(Hint: Atomic Number Mg = 12 Cl = 17)
Mg + Cl2 → MgCl2
Mg → Mg2+ + ……..
Cl + 1e → ………..
a) Complete the chemical equations.
b) Which is the cation? Which is the anion?
c) Which type of chemical bond is present in MgCl2?
Answer:
a) Mg → Mg2- + 2e
Cl + 1e → Cl
b) The cation or positively charged ion is Mg2+ and the anion or negatively charged ion is Cl
c) Ionic Bonding

Extended Activities

Question 1.
Draw the electron dot diagram of chemical bonds in methane (CH4) and ethane (C2H6).
Answer:
Kerala Syllabus 9th Standard Chemistry Solutions Chapter 2 Chemical Bonding 22

Question 2.
P, Q, R, S are four elements. Their atomic numbers are 8, 17, 12 and 16 respectively. Find the type of chemical bond in these compounds formed by combining the following pairs of elements. Construct and exhibit the type of bonds using different. substances (eg. pearls) (Electronegativity values:
P = 3.44, Q = 3.16, R = 1.31, S = 2.5)
1) P, R
2) P, S
3) Q, R
Answer:
1. PR
Electronegativity difference = 3.44 – 1.31 = 2.13
The electronegativity difference is greater 1.7 it shows ionic compound.
2. PS
Electronegativity difference = 3.44 – 2.58 = 0.86
The electronegativity difference is less than 1.7 it shows covalent compound.
3. QR
The electronegativity difference = 3.16 – 1.31 = 1.85
The electronegativity difference is greater than 1.7 it shows ionic compound.

Question 3.
Perform the experiments arranging the apparatus as shown in figure.
Kerala Syllabus 9th Standard Chemistry Solutions Chapter 2 Chemical Bonding 23
Record your observations and identify what type of compounds sodium chloride and sugar are
Answer:
When electricity is passed through sodium chloride solution in which carbon rod is immersed hydrogen and chlorine gas are produced. It is an ionic compound. In the second when electricity is passed through sugar solution there is no charge. Hence it belongs to covalent compound.

Kerala Syllabus 9th Standard Maths Solutions Chapter 6 Parallel Lines

You can Download Parallel Lines Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 6 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 6 Parallel Lines

Kerala Syllabus 9th Standard Maths Parallel Lines Text Book Questions and Answers

Textbook Page No. 88

Parallel Lines Class 9 Kerala Syllabus Question 1.
Draw an 8 cm long line and divide it in the ratio 2 : 3.
Answer:
Draw AB whose length is 8 cm. Draw AC with length 5 cm. Join BC. Let the point D divides AC in the ratio 2 : 3. Draw a line through D parallel to BC. This line divides AB in the ratio 2 : 3.
Parallel Lines Class 9 Kerala Syllabus

Kerala Syllabus 9th Standard Maths Chapter 6 Question 2.
Draw a rectangle of perimeter 15 cm and sides in the ratio 3 : 4
Answer:
Perimeter is 15 cm.
So length + breadth = 7.5 cm
Draw AB, a line segment of length 7.5 cm.
Draw AC from A, 4 + 3 = 7cm long.
Mark D on this line such that AD = 4 cm. Draw BC.
Draw a line parallel to BC and passing through D, which meets AB at E. Draw rectangle AEGF whose length as AE and breadth as EB.
Kerala Syllabus 9th Standard Maths Chapter 6

Class 9 Maths Parallel Lines Kerala Syllabus Question 3.
Draw rectangles specified below each of perimeter 10 cm.
i. Equilateral triangle.
ii. Sides in the ratio 3 : 4 : 5
iii. Sides in the ratio 2 : 3 : 4
Answer:
i. Draw a line AB of length 10 cm.
Draw AC with length 9 cm. Join BC.
AD = DE = EC = 3
AF : FG : GB = 3 : 3 : 3 = 1 : 1 : 1
Divide it in the ratio 1 : 1 : 1 and draw a triangle.
Class 9 Maths Parallel Lines Kerala Syllabus
Draw ΔFHG with FG as its side.

ii. Draw a line segment AB of length 10 cm. Draw AC = 12 cm, join BC. Divide it in the ratio 3 : 4 : 5.
AD : DE : EC = 3 : 4 : 5
AF : FG : GB = 3 : 4 : 5
Draw ΔGHB with sides AF, FG and GB.
Parallel Lines 9th Standard Kerala Syllabus

iii. Draw a line segment AB of length 10 cm. Draw AC with length 9 cm, join BC. Divide it in the ratio 2 : 3 : 4.
Parallel Lines Questions For Class 9 Kerala Syllabus
AD : DE : EC = 2 : 3 : 4
AF : FG : GB = 2 : 3 : 4
Draw ΔGHB with sides AF, FG and GB.

Parallel Lines 9th Standard Kerala Syllabus Question 4.
In the picture below, the diagonals of the trapezium ABCD intersect at P. Prove that PA × PD = PB × PC.
Parallel Lines Class 9 Kerala Syllabus
Answer:
Parallel Lines Chapter Class 9 Kerala Syllabus
Sides AB, CD are parallel.
Draw EF parallel to AB and passing through P.
Now lines AB, EF and DC are parallel.
These lines cut the lines AC and BD in the same ratio.
So, PA : PC = PB : PD
i.e., \(\frac {PA}{PC} = \frac {PB}{PD}\)
From this by cross multiplication we get PA × PD = PC × PB

Textbook Page No. 93

Parallel Lines Class 9 Kerala Syllabus Question 1.
In the picture, the perpendicular is drawn from the midpoint of the hypotenuse of a right triangle to the base. Calculate the length of the third side of the large right triangle and the lengths of all three sides of small right triangle.
9th Standard Maths Notes Kerala Syllabus
Answer:
AC = 10 cm
∴ AM = 5 cm
Hsslive Guru 9th Maths Kerala Syllabus
MN, CB are perpendicular to AB
MN, CB are parallel lines.
∴ MN divides AB and AC in the same ratio.
∴ AN = BN = 4cm
BC = \(\sqrt{10^{2} – 8^{2}} = \sqrt {100 – 64} = \sqrt {36}\) = 6 cm
M is the midpoint of the AC. N is the midpoint of AB.
The length of the line joining the midpoints of two sides of a triangle is half the length of the third side. So MN is half of BC.
∴ MN = 3cm

Parallel Lines Chapter Class 9 Kerala Syllabus Question 2.
Draw a right triangle and the perpendicular from the midpoint of the hypotenuse to the base.
9th Standard Maths Kerala Syllabus
i. Prove that this perpendicular is half the perpendicular side of the large triangle.
ii. Prove that perpendicular bisects the bottom side of the larger triangle.
iii. Prove that in the large triangle the distances from the mid point of the hypotenuse to all the vertices are equal.
iv. Prove that the circumcentre of a right triangle is the mid point of its hypotenuse.
Answer:
MN is perpendicular to AB.
MN and CB are parallel lines. MN divides AC and AB in the same ratio.
Class 9 Maths Chapter 6 Kerala Syllabus

i. AM = \(\frac{1}{2}\)AC, AN = \(\frac{1}{2}\)AB
∴ MN = \(\frac{1}{2}\)CB

ii. We get two right triangles of same base and perpendicular. So their hypotenuses are also equal.
∴ MA = MC = MB
AN = NB

iii. ∠ANM = ∠BNM = 90°
MN = MN
ΔANM ≅ ΔBNM
AM = MB …….(1)
M is the midpoint of AC
AM = MC …….(2)
(1) = (2) ⇒ AM = MB = MC
Chapter 6 Maths Class 9 Kerala Syllabus

iv. The points A, B and C are at same distance from M. So a circle can be drawn with M as centre and pass through there three points. So this circle is the circum circle of ΔABC.

9th Standard Maths Notes Kerala Syllabus Question 3.
In the parallelogram ABCD, the line drawn through a point P on AB, parallel to BC, meets AC at Q. The line through Q, parallel to AB meets AD at R.
Hsslive Guru Maths 9th Kerala Syllabus
Prove that \(\frac {AP}{PB} = \frac {AR}{RD}\)
Answer:
In ΔABC, RQ the parallel line of BC, divides AB and BC in the same ratio When we consider the relation
\( \frac {AR}{RD} = \frac {AQ}{QC}, \frac {AP}{PB} = \frac {AQ}{AC} = \frac {AR}{RD}\)
Hsslive Guru 9 Maths Kerala Syllabus
\( \frac {AR}{RD} = \frac {AQ}{QC}\)
\( \frac {AP}{PB} = \frac {AQ}{AC} = \frac {AR}{RD}\)
∴ \( \frac {AP}{PB} = \frac {AR}{RD}\)

Hsslive Guru 9th Maths Kerala Syllabus Question 4.
In the picture below, two vertices of a parallelogram are joined to the mid points of two sides. Prove that these lines divide the diagonal in the picture into three equal parts.
Hss Live Guru Maths Kerala Syllabus
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 6 Parallel Lines 16
In the parallelogram ABCD, AB = CD and AB parallel CD.
P is the mid point of AB and Q is the midpoint of DC.
Since P and Q are the midpoints of AB and CD respectively.
PB = ½AB
DQ = ½CD = ½AB
∴ PB = DQ
i.e., PB = DQ and PB parallel DQ.
∴ PBQD is a parallelogram.
Since the parallel lines PD and BQ cut AB into equal parts, they cut AY also into equal parts.
AX = XY
The parallel lines PD and BQ cut CD into equal parts. So they cut CX also into equal parts.
YC = XY
∴ AX = XY = YC

9th Standard Maths Kerala Syllabus Question 5.
Prove that the quadrilateral formed by joining the mid points of a quadrilateral is a parallelogram. What if the original quadrilateral is a rectangle? What if it is a rhombus?
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 6 Parallel Lines 17
P, Q, R and S are the midpoints of the sides of the quadrilateral ABCD. Draw the diagonals AC and BD.
The length of the line joining the mid-points of two sides of a trianglle is half the length of the third side.
In ΔABC, PQ = ½AC
In ΔADC, SR = ½AC
∴ PQ = SR
In ΔABD, PS = ½BD
In ΔBCD, QR = ½BD
∴ PS = QR
In quadrilateral PQRS, both pairs of opposite sides are equal. So it is a parallelogram.
PQ = ½AC, RS = ½AC
∴ RS = PQ = ½AC
PS = ½BD, QR = ½BD
∴ PS = QR = ½BD = ½AC
∴ PQ = RS = PS = QR
All sides of the quadrilateral PQRS are equal. So it is a rhombus.
Since ABCD is a rhombus. AC perpendicular to BD.
PQ and SP are perpendicular, i.e., ∠SPQ = 90°
One angle of the parallelogram PQRS is 90° and therefore it is a rectangle.

Kerala Syllabus 9th Standard Maths Parallel Lines Exam Oriented Text Book Questions and Answers

Class 9 Maths Chapter 6 Kerala Syllabus Question 1.
Draw ΔABC with AB = 5cm, BC = 4cm, AC = 6cm, ΔABC. In BC mark P such that BP = 1cm, PC = 3cm. What is the relation between the areas of ΔABP and ΔAPC?
Answer:
Area of ΔAPC is 3 times the area of ΔABP.
Kerala Syllabus 9th Standard Maths Solutions Chapter 6 Parallel Lines 18
If two triangles have the common vertex and their bases are along the same straight line, the ratio between their areas is equal to the ratio between the lengths of their bases.
⇒ \(\frac{Area of ΔABP}{Area of ΔAPC} = \frac{BP}{PC} = \frac{1}{3}\)

Chapter 6 Maths Class 9 Kerala Syllabus Question 2.
In ΔABC, the line parallel to BC meet AB and AC at D and E respectively. AE = 4.5cm, \(\frac{AD}{DB} = \frac{2}{5}\) = then find EC.
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 6 Parallel Lines 19
\(\frac{AD}{DB} = \frac{AE}{EC}\)
\(\frac{2}{5} = \frac{4.5}{EC}\) = EC = \(\frac{4.5 \times 5}{2} = \frac{22.5}{2}\)
= 11.25 cm

Hsslive Guru Maths 9th Kerala Syllabus Question 3.
In ΔPQR, PQ = 9cm, QR = 14 cm, PR = 12 cm The bisector of ∠PRS meets QR in S. Find QS and RS.
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 6 Parallel Lines 20
\(\frac{PQ}{PR} = \frac{QS}{SR}\);
\(\frac{18}{24} = \frac{6}{RS}\)
= RS = \(\frac{24 \times 6}{18}\) = 8 cm

Hsslive Guru 9 Maths Kerala Syllabus Question 4.
In ΔABC, the point P is on AB such that the length of AP is double the length of PB. The line through P parallel to BC meets AC at Q and the length of AQ is 1 cm more than that of QC, what is the length of AC?
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 6 Parallel Lines 21
AP = 2 × PB
AP : PB = 2 : 1
\(\frac{AP}{PB} = \frac{AQ}{QC} = \frac{2}{1}\) ⇒
AQ = 2QC
Consider QC = x cm
x + 1 = 2x ⇒ x = 1 ⇒ QC = 1
AQ = 2 × 1 = 2
AC = AQ + QC = 2 + 1 = 3 cm

Hss Live Guru Maths Kerala Syllabus Question 5.
In ΔABC, a line parallel to BC cuts a B and AC at X and Y
Kerala Syllabus 9th Standard Maths Solutions Chapter 6 Parallel Lines 22
(a) If AB = 3.6cm, AC = 2.4cm, AX = 2.1 cm, what is the length of AY?
(b) If AB = 2cm, AC = 1.5cm, AY = 0.9cm
what is the length of BX?
Answer:
a. \(\frac{AB}{AX} = \frac{AC}{AY} ⇒ \frac{3.6}{2.1} = \frac{2.4}{AY}\);
⇒ 3.6 × AY = 2.4 × 2.1 ⇒ AY = 1.4 cm

b. Let length of AX = xcm
\(\frac{AX}{AB} = \frac{AY}{AC} ⇒ \frac{x}{2} = \frac{0.9}{1.5}\);
1.5 × x = 1.8 ⇒ x = 1.2;
BX = AB – AX; BX = 2 – 1.2 = 0.8 cm

Hss Live Guru Maths 9 Kerala Syllabus Question 6.
In the figure, ABCD is a trapezium, and AB || CD. P is the midpoint of AD. If PQ || AB, then prove that Q is the midpoint of BC.
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 6 Parallel Lines 23
Given PQ || AB,
therefore
PQ || DC also. Line segments AD and BC are dividing 3 parallel lines EC, PQ and AB. Given P is the midpoint of AD
⇒ DP = PA ∴ \( \frac{DP}{PA}\) = 1, ∴ \( \frac{CQ}{QB}\) = 1
∴ CQ = QB ∴ Q is the midpoint of BC.

Ch 6 Maths Class 9 Kerala Syllabus Question 7.
In the figure AB || EH || DC, Answer the following questions given below.
Kerala Syllabus 9th Standard Maths Solutions Chapter 6 Parallel Lines 24
a. If DE = 3cm, EA = 4cm and DG = 5cm, find GB?
b. If BH = 6cm,CB = 16cm and DB = 20cm, find DG?
c. If DA = 36cm, EA = 22cm and DG = 15cm find DB?
d. If DE = 20cm, EA = 25cm and HB = 22cm, find CH?
Answer:
a. In ΔDAB, AB || EG
∴ \(\frac{DE}{EA} = \frac{DG}{GB}; \frac{3}{4} = \frac{5}{GB} \Rightarrow GB = \frac{20}{3} = 6 \frac{2}{3}\)cm

b. In ΔBDC, DC || GH
∴ \(\frac{BH}{BC} = \frac{BG}{BD}; \frac{6}{16} = \frac{BG}{20}\);
BG = \(\frac{120}{16} = 7\frac{1}{2}\)cm
DG = BD – BG = 20 – \( 7\frac{1}{2}\) = \( 12\frac{1}{2}\)cm

c. In ΔABD, AB || EG
\(\frac{DA}{EA} = \frac{DB}{BG}; \frac{36}{22} = \frac{DB}{BG}\);
\(\frac{36}{22} \frac{DB}{DB – DG}\)
⇒ 36D(DB – DG) = DB × 22
14 DB = 540 DB = \(\frac{540}{14}\) = 38\(\frac{4}{7}\)cm

d. AB, EH and DC are 3 parallel lines.
\(\frac{DE}{EA} = \frac{CH}{HB}; \frac{20}{25} = \frac{CH}{22}\) ⇒ CH = \(\frac{88}{5}\) = 17\(\frac{3}{5}\)cm

Hss Live Guru Class 9 Maths Kerala Syllabus Question 8.
In ABC, a line parallel to BC cuts AB and AC at P and Q. Show that \(\frac{AP}{AB} = \frac{AQ}{AC}\)
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 6 Parallel Lines 25
\(\frac{AB}{PB} = \frac{AQ}{QC}\) (BC || PQ)
\(\frac{AP}{PB}+1=\frac{AQ}{QC}+1; \frac{AP+PB}{PB} = \frac{AQ + QC}{QC}\)
\(\frac{AB}{PB} = \frac{AQ}{QC}\);
∴ \(\frac{AP}{AB} = \frac{AQ}{AC}\)

Question 9.
In the figure below, ABC is a right angled triangle. AB = 10cm, AC = 6cm, BC = 8cm, The midpoint of AB is M. Compute the lengths of the sides of ΔMBN
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 6 Parallel Lines 26
(MN || AC)
∴ \(\frac{AM}{MB} = \frac{NC}{NB}\)
1 = \(\frac{NC}{NB}\) ∴ CN = NB = 4
∴ MB = 5, NB = 4, MN = 3

Question 10.
In the figure ABC is a right angled triangle and M is the midpoint of AB. Prove that MN = V2AC and also prove that MN = MA = MB.
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 6 Parallel Lines 27
MN = \(\sqrt{x^{2} – y^{2}}\)
AC = \(\sqrt{4x^{2} – 4y^{2}} = 2\sqrt {x^{2} – y^{2}}\)
AC = 2 × MN or MN = \(\frac{1}{2}\)AC
MC = \(\sqrt{x^{2} – 4^{2} + 4^{2}}\); MC = X
∴ MC = MA = MB

Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures

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Kerala State Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures

Circle Measures Textual Questions and Answers

Textbook Page No. 131

Circle Measures Class 9 Kerala Syllabus Question 1.
Prove that the circumcentre of an equilateral triangle is the same as its centroid.
i. Calculate the length of a side of an equilateral triangle with vertices on a circle of diameter 1 centimetre.
ii. Calculate the perimeter of such a triangle.
Answer:
Perpendicular bisector of sides of a triangle that can meet at a point is its circumcenter
Circle Measures Class 9 Kerala Syllabus
Since the triangle is equilateral the perpendicular bisectors of the sides are also median. Since the triangle is equilateral the perpendicular bisectors of the sides are also centroid. That is circumcentre of an equilateral triangle is the same as its centroid.
i. Length of one side of an equilateral triangle is.
Circle Measures Class 9 Scert Chapter 9 Kerala Syllabus

Circle Measures Class 9 Scert Chapter 9 Kerala Syllabus Question 2.
Calculate the perimeter of a square with vertices on a circle of diameter 1 centimetre.
Answer:
In the square ABCD
AO = 1/2 cm
In A AEO, ∠EAO = 45°
∠AEO = 90°
Since AO = 1/2 cm
AE = \(\frac{1}{2 \sqrt{2}} \mathrm{cm}\)
(Since Δ AEO is a isosceles right triangle, hypotenuse is √2 times of a perpendicular side.)
Circle Measures Class 9 Chapter 9 Kerala Syllabus
Perimeter = \(=4 \times \frac{\sqrt{2}}{2}=2 \sqrt{2} \mathrm{cm}\)

Circle Measures Class 9 Chapter 9 Kerala Syllabus Question 3.
Calculate the perimeter of a regu¬lar hexagon with vertices on a circle of diameter 1 centimetre.
Answer:
If we draw diagonals through centre of circle inside the regular hexagon it divides the regular hexagon into 6 equilateral triangles.
Let diameter be 1 cm
OA = 1/2 cm
OA = OB = AB.
therefore
One side = 1/2 cm
Perimeter of regular hexagon = 6 × 1/2 = 3 cm
Hss Live Guru 9th Maths Chapter 9 Kerala Syllabus

Textbook Page No. 134

Hss Live Guru 9th Maths Chapter 9 Kerala Syllabus Question 1.
The perimeter of a regular hexagon with vertices on a circle is 24 centimetres.
i. What is the perimeter of a square with vertices on this circle?
ii What is the perimeter of a square with vertices on a circle of double the diameter?
iii. What is the perimeter of an equilateral triangle with vertices on a circle of diameter half that of the first circle?
Answer:
Perimeter of a regular hexagon = 24 cm
Length of one side of a regular hexagon = 24/6 = 4 cm
Length of one side of a regular hexagon is equal to the radius of the circle.
i. Diagonal of a square = 8 cm Diagonal of a square is equal to the diameter of the circle.
Let a be the side of the square
a2 + a2 = 82
2a2 = 64
a2 = 64/2 = 32
a= 4√2
∴ One side of a square = 4√2 cm
Perimeter of a square = 4 × 4√2 cm = 16√2 cm
Circles Class 9 State Syllabus Chapter 9 Kerala Syllabus

ii. The perimeter of a square with verti¬ces on a circle of double the diameter 2 × 16√2 = 32 √2 cm
(The perimeters of circles are scaled by the same factor as their diameters.)

iii. One side of an equilateral triangle with vertices on a circle of half the diameter of the first circle
\(=2 \sqrt{2^{2}-1^{2}}=2 \sqrt{3} \mathrm{cm}\)
Perimeter = 3 × 2√3 = 6√3 cm

Circles Class 9 State Syllabus Chapter 9 Kerala Syllabus Question 2.
A wire was bent into a circle of diameter 4 centimetres. What would be the diameter of a circle made by bending a wire of half the length?
Answer:
The ratio between the perimeters are equal to the ratio between their diameters. The perimeter of the first circle is twice the perimeter of the second circle.
Therefore diameter of the second circle is half of the diameter of the first circle. Diameter of the second circle.
= 4/2 = 2 cm
Kerala Syllabus 9th Standard Maths Solutions

Kerala Syllabus 9th Standard Maths Solutions  Question 3.
The perimeter of a circle of diameter 2 metres was measured and found to be about 6.28 metres. How do we compute the perimeter of a circle of diameter 3 metres, without measuring?
Answer:
If diameter is 2 meters, perimeter is 6.28 meter.
If the diameter is 1 metre, perimeter is 6.28/2 meter.
If the diameter is 3 metres, perimeter = \(\frac{6.28}{2} \times 3=9.42 m\)

Textbook Page No. 137

Hss Live Guru Maths 9 Chapter 9 Kerala Syllabus Question 1.
In the pictures below, a regular hexagon, square and a rectangle are drawn with their vertices on a circle. Calculate the perimeter of each circle.
Hss Live Guru Maths 9 Chapter 9 Kerala Syllabus
Answer:
a. AB = 2 cm
In the figure triangle are
equilateral triangles,
therefore
radius OA = 2 cm
Perimeter of circle = 2 πr
= 2 × π × 2 = 4π cm
Kerala Syllabus 9th Standard Maths Notes

b. ABCD is a square
AB = BC = 2 cm, ∠5 = 90°
AC = \(\sqrt{2^{2}+2^{2}}=\sqrt{8}=2 \sqrt{2}\)
Circles Class 9 Kerala Syllabus Chapter 9
Radius of circle = 1/2 × 2√2
= √2 cm
Perimeter of circle = 2π × √2 cm
= 2√2 π cm

c. PR = \(\sqrt{2^{2}+(1.5)^{2}}\)
= \(\sqrt{6.25}=2.5 \mathrm{cm}\)
Radius of circle = 1/2 × 2.5 = 1.25 cm
Perimeter of circle = 2 × π × 1.25
= 2.5 π cm

Kerala Syllabus 9th Standard Maths Notes Question 2.
An isosceles triangle with its vertices on a circle is shown in this picture.
Hss Live Class 9 Maths Chapter 9 Kerala Syllabus
What is the perimeter of the circle
Answer:
Consider the centre of circle as O and triangle as ABC
OC = Radius of circle = r
OD = 4 – r
AD = 2 cm
AO = r
Hss Live Guru 9 Maths Chapter 9 Kerala Syllabus
Therefore, in triangle AOD
(AO)2 = (AD)2 + (OD)2
r2=22 + (4 – r)2
r2= 4 +16 – 8r + r2
8r=20;
r = 20/8 = 5/2 = 2.5 cm
∴ Perimeter = 2π × r = 2π × 2.5 cm
= 5π cm

Circles Class 9 Kerala Syllabus Chapter 9 Question 3.
In all the pictures below, the centres of the circles are on the same line. In the first two pictures, the small circles are of the same diameter.
9th Std Kerala Syllabus Maths Solutions
Prove that in all pictures, the perimeters of the large circle is the sum of the perimeters of the small circles.
Answer:
a. Smaller circles have same diameters. Consider the diameter as d, perimeter of smaller circle
= π × diameter = π d
Perimeter of two small circles = 2πd
Hsslive Guru 9th Maths Chapter 9 Kerala Syllabus
Diameter of the large circle = d + d = 2d
Perimeter of the large circle
= π × 2d = 2πd
Therefore perimeters of the large circle is the sum of the perimeters of the small circles.

b. Let d be the diameter of one small circle, then perimeter = π d
Sum of perimeter of three small circles =3 π d
Diameter of the large circle = d + d + d = 3d
Perimeter of the large circle = π × 3d = 3 π d
Therefore perimeters of the large circle is the sum of the perimeters of the small circles.

c. In figure diameter of three circless are different, let consider the diameters of small circles are p, q and r.
Hsslive 9th Maths Chapter 9 Kerala Syllabus
Perimeter of first small circle = πp
Perimeter of second small circle = πq
Perimeter of third small circle = πr
Sum of perimeters of three small circles
= πp + πq + πr = π (p + q + r)
Diameter of large circle = p + q + r
Perimeter of large circle = π (p + q + r)
Therefore also here the perimeters of the large circle is the sum of the perimeters of the small circles.

Hss Live Class 9 Maths Chapter 9 Kerala Syllabus Question 4.
In this picture, the circles have the same centre and the line drawn is a diameter of the large circle. How much more is the perimeter of the large circle than the perimeter of the small circle
Hsslive Guru 9 Maths Chapter 9 Kerala Syllabus
Answer:
If ‘r’ is the radius of the small circle,
radius of the large circle = r + 1
Perimeter of the small circle = 2 π r
Perimeter of the large circle = 2 π (r + 1) = 2 π r + 2 π
i.e., perimeter of the large circle is 2 π units more than the perimeter of the small circle.

Textbook Page No. 141

Hss Live Guru 9 Maths Chapter 9 Kerala Syllabus Question 1.
In the pictures below, find the difference between the areas of the circle and the polygon, up to two decimal places.
Hss Guru 9 Maths Chapter 9 Kerala Syllabus
Answer:
i. Radius of the small circle = 2cm
Area = π × 22 = 3.14 × 4
= 12.56 cm2
Daigonal of the square = 4 cm
One side of the square = 4/√2 cm
Area of the square = \(\frac{4}{\sqrt{2}} \times \frac{4}{\sqrt{2}}=\frac{16}{2}=8 \mathrm{cm}\)
Differences between the areas = 12.56 – 8 = 4.56 cm

ii. Radius of the circle = 2 cm
Area= π × 22 = 3.14 × 4 = 12.56 cm2
One side of the regular hexagon=2 cm
Area of the regular hexagon = \(6 \times \frac{\sqrt{3} \times 2^{2}}{4}=6 \sqrt{3}\)
= 6 × 1.73 = 10.38 cm2
Differences between the areas = 12.56 – 10.38 = 2.18 cm2

9th Std Kerala Syllabus Maths Solutions Question 2.
The pictures below show circles through the vertices of a square and a rectangle
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 17
Calculate the areas of the circles
Answer:
i. One side of a square is 3 cm, therefore its diagonal is 3√2 cm
Diameter of the circle = 3√2 cm
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 18

ii. Rectangle inside the circle having length 4 cm and breadth 2 cm.
Diagonal = \(\sqrt{4^{2}+2^{2}}=\sqrt{16+4}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 19
Diameter of the circle = √20 cm
Radius = \(\frac{\sqrt{20}}{2} \mathrm{cm}\)
Area = \(\pi \times\left(\frac{\sqrt{20}}{2}\right)^{2}=\pi \times \frac{20}{4}\)
= 5 π cm2

Hsslive Guru 9th Maths Chapter 9 Kerala Syllabus Question 3.
Draw a square and draw circles cen¬tered on the corners, of radius half the side of the square. Draw another square formed by four of the first square and a circle just fitting into it.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 20
Prove that the area of the large circle is equal to the sum of the areas of the four small circles
Answer:
In the figure length of one side of the square is 2r
Radius of one small circle = r
Perimeter of one small circle = π r2
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 21
Radius of four small circles = 4 π r2
One side of a square in the second figure = 2r + 2r = 4r
Radius of the circle in the figure
= 4r/2 = 2r
perimeter of the circles in the figure = π × (2r)2
= π × 2r × 2r = 4πr2
The area of the large circle is equal to the sum of the areas of the four small circles

Hsslive 9th Maths Chapter 9 Kerala Syllabus Question 4.
In the two pictures below, the squares are of the same size.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 22
Prove that the green regions are of the same area.
Answer:
Let ‘a’ be the side of the square in the picture.
Area of the square = a2
If the four sectors in the vertices are joined together, a circle is formed because the radius of each sector is a/2.
Area of the shaded part = Area of the square – area of the circle.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 23

ii. In the second picture diameter of the
circle = a Radius = a/2
Area of the shaded part = Area of the square – area of the circle.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 24
i.e., Areas of the shaded portions are equal.

Hsslive Guru 9 Maths Chapter 9 Kerala Syllabus Question 5.
Parts of circles are drawn inside a square as shown in the picture below. Prove that the area of the blue region is half the area of the square.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 25
Answer:
Let ‘a’ be the side of the square
Area of the blue part in the half portion of the square is equal to half of the area of the circle having diameter ‘a’.
Area of the blue part in the half portion of the square \(=\frac{1}{2} \times \pi\left(\frac{a}{2}\right)^{2}=\frac{\pi a^{2}}{8}\)
We must subtract area of two circles
having diameter a/2 from the half the
area of the square to get the area of remaining blue shaded part.
Area of remaining blue shaded part.
= \(\frac{a^{2}}{2}-2 \times \frac{1}{4} \times \pi\left(\frac{a}{2}\right)^{2}\)
Area of blue shaded part.
= \(\frac{\pi a^{2}}{8}+\frac{a^{2}}{2}-\frac{\pi a^{2}}{8}=\frac{a^{2}}{2}\)
i.e., area of the blue part is equal to half the area of the square.

Hss Guru 9 Maths Chapter 9 Kerala Syllabus Question 6.
In the figure, semicircles are drawn with the sides of a right triangle as diameter.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 26
Prove that the area of the largest semicircle is the sum of the areas of the smaller ones.
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 27
In ΔABC, ∠5 = 90°
According to Pythagoras principle,
AB2 + BC2 = AC2 ………. (1)
Radius of the semicircle with diameter
AB = AB/2
Area of the semicircle with diameter AB
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 28
= \(\frac{1}{2} \times \pi \times \frac{A C^{2}}{4}=\frac{\pi}{8} A C^{2}\)
Sum of the areas of the smaller semicircles = \(\frac{\pi}{8} A B^{2}+\frac{\pi}{8} B C^{2}=\frac{\pi}{8}\left(A B^{2}+B C^{2}\right)\)
= π/8 AC2 = Area of the largest semicircle

Textbook Page No. 148

Hsslive Maths Class 9 Chapter 9 Kerala Syllabus Question 1.
In a circle, the length of an arc of central angle 40° is 3 π centimetres. What is the perimeter of the circle? What is its radius?
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 29

Question 2.
In a circle, the length of an arc of central angle 25° is 4 centimetres.
i. In the same circle, what is the length of an arc of central angle 75°?
ii. In a circle of radius one and a half times the radius of this circle, what is the length of an arc of central angle 75°?
Answer:
i. Length of the arc having central angle 25° = 4 cm
Three times of 25 is 75.
Length of the arc having central angle 75° = 4 x 3 = 12 cm

ii. Length of the arc having central angle 75° and radius r = 12 cm
Length of the arc having central angle 75° and radius 1 1/2 r
= \(12 \times 1 \frac{1}{2}=18 \mathrm{cm}\)

Question 3.
From a bangle of radius 3 centimetres, a piece is to be cut out to make a ring of radius ^ centimetres.
i. What should be the central angle of the piece to be cut out?
ii. The remaining part of the bangle was bent to make a smaller bangle. What is its radius?
Answer:
Perimeter of the bangle having radius 3 cm = 6 π cm.
Perimeter of the bangle having radius 1/2 cm = π cm.
π is the 1/6 part of 6 π.
Therefore the central angle of the piece to be cut out = 360 × 1/6 = 60°
ii. Length of the remaining part of the bangle = 6π – π = 5π cm
Radius of the other small bangle = 5 π / 2π = 2.5 cm

Question 4.
The picture shows the parts of a circle centred at each vertex of an equilateral triangle and passing through the other two vertices.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 30
What is the perimeter of this figure?
Answer:
Since the triangle is equilateral, each angle is 60°. There are in each side is in a circle of radius 4 cm and the central angle is 60°.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 31

Question 5.
Parts of circles are drawn, centred at each vertex of a regular octagon and a figure is cut out as show below:
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 32
Calculate the perimeter of the fig¬ure.
Answer:
Sum of the angles in an octagon
= (n-2) × 18o° = 6x 180°= 1080° One angle of the regular octagon
= 1080 – 8 = 135°
One side of the regular octagon = 2 cm. Radius if the sectors having centre is each vertices of the circle= 1 cm The second picture shows the cut-down form of 8 sectors having centre angle 135° and radius 1 cm.
The perimeter is found by calculating the length of 8 arc having radius 1 cm and central angle 135°.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 33

Textbook Page No. 151

Question 1.
What is the area of a sector of central angle 120° in a circle of radius 3 centimetres? What is the area of a sector of the same central angle in a circle of radius 6 centimetres?
Answer:
Area of the sector in the circle having radius 3 cm and angle of center 120°
π × 32 × \(\frac { 120 }{ 360 }\) = 3π cm2
Area of the sector in the circle having radius 6 cm and angle of center 120°
= π × 62 × \(\frac { 120 }{ 360 }\) = 12π cm2

Question 2.
Calculate the area of the green coloured part of this picture.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 34
Answer:
In the picture area of the shaded part is the difference between the area of the two sectors.
Area of the large sector
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 35
Area of the shaded part = 9.42 – 4.19 = 5.23 cm2

Question 3.
Centred at each corner of a regular hexagon, a part of a circle is drawn and a figure is cut out as shown below:
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 336
What is the area of this figure?
Answer:
The area of the cut-down portion = Area of the regular hexagon – Area of 6 sectors Area of the regular hexagon
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 36
Area of the sector = 120°/360° part of area of the circle
(One angle of a regular hexagon is 120°)
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 37
Area of the 6 sectors = \(\frac{6 \times \pi}{3}=2 \pi \mathrm{cm}^{2}\)
Area of cut down portion = 6 √3 – 2 π cm2

Question 4.
The picture below shows two circles, each passing through the centre of the other:
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 38
Calculate the area of the region common to both.
Answer:
Consider the picture given below, we can divide the circle into two part by using the line AB, the area of the two portions are same.
That is we get the total area by multi¬plying area of the sector by 2.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 39
We can find out the area of part above the line AB .
Given AB = 2cm
Circles having equal
radius. So,
AC = BC = 2 cm.
AABC is an equilateral triangle so angles are 60° each.
Add the area of sectors having centre A and B .
The area of ΔABC include twice, so we will subtract it once.
Area of sectors having centre A
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 40
Area of sectors having centre B
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 41
Area of ΔABC
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 42
Area of the part above the line AB = 2.09 + 2.09 – 1.7 = 2.45 cm2
The area of the region common to both = 2 × 2.45 = 4.90 cm2

Question 5.
The figure shows three circles drawn with their centres on each vertex of an equilateral triangle and passing through the other two ver¬tices;
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 43
Find the area of the region common to all three.
Answer:
The area of the common part = Area of three sectors – 2 × area of an equilateral triangle having side 2 cm
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 44
= 2 × 3.14 – 2 × 1.73
= 6.28 – 3.46
= 2.82 cm2
The area of the region common to all three = 2.82 cm2

Circle Measures Exam Oriented Questions And Answers

Question 1.
In the picture PQRS is a square of side 10 cm. A, B, C and D are midpoints of the sides of the square
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 45
Semi circles are drawn inside the square.
a. Compute the area of the square.
b. Compute the area of the semi-circle.
c. Compute the area of the shaded part.
Answer:
a. Since one side of the square is 10 cm,
Area = side x side = 10 × 10= 100 cm2

b. The diameter of one semicircle = half of the side of the square
Diameter = 5 cm
∴ Radius = 5/2 cm
Since the four semicircles are equal. Area of the 4 semicircles
= Area of 2 circles = 2 × πr2
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 46

c. Area of the shaded part = Area of the square – Area of four semicircles.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 47

Question 2.
If a circular shaped dining table has an area of 31400 cm2, find its radius. What will be its perimeter ?
Answer:
Area of the table = 31400 cm2
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 48

Question 3.
In the given picture shown two semicircular shaped iron bar can be cut down from a rectangular shaped iron bar. Calculate the area of the remaining shaded portion.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 49
Answer:
Area of the rectangle = 24 × 14 = 336 cm2
Area of two semicircle = Area of a complete circle
Diameter of the circle = 14 cm
Radius = 7 cm
Area of the circle = π r2
= π × 72 =49
π = 153.86 cm2 Area of the remaining portion
= 336 – 153.86= 182.14 cm2

Question 4.
In the figure A, B, C and Dare the points on the square which touches the circle. If the radius of the circle is 6.
a. What is the length of one side of the square?
b. Find the area of the circle.
c. What will be the area of the shaded portion?
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 50
Answer:
a.Diameter of the circle = 6 × 2 = 12 cm
Side of the square = 12 cm
b. Area of the circle = n r2
= n × 6 × 6 = 36n =36 × 3.14 =113.04 cm2
c. Area of the square = 12 × 12 = 144 cm2
Area of the shaded portion = 144 – 113.04 = 30.96 cm2

Question 5.
In the ACB is the arc drawn by taking O as the centre and OA as the radius. Then find the area of the shaded region.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 51
Answer:
The area of sector which has 7 cm radius and 90° central angle =
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 52
Area of the right angled triangle AOB
\(=\frac{7 \times 7}{2}=24.5\)
Area of the shaded part = 38.455 – 24.5 = 13.965 cm2

Question 6.
In the pictures given below, find the area of the shaded part?
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 53
Answer:
’The radius of the sector in the picture (1) is 5cm and its central angle is 40°.
Area of the shaded part \(=\pi \times 5^{2} \times \frac{40}{360}=\frac{25}{9} \pi \mathrm{cm}^{2}\)
The radius of the sector in the picture (2) is 6cm and its central angle is 300° (360 – 60).
Area = \(\pi \times 6^{2} \times \frac{300}{360}=36 \pi \times \frac{5}{6}\)
= 30 π cm2

Question 7.
What amount of reeper is needed to enclose a circular shaped dining table of area 6.28 cm2?
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 54

Question 8.
A wheel which has 20 cm radius is rotating forward. After 10 rotations what distance will the wheel travel forward?
Answer:
Radius of the wheel = 20 cm
When the wheel rotates once it will travel the distance same as its area
Perimeter of the wheel = 2 × π × radius = 2 × 3.14 × 20 = 125.64cm
The distance travelled forward when the wheel rotates once = 125.64 cm
The distance travelled forward when the wheel rotates 10 times
= 125.64 × 10 = 1256.4 cm

Question 9.
In the picture the central angles of both the sectors are equal. Sum of the radii of the sectors is 18 cm.
Area of the shaded part is 18 π cm2. Find the radii of the sector.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 55
Answer:
‘Let ‘r’ be the radius of the small sector and ‘R’ be the radius of the large sector.
R + r = 18 …………. (1)
Area of the shaded portion = Area of the large sector – area of the small sector
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 56

Question 10.
In the figure O is the radius of the circle and OABC is a rectangle. OA = 8 cm, OC = 15 cm. Hence find the Area of the shaded portion
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 57
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 58
Area of the shaded portion
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 59

Question 11.
The wheel of a vehicle has a diameter 60 cm. For this vehicle travels a distance of 200 m, how many times must this wheel rotates.
Answer:
‘The distance travelled when the wheel is rotated once =2 π r = 2 × π × 30
= 188.4 cm = 1.884 m.
The time required for the wheel to travel a distance of 200m = 200/1.884 = 106.16 = 106

Question 12.
In the below-given figures, there are two circles with the same centre. Then find the area of the second region.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 60
Answer:
a. Area of the shaded portion =Area of the outer circle – Area of the inner circle
= π R2 – π F= π × 102 – π × 82
= 100π – 64π = 36π
=36 × 3.14 = 113.04 cm2

b. Outer radius = 7 + 2 = 9 cm Inner radius = 7 cm
Area of the shaded portion = π × 92- π × 72 = 81π – 49π = 32π = 32 × 3.14 =100.48 cm2

c. Outerradius = 10.5
Inner radius = 10.5 – 1 = 9.5
Area of the shaded portion = π × (10.5)2 – π × (9.5)2
= π × (10.52 – 9.5)2
= π × (10.52 – 9.52)
= π × (10.5 + 9.5) (10.5 – 9.5) = π × 20 × 1 = 62.8 cm2

d. Outer radius = \(\frac { 16π }{ 2π }\) = 8
Inner radius = \(\frac { 14π }{ 2π }\) = 7
Area of the shaded portion
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 61

Question 13.
Two semicircular pieces are cut out from a rectangular sheet. Find the area of the remaining portion.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 62
Answer:
Area, of the rectangle = 50 × 20 = 1000 cm2 If the two semicircles cut out are joined it becomes a circle
Its radius = 20/2 = 10 cm
Area of the portion cut out
= πr2 = π × 10 × 10 = 3.14 × 10 × 10 = 314 cm2
Area of the remaining portion
= 1000 – 314 = 686 cm2

Question 14.
If a square, equilateral triangle, regular hexagon and circle have the same perimeter. Which of these has the largest area ?
Answer:
If a square, equilateral triangle, regular hexagon and circle has a perimeter of 12 cm.
Equilateral triangle
One side = 12/3 = 4
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 63
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 64
Circle has the largest area.

Question 15.
In the figure, if ACB is the arc of circle having O as the centre and OA as the O, radius. Then find the area of the shaded portion
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 65
Answer:
Radius of the sector = 8cm Central angle = 90°
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 66
OBA is a right angled triangle
∴ Area of ΔOBA = 1/2 × 8 × 8 = 32 cm2
Area of shaded portion = 50.24 – 32 = 18.32 cm2

Question 16.
The area of an equilateral triangle is 17300 cm2. Draw circles with radius half the length of one side of the triangle and vertices as the centre of the circle. Calculate the area of the shaded portion.
Answer:
Area of the equilateral triangle
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 67
One side = 200 cm,
Radius of the circle = 100 cm
Central angle of each sector = 60°
Area of one sector = \(\frac{\pi r^{2} \times 60^{\circ}}{360}\)
Area of the three sectors
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 68

Question 17.
In the figure, a side of the regular hexagon has length 20 cm. Find the area of the shaded portion.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 69
Answer:
We have learnt that length of a side of the regular hexagon drawn inside a circle will be equal to the radius of the circle.
So the radius of the circle will be 20 cm.
Area of the circle = πr2 = π × 20 × 20 = 400π = 1256 cm2
Area of the regular hexagon \(=\frac{6 \times \sqrt{3} \times a^{2}}{4}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 70
Area of shaded portion = 1256 – 1039.2 = 216.8 cm2

Question 18.
If radius of a sector is 7 cm and its perimeter is 2.5 cm. Then find its area.
Answer:
Area of sector
= 2 × radius + length of the arc = 25
Length of the arc = 25 – 2 × radius = 25 – 14 = 11 cm
Let us check the ratio between the length of the arc and its area
The length of the arc: Area of the sector
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 71
11 : area of the sector = 2:7
Area of two sectors =7 × 11
Area of the. sector = \(\frac{7 \times 11}{2}\) = 38.5 cm2

Question 19.
In the figure if the length of one side of a square is 12 cm, then find the area of the shaded portion.
Kerala Syllabus 9th Standard Maths Solutions Chapter 9 Circle Measures 72
Answer:
Area of the square = 122 = 144
Diameter of the circle = 12; r = 6 Area of the circle
= πr2 = π × 6 × 6 = 3.14 × 6 × 6 = 113.04 cm2
Area of shaded portion = 144 – 113.04 = 30.96 cm2

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