Class 9

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

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.

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.

(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.

(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.

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:

(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.

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.

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.

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

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.

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.

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.

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

You can Download The Last Leaf (Story) Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Physics Solutions Part 2 Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State State Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power

A The Trio (Story) Textual Questions and Answers

Hss Live Guru 9th Physics Kerala Syllabus Chapter 5 Question 1.
Observe figure try to write down the activities shown in them.
Answer:

1. A man pushes a trolly.
2. Batting of a cricket ball.
3. Pushing a wall.

Kerala Syllabus 9th Standard Physics Notes Chapter 5 Question 2.
Write down more activities familiar to you
Answer:

1. A man carrying a load
2. Throwing a ball
3. Lifting the bag on to the shoulder
4. Pushing a car into motion

Hsslive Guru 9th Physics Kerala Syllabus Chapter 5 Question 3.
You have understood that a force is to be applied on a body to do an activity. Find out the source of applied force for every activity and note them down in the table.


Answer:

Activity	Source of applied force
Falling of a mango	The earth
A trolley being pushed	The person pushing
Batting of a cricket ball	Batsman
Pushing a wall	The person pushing
Lifting a load	The person lifting

Objects undergo displacement only when the force is applied on it them.

Displacement takes place in the direction of force applied	No displacement for the body in the direction of force applied
1. A cricket ball when hit by a bat	1. A wall is pushed
2. A trolley is being pulled.	2. A car is pushed by sitting inside the car
3. Climbing a ladder with a load on head.
4. Falling of a mango from mango tree.

Work :

Work is said to be done only when a body under¬goes displacement in the direction of the applied force.

work energy and power Question 4. Observe figure and write down the situation where work is said to be done
Answer:
The person loaded on head and goes upward.

Kerala Syllabus 9th Physics Notes Chapter 5 Question 5.
A boy pushed an object of mass 30Kg horizontally across the floor through 50m. Another boy pushed an object of mass 50kg across the same floor through 50m. Both of them gave the same speed for moving the objects.
a) Who applied greater force here?
b) In which case was the work greater?
c) Write down a factor influencing work.
Answer:
a) Second boy applied greater force
b) More work is done in the second situation
c) The Factor influencing work is force.

Learn Mass Transfer MCQ questions & answers are available for a Chemical Engineering students.

Hss Live Guru Physics 9th Kerala Syllabus Chapter 5 Question 6.
A boy pushed an object of mass 30kg across a horizontal floor through 20m. Another boy pushed the same body through 30m on the same floor with the same speed.
a) Who pushed a greater distance here?
b) What about the force applied
c) Who did the greater work?
d) Which is the factor influencing work here?
Answer:
a) Second boy
b) Force is same
c) Second boy done greater work
d) The Factor influencing work is displacement

The factor affecting work done are force (F) and displacement (S)
Equation for calculating work done Work W = Fs
Unit of work is Joule (J), Kilojoule (KJ)
1KJ = 1000 J

If a force of F newton is applied continuously on a body and the body undergoes a displacement of s metre in the direction of the force, then the work done by the applied force is W = Fs

Get the free “Nuclear Equation Calculator” widget for your website, blog, WordPress, Blogger, or iGoogle.

Physics Notes For Class 9 Kerala Syllabus Chapter 5 Question 7.
When a force of 10N is applied continuously on a body it undergoes a displacement of 2m find out the magnitude of the work done?
Answer:
F = 10N, s = 2m
Work W= Fs = 10 × 2 = 20 Nm
Look at the Figure

Hsslive Guru Physics Class 9 Kerala Syllabus Chapter 5 Question 8.
A body of mass m kg is placed on a table. What are the forces experienced by this body?
Answer:
Weight of the body applies downwards and the table applies an equal force upwards.

Hss Live 9th Physics Kerala Syllabus Chapter 5 Question 9
In which directions do these forces act?
Answer:
Forces acting on both directions, upward direction and downward direction.

9th Physics Notes Kerala Syllabus Chapter 5 Question 10.
A book of mass 100g is raised to the top of a table of height 1 m. Find the magnitude of the work done by the force applied against the gravitational force(g = 10m/s²)
Answer:
m = 100g = 0.1kg
g = 10m/s²
h = 1m
W = mgh
= 0.1 × 10 × 1 = 1J
1 J is the amount of work done to raise a body of mass 100 g through a height of 1m.

Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Question 11.
If a force of 50N is applied on a body and it under¬goes a displacement of 2m in the direction of the force, calculate the amount of work done.
Answer:
F = 50N
s = 2m
W = Fs = 50 x 2 = 100J

Hss Live Guru 9 Physics Kerala Syllabus Chapter 5 Question 12.
a) If a force of 200N is applied on a table of mass 50kg, it undergoes a displacement of 0.5m in the direction of force. Calculate the amount of work done.
b) If the same table is raised by 3 m, what would be the work done against the gravitational force?
Answer:
a) F= 200N, s = 0.5m
W = Fs = 200 x 0.5 = 100J
b) m = 50 kg, g = 10m/s²
h = 3 m
W = mgh = 50 x 10 x 3 = 1500J
Observe Figure

Let a body mass m be pulled by a force F. If the body has a displacement s in the direction of the force, then the work done by the force F, Wf = Fs Here the displacement produced is in the direction of the force itself.

Kerala Syllabus Class 9 Physics Solutions Chapter 5 Question 13.
Write whether this work is negative or positive.
Answer:
Work is positive

9th Class Physics Notes Kerala Syllabus Chapter 5 Question 14.
The displacement is opposite to the frictional force, is the work done by the frictional force positive or negative
Answer:
If the displacement is in opposite direction work done by frictional force is negative.

Class 9 Physics Kerala Syllabus Chapter 5 Question 15.
In which direction is the force of gravity on the body?
Answer:
Direction of the gravitational force will be in the downward direction

9th Standard Physics Notes Kerala Syllabus Chapter 5 Question 16.
Is there a displacement for the body in the direction of the gravitational force?
Answer:
No displacement occurs in the direction of the gravitational force.
When a body on a floor is pulled and if it is displaced in the direction of the applied force, the work done by the applied force will be positive and the work done by the frictional force exerted by the floor will be negative.

Energy

Question 17.
What is the work to be done to raise a body of mass m kg through h meter?
Physics chapter 5 work and energy Answer:
w = mgh
Energy is the capacity to do work. Unit is Joule (J)

work energy book Question 18. In daily life, we use different forms of energy for various activities. List the forms of energy familiar to you
Answer:

1. Mechanical energy
2. Heat energy
3. Electrical energy
4. Chemical energy
5. Light energy
6. Nuclear energy

There are two type of Mechanical energy Kinetic energy and potential energy

Kinetic Energy

Pulling the toy car backwards a little and allow it to hit the plastic ball

Question 19.
What happened to the ball when the moving car hit it?
Answer:
Ball moves forward

10 to the 9th power Question 20.
How did the car get the energy to move the ball forward?
Answer:
The energy is obtained from the motion of toy car
Conclusion:

• Moving objects possess energy
• The energy possessed by a body by virtue of its motion is the Kinetic energy

Experiment:
Let’s do another activity. Allow a powder tin to slide down a polished, inclined plane as shown in the figure and let it hit a toy car. Try to measure the displacement of the toy car. Repeat the experiment by increasing the height of the inclined plane and filling the tin with sand.

Observation:

• The displacement of the toy car is greater When it is hit by the powder filled with sand
• Also the displacement is greater when it is dropped from a greater height

Conclusion: Kinetic energy depends on mass (m) and velocity (v)
Derivation of the equation for calculating K.E
Work W = Fs
As per II law of motion
F = ma
∴ W = mas
According to third equation of motion
v² = u² + 2as = 0 + 2as (u = 0) = 2as
∴ as = \(\frac{v^{2}}{2}\)
W = mas , if we put \(\frac{v^{2}}{2}\) instead of as

Work is equal to the magnitude of kinetic energy
i.e, Kinetic energy K = \(\frac { 1 }{ 2 }\) mv2,
M – mass, V – Velocity
When a body of mass m moves with a velocity v, its kinetic energy will be K = \(\frac { 1 }{ 2 }\) mv2

Question 21.
A man having a mass of 70kg is riding a scooter of mass 80kg. What is the total kinetic energy if the velocity of the scooter is 10m/s?
Answer:
m = 70kg + 80kg = 150kg
v = 10m/s

Question 22.
A car of mass 1500 kg is moving at a velocity of 20m/s. What is its kinetic energy?
Answer:
m = 1500 kg
v = 20m/s
K = \(\frac { 1 }{ 2 }\) mv²
= 1/2 × 1500 × 20²
= 300000J = 300kJ

Question 23.
A boy of mass 50kg is riding a bicycle with a speed 2m/s. The bicycle has a mass of 10kg. Calculate the total kinetic energy?
Answer:
m = 50kg + 10kg = 60kg
v = 2m/s
K = \(\frac { 1 }{ 2 }\) mv²
1/2 × 60 × 2² = 120J

Potential Energy

When we lift a body in perpendicular direction, the work is said to be done against gravitational force. According to figure, maximum work is done when it reaches at a height fb from the group, That means as the height increases work done also increases.

• The energy received by the body is equal to the work done on it.
• The body attains more energy when the height from the ground level increase

The energy possessed by a body due to its position is the potential energy.
ie, Potential energy
U = mgh
m – mass, g – acceleration due to gravity
h = height from the ground

Question 24.
Identify more situation in which potential energy is acquired by virtue of position
Answer:

• Coconut in a coconut tree
• Water stores in huge reservoirs
• Objects placed above the buildings,

Inference:
Height increases, potential energy increases

Question 25.
Write down situations in which potential energy varies.
Answer:

• Falling of a coconut from a coconut tree
• Pumping water to a tank at a height
• Climbing a ladder

Question 26.
Calculate the potential energy of a body of mass 1 kg at a height of 6m from the ground?
Answer:
Mass m = 1 kg, Acceleration due to gravity
g = 10m/s²
h = 6 m,
U = mgh = 1 × 10 × 6 = 60J

Question 27.
A bird of mass 0.5 kg is flying at the same speed at the same height of 5m. In this state, if its Kinetic energy and potential energy are equal.
a) What is the potential energy of the bird?
b) What is the velocity of the bird?
Answer:
a) m = 0.5kg
g = 10m/s²
h = 5m
u = mgh
= 0.5 × 10 × 5 = 25J
b) Kinetic energy K = 25J (∴U = K)

Question 28.
Write other examples for getting potential energy due to strain
Answer:

• A Stretched bow
• An elongated rubber band
• Compressed spring
• Compressed spring in a toy car

Conclusion: The factor which gives potential energy are position and strain

Law Of Conservation Of Energy

Question 29.
What form of energy does the flower pot have when it is on the sunshade of a building?
Answer:
Potential energy

Question 30.
While the flower pot is falling down, what forms of energy does it possess?
Answer:
potential energy and kinetic energy

Question 31.
Answer:
Does its potential energy increase/decrease when the pot falls down.
Answer:
Potential energy decreases

Question 31.
Will the kinetic energy increase/decrease at that time?
Answer:
Kinetic energy increases

Question 33.
What energy transformation takes place just before the flower pot reaches the ground?
Answer:
Potential energy is converted completely into kinetic energy

Question 34.
Let the mass of the flower pot 15kg and the height of the sunshade 4m.
a) When the flower pot is on the sunshade, what is its potential energy? (g = 10m/s2).
b) When it is on the sunshade, what is its kinetic energy?
c) If so, what is its total energy?
Answer:
U = mgh
= 15 × 10 × 4
= 600J
b) Kinetic energy will be zero
c) Total energy = PE + KE
= 600J + 0 = 600J

Question 35.
While falling, when the flower pot is at a height of 2m from the ground, what will be its kinetic energy?
Answer:
K = \(\frac { 1 }{ 2 }\) mv²
u = 0, g = 10m/s² ,
s = 4 – 2 = 2m
v² = u2² + 2as
=0 + 2 × 10 × 2 = 40
K = 1/2 × 15 × 40
= 300J

Question 36.
What is the potential energy when it is at a height of 2m from the ground? What is the total energy now?
Answer:
Potential energy at 2m height
U = mgh = 15 × 10 × 2 = 300J
∴ Total energy = 300 + 300 = 600J

Question 37.
What is the kinetic energy of the flower pot just before it touches the ground?
Answer:
K = \(\frac { 1 }{ 2 }\) mv²
v² = u² + 2as
= 0 + 2 x 10 x 4 = 80
Kinetic energy K = \(\frac { 1 }{ 2 }\) mv²
= 1/2 × 15 × 80 = 600J

Question 38.
The potential energy
U = mgh
= 15 x 10 x 10 x = 0.
What will be the total energy?
Answer:
Total energy = 600 + 0 = 600J

Question 39.
To sum up the amount of energy at each situation:
a) When on the sunshade
b) When at a height of 2m from the ground
c) Just before hitting the ground
Answer:
a) When on the sunshade = 600J
b) At a height of 2m from the ground = 600J
c) Just before hitting the ground = 600J
Energy can neither be created nor be destroyed. It can be transformed from one from to another

Sun is the major source of energy. We utilize solar energy in different ways. It is renewable source of energy Plants prepare food by using sunlight. From these fossil fuels, firewood are formed. Windmills works by utilizing kinetic energy of wind energy from infra¬red rays. Tidal energy is obtained due to gravitation. When we consider any energy sources, they were originated from solar energy.

Power

Question 40.
Given below is the information regarding the working of pumps in three neighboring houses. Complete the table (g = 10 m/s²).

Answer:
b) 150000J
c) 150000J

Question 41.
Is the amount of work done by the pump to fill water in the three tanks equal?
Answer:
Amount of work done is equal

Question 42.
Find the amount of work done per second by each pump.

Answer:

Pump	Work done	Time (s) (J)	Work done per second (J/S)
A	150000	100	1500
B	150000	200	750
C	150000	400	375

Amount of work done per second is referred as power of the pump.
Work done per unit time or rate of doing work is power
Power = \(\frac{\text { work }}{\text { time }}, P=\frac{w}{t}\)
Units of power
watt (W), kilowatt (KW), Horse power (H.P)
1 kW= 1000W
1 HP = 746W

Question 43.
If a man of mass 70kg climbs up a mountain of height 30m in 5 minutes, what is his power?
Answer:
m = 70kg, g = 10m/s²
h = 30m
Work W = mgh = 70 × 10 × 30 = 21000J
Time t = 5mt = 5 × 60 = 300s
Power p = \(\frac { w }{ t }\) = \(\frac { 21000 }{ 300 }\) = 70W

Question 44.
If a man of mass 50kg takes 60s to climb up 20 steps, each 15cm high, calculate his power.
Answer:
m = 50kg, g = 10m/s²
h = 15cm × 20 = 300cm = 8m
t = 60s
work W = mgh = 50 × 10 × 3 = 1500J
Power p = \(\frac { w }{ t }\) = \(\frac { 1500 }{ 60 }\) = 25W

Let Us Assess

Question 1.
A boy is trying to push the concrete pillar of the building using a force of 300N. Calculate the amount of work done by the boy.
Answer:
F= 300N, s = 0
w = Fs = 300 × 0 = 0

Question 2.
From what you have learnt of potential energy and kinetic energy write down the from of energy possessed by the bodies given below.
a) water in a dam
b) Stretched rubber band
c) Mango falling from a tree
Answer:
a) Potential energy
b) Potential energy
c) Potential energy lesser. Kinetic energy greater

Question 3.
Calculate the kinetic energy of an athlete of mass 60kg running with a velocity 10 m/s
m = 60kg
v= 10m/s

Question 4.
A stone of mass 2kg is thrown upwards from the ground with a velocity of 3m/s. When it reaches maximum height, calculate its potential energy.
Answer:
u = 3m/s v = 0
a = g = -10m/s² (Velocity of a body moving upwards is decreasing so acceleration is negative)
v² = u² + 2as
0 = 32 + 2 × -10 × s
0 = 9 – 20s
20s = 9, s = \(\frac { 9 }{ 20 }\)
U = mgh = 20 × 10 × \(\frac { 9 }{ 20 }\) = 90J

Question 5.
The heart of a healthy person beats 72 times per minute and each beat uses up about 1J of energy. Calculate the power of the heart.
Answer:
W= 1 x 72 = 72J
t = 1mt = 60s
P = \(\frac { W }{ t }\) = \(\frac { 72 }{ 60 }\) = 1.2w

Question 6.
Which among the following is a vector quantity?
a) work
b) momentum
c) power
d) energy
Answer:
momentum

Question 7.
If the velocity of an object is doubled, its kinetic energy becomes
a) 2 times
b ) 1/2
c) 4 times
e) 1/4
Answer:
4 times

Question 8.
An object of mass 1 kg is falling from a height 10 m. What be the work done while falling?
a) 10 J
b) 1J
c) 100 J
d)1000 J
Answer:
w = mgh = 1 × 10 × 10 = 100J

Question 9.
Which one among the following is correct?

Answer:
W = p × t

Question 10.
A roller weighing 1 tonne is being dragged along a road. What is the work done against gravity? Why?
Answer:
Zero, No displacement in the direction of force.

Question 11.
Is it possible for an object to possess energy without momentum? Give one example for such a situation.
Answer:
Yes, possible. A coconut on a coconut tree has potential energy, but it has no momentum.

Question 12.
Say whether the following are positive work or negative work.
1) The work done by a person drawing water from a well using a rope without a pulley.
2) Work done by gravitational force in this situation
3) Work done by the frictional force while and object is sliding down along an inclined plane.
4) Work done by the force is moving along a plane surface.
Answer:

1. Positive work
2. Negative work
3. Negative work
4. Positive work

Question 13.
How much joule is 1 kWh?
Answer:
1 kWh = 1000 × 60 × 60 = 3600000J

Question 14.
Find out the work done against the gravitational force in the situations given below.
1. A child is standing still with a bundle of books of mass 5 kg
2. With the same bundle of books, she travels 1m along a plane surface with a speed 5 m/s
3. The bundle of books is lifted onto the top of a cupboard having 1m height (g=10m/s2).
Answer:
1. 0
2. 0
3. W = mg = 5 × 10 × 1 = 50 J

Question 15.
A ball of mass 0.4 kg is thrown vertically upward with a velocity 14 m/s. Calculate its kinetic energy and potential energy after 1 s.
(Hint: v + at, s = ut + 1/2at²)
Answer:
v = u + at
= 14 + 10 × 1
= 24 m/s
KE = \(\frac { 1 }{ 2 }\) mv²
= 1/2 × 0.4 × 24
= 115.2 J
S = ut + 1/2 at²
= 14 × 1 + 1/2 × 10 × 1²
= 14 + 5 = 19 m
E = mgh = 0.4 × 10 × 19
= 76 J

Question 16.
An object of mass 1000 kg is travelling with a velocity 72 km/h. Calculate the work done to bring it rest.
Answer:
m = 1000 kg
v = 72 km/h = 20 m/s
Work = difference of kinetic energy
= \(\frac { 1 }{ 2 }\) mv²
= \(\frac { 1 }{ 2 }\) × 1000 × 20 × 20 = 400000J

Question 17.
Estimate the work done on a object of mass 80 kg to change its velocity from 5 m/s to 10 m/s
Answer:
m = 80 kg
Work = difference of kinetic energy

Work, Energy and Power More Questions

Question 1.
Fill in the blanks
a) Work done when a body of mass 100g is lifted up to a height of 1 meter is ……….
b) The two factors related to potential energy to body are………….
c) 1HP = ……… Watt
Answer:
1 Joule
b) Position, Strain
c) 746W

Question 2.
Classify the following into work done and work is not done
1. A mango falling from a mango tree
2. Pushing a table while sitting on it
3. Pushing a wall
4. Kicking a football
5. A trolley is moving forward
6. Standing with a load above the head
Answer:
Work done:
1. A mango falling from a mango tree
2. Kicking a football
3. A trolling is moving forward
Work is not done:
1. Pushing a table while sitting on it
2. Pushing a wall
3. Standing with a load on the head

Question 3.
When Lekshmi is applied a force of 50N on an object it undergoes a displacement of 2m. When Vinitha applied the same force on itundergoesadisplacementof3m if so
a) Which person done more work?
b) Give reason
c) Calculate the work done by each person?
Answer:
a) Vinitha
b) Displacement is greater
c) Work done by Lekshmi
W = Fs = 50N × 2m = 100J
Work done by Vinitha
W = Fs = 50N × 3m = 150J

Question 4.
Calculate work done when a boy of mass 40kg climb¬ing a staircase of height 50cm, g = 10m/s2
Answer:
W= Fs
F = mg = 40 × 10 = 400N
S = h = 0.5m
W= 400 × 0.5 = 200J

Question 5.
a) What is mean by kinetic energy?
b) How mass and velocity affect kinetic energy?
c) A body of mass 20kg is moving with a velocity of 5m/s. Calculate the kinetic energy?
Answer:
a) The energy possessed by a body by virtue of its motion is called kinetic energy
b) Mass increases, kinetic energy increases mass . is doubled kinetic energy also doubled. Velocity increases, kinetic energy increases, Velocity is doubled kinetic energy becomes four times.
c) Kinetic energy KE = \(\frac { 1 }{ 2 }\) mv²
= 1/2 × 20 × 5² = 1/2 × 20 × 25 = 250J

Question 6.
a) Write the equation for finding potential energy?
b) indicate the representation of each letter
Answer:
a) U = mgh
b) M – mass,
g – acceleration
h – height from the ground

Question 7.
State law of conservation of energy?
Answer:
Energy can neither be created nor be destroyed. It can be transformed from one form to another.

Question 8.
A store of mass 5kg was raised from the ground to the second floor of height 7m and from there to the third floor of height 3m from the second floor. Calcu¬late the potential energy of the store with respect to the ground floor and the second floor?
Answer:
Mass m = 5kg
Height from ground to third floor
h = 7m + 3m
g = 10m/s²
Potential energy with respect to the ground
U = mgh = 5kg × 10m/s² × 10 = 500J
Height from second floor to third floor h = 3m
Potential energy with respect to the second Floor
U = mgh = 5kg × 10m/s × 3 = 150J

Question 9.
a) What is power?
b) The time takes to move an object at a distance of 5m is 10s is the force applied is 30N, what is the power?
Answer:
a) Rate of doing work is power
b) Power P = E
W = Fs = 30 × 5 = 150J
t = 10sec.
Power P = \(\frac { 150 }{ 10s }\) = 15W

You can Download Area Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 1 Area

Kerala Syllabus 9th Standard Maths Area Text Book Questions and Answers

Textbook Page No. 11

9th Standard Maths Notes Kerala Syllabus Question 1.
Draw a triangle of sides 3,4 and 6 centimeters. Draw three different right triangles of the same area.
Answer:
Draw ∆ ABC having sides 3 cm, 4 cm, and 6 cm.

Draw a line parallel to AB through C and a line from B perpendicular to the parallel line, both lines intersect at point E. Now ∆ABE is a right angled triangle. Also draw a line perpendicular to the parallel line through A, which makes a point Don parallel line, join BD. ∆ ABD is a right angled triangle.

Similarly make right-angled triangle by drawing line parallel to AC through B, and a line from A perpendicular to the parallel line, both lines intersect at point F.

The right traingle ∆ DAB, ∆ ABE, ∆ CAF have the same area as that of ∆ ABC.

Kerala Syllabus 9th Standard Maths Solution Chapter 1 Area In English Question 2.
Draw the triangle shown below in your notebook. Draw triangles ABP, BCQ and CAR of the same area with measurements given below.

1. ∠BAP = 90°
2. ∠BCQ = 60°
3. ∠ACR = 30°

Answer:
1.

Draw a line parallel to AB through C. Draw a perpendicular from A and the point of intersection of perpen-dicular and parallel line is P. Then draw ∆ABP then ∠BAP = 90°. Areas of ∆ABC and ∆ABP are equal.

2.

Draw a line parallel to BC through A from C make angle 60° both lines intersect at point Q. Then draw ∆ BCQ. Areas of ∆ ABC and ∆ BCQ are equal.

3.

Draw a line parallel to AC through B from C make angle 30° both lines intersect at point R. Then draw ∆ACR. Areas of ∆ABC and ∆ ACR are equal.

Download these Free R Programming MCQ Quiz Pdf and prepare for your upcoming exams.

Maths 9th Class Chapter 1 Area English Medium Kerala Syllabus Question 3.
Draw a circle and a triangle with one vertex at the centre of the circle and the other two on the circle. Draw another triangle of the same area with all three vertices on the circle.
Answer:
Draw a line with centre C and AB is the chord of the circle. Draw ∆ ABC then draw a line parallel to AB through C this line touches the circumference at a point D then complete ∆ ABD. Area of ∆ ABC = Area of ∆ ABD

9th Standard Maths Notes Malayalam Medium Kerala Syllabus Question 4.
How many different (non-congruent) triangles can you draw with two sides 8 and 6 centimetres and area 12 square centimetres? What if the area is to be 24 square centimetres?
Answer:
For the triangle of area of 12 sq.cm., the height from the 8 cm side should be 3cm.
Then the area = \(\frac{1}{2}\) x 8 x 3 =12 sq.cm
Draw a line 8 cm long. Draw another line parallel to it at a distance of 3 cm. Draw a circle of radius 6 cm with one end of the first line as the center. The points where this circle cuts the second line are the third vertex of the triangle.

We can draw two triangles ∆ OBC and ∆ OCA having area 12 sq.cm.
For the triangle of area 24 sq cm, the height from the 8 cm side should be 6 cm.

We can draw only one triangle of this type.

Draw A Square Of Area 10 Square Centimeters Malayalam Question 5.
In the picture below, the lines parallel to each side of the blue triangle through the opposite vertex are drawn to make the big triangle.

How many triangles in the picture have the same area as that of the blue triangle? How many of them have the same measures of the blue triangle?
Answer:

AB is parallel to PQ. So ∆ PAQ and ∆ QPB have the same area as that of ∆ PQR.
BC is parallel to RP, So ∆ CPB and ∆ CRB have the same area as that of ∆ PQR.
AC is parallel to RQ, So A CRA and ∆ CAQ have the same area as that of ∆ PQR.
Also ∆ PCQ, ∆ PAR and ∆ BRQ have the same area as that of ∆ PQR.
Total = 9 triangles
Also ∆ PCQ, ∆ PAR and ∆ BRQ have the same measurement as that of ∆ PQR.
Total = 3 triangles.

Kerala Syllabus 9th Standard Notes Maths Question 6.
Prove that the two triangles shown below have the same area:

Answer:

∠ A = 50°
∠B = 130°
∠A + ∠B = 50 + 130 = 180°
These angles are supplementary. So lines AD and BC are parallel. Their lengths are equal.
Since these triangles have the same base and their third vertices on a line parallel to the base, their areas are equal.

Textbook Page No. 15

Maths Notes Class 9 Kerala Syllabus Question 1.
Draw the two quadrilaterals shown below, in your note book. Draw tri-angles of the same area and calcu-late the areas (The lengths needed may be measured).

Answer:
1.

In the given figure, a line parallel to BD through C and AB produced intersect at E. Now area of ∆AED is equal to the area of quadrilateral. Measure the required lengths and then find the area.
AE = 10.8 cm
Height from D to AE = 2.9 cm
Area of ∆ AED = \(\frac{1}{2}\) x 10.6 x 2.9
= 15.66 sq.cm

2.

In the given figure, a line parallel to BD through C and AB produced intersect at E. Now area of ∆ AED is equal to the area of quadrilateral. Measure the required lengths and then find the area. AE = 12.6 cm DP = 5.2 cm Area of ∆ AED = \(\frac{1}{2}\) x AE X PD
= \(\frac{1}{2}\) x 12.6 x 5.2 = 32.76 sq cm

Kerala Syllabus Class 9 Maths Solutions Question 2.
Draw a rhombus of sides 6 centimetres and one angle 60°; then draw a right triangle of the same area.
Answer:

Draw rhombus ABCD.
∠A = ∠C = 60° and ∠B = ∠D = 120°
Draw diagonal BD. Draw line parallel to BD through C and AB produced intersect at point E. Now area of rhombus is same as the area of ∆ ADE and ∆ ACE. Lines AC bisect ∠A and ∠C.
∴ Angles of ∆ADE is equal to 30°, 60°, 90°
∴ The triangles drawn are right-angled triangles.

Kerala Syllabus 9th Standard Maths Guide Pdf Question 3.
Draw a regular pentagon and then a triangle of the same area. Calculate the area.
Answer:

Draw a regular pentagon ABCDE. Draw a line parallel to BD through C, line parallel to AD through E, AB produced to both sides intersects points F and G. Now area of AFDG is same as area of pentagon. Measure the required lengths and then find the area.

Kerala Syllabus 9th Standard Maths Notes Question 4.
The picture shows a rectangle divided into two parts. Instead of the broken line separating these parts, draw a straight line to divide the rectangle into two other parts of the same area. Calculate the areas of these parts.

Answer:

Join PQ, draw a line parallel to PQ through O intersecting points are M and N line PN or MQ can make 2 parts with same area.

Proof:
Area of rectangle ABCD = 5 x 3 = 15 sq.cm
Area of trapezium JOPD = \(\frac{1}{2}\) x 1 x (1+3) = 2 sq cm
Area of trapezium AQOJ = \(\frac{1}{2}\) x 2 x (2 + 3) =5 sqcm
Area of part AQOPD = Area of trapezium JOPD + Area of trapezium AQOJ
= 2 + 5 = 7 sqcm
Area of the remaining part = 15 – 7 = 8 sq cm
Now area of trapezium ANPD is equal to the area of pentagon AQOPD.

Textbook Page No. 19

9th Class Maths Notes Malayalam Medium Kerala Syllabus Question 1.
In the picture below, two lines are drawn from the top vertex of a tri-angle to the bottom side:

Prove that the ratio in which these lines divide the length of the bot¬tom side is equal to the ratio of the area of the three smaller triangles in the picture.
Answer:

If AB : BC : CD
x : y : z and height = h then area of triangles are \(\frac{1}{2}\) x h, \(\frac{1}{2}\) yh, \(\frac{1}{2}\)zh
then ratio of areas are \(\frac{1}{2}\) xh : \(\frac{1}{2}\) yh \(\frac{1}{2}\) zh = x : y : z

Maths Class 9 Kerala Syllabus Question 2.
In the picture below, the top vertex of a triangle is joined to the mid point of the bottom side of the tri-angle and then the mid point of this line is joined to the other two vertices.

Prove that the areas of all four tri-angles obtained thus are equal to a fourth of the area of the whole triangle.
Answer:

Mid point of BC is D.
Line AD divides ∆ ABC into 2 triangles with equal areas.
E is the midpoint of AD.
Line BE divide ∆ ADB into 2 triangles with equal areas. Similarly CE divides ∆ ADC.
4 triangles of equal area.

Class 9 Maths Chapter 1 Kerala Syllabus Question 3.
In the picture below, the top vertex of a triangle is joined to the mid point of the opposite side and then the point dividing this line in the ratio 2 :1 is joined to the other two vertices:

Prove that the areas of all three triangles in the picture on the right are equal to a third of the area of the whole triangle.
Answer:

Mid point of AB is Q
∴ ∆ AQC and ∆ BQC have equal area’s P divides AQ in the ratio 2 : 1.
Area of ∆ APC is twice the area of ∆ AQP.
∴ Sum of the areas of ∆ AQP and ∆ BQP is equal to the sum of the area’s of ∆ APC and ∆ BPC.
The area of the APB, PCB and CPA are same. The area of three triangles is equal to the third of area of the whole triangle.

Question 4.
Prove that the lengths of the per-pendiculars from any point on the bisector of an angle to the sides are equal.
Answer:

OC is the bisector of ∠AOB
C ia a point on this bisector.
CA is perpendicular to OA.
CB is perpendicular to OB.
We have to prove that CA = CB Consider the two triangle ∆ OAC and ∆ OBC.
OC = OC (Common side)
∠OAC = ∠OBC = 90°
∠AOC = ∠BOC (OC is the bisector of AOB)
One side and two angles on it of OAC are equal o one side and two angles on it of OBC.
So these triangles are equal.
Then sides opposite to equal angles are equal.
CA = CB
∴ The lengths of the perpendiculars from any point on the bisector of an angle to the sides are equal.

Question 5.
In the picture below, the side AC of the triangle ABC is extended to D, by adding the length of the side CB. Then the line through C parallel to DB is drawn to meet AB at E.

1. Prove that CE bisects ∠C.
2. Describe how this can be used to divide an 8 centimeters long line in the ratio 4 : 5.
3. Can we use it to divide an 8 centimeters long line in the ratio 3 : 4? How?

Answer:
1. ∆ BCD is an equilateral triangle If∠CBD = x then ∠CDB = x
∠BCD = 180 – 2x
∠BCE = x
(line BC passing through parallel lines EC and BD makes equal angle)
∴ ∠ACE = x
∴ line CE bisects ∠C.

2. If AB = 8, AC = 4, and CB = 5
then the line CE divides AB in the ratio 4 : 5.

3. No.
We can’t draw a triangle with sides 8cm, 3cm and 4cm. Since 3 : 4 = 6 : 8, take 6cm and 8cm as two sides then find third side.

Question 6.
In the figure, the diagonals of a quadrilateral split it into four tri-angles. The areas of three of them are shown in the picture:

Calculate the area of the whole quadrilateral.
Answer:
A line from the vertex of a triangle divides the length of the opposite side and the area of triangle in the same ratio.

In ∆ ACD, a line DE is drawn from D to AC.
∴ AE : EC = Area ∆ AED : Area of ∆ CED
AE : EC = 60 : 30 = 2 : 1
In ∆ ACB, a line BE is drawn from B to AC.
∴ AE : EC = Area of ∆ AEB : Area of ∆ CEB
2 : 1= Area of ∆ AEB : 40 = 2 : 1
∴ Area of ∆ AEB = \(\frac{40×20}{1}\) =80 sq.cm
∴ Total area of the quadrilateral = 60 + 30 + 80 + 40 = 210 sq.cm

Question 7.
In this picture the hori∠ontal lines at the top and bottom are parallel.

Prove that the yellow and red tri-angles are of the same area.
Answer:

AB ∥ DC
∴ Area of ∆ ABC = Area of ∆ ABD (Same base and the third vertex on a line parallel to the base)
Area of ∆ ABC – Area of ∆ ABE = Area of ∆ ABD – Area of ∆ ABE
∴ Area of ABEC = Area of ∆ AED ∆ AED ↔ ∆ BEC

Question 8.
In the figure, the diagonals of a trapezium split it into four triangles.

The area of the yellow triangle is 10 square centimetres and the area of the green triangle is 20 square centi – metres. What is the area of the whole trapezium?
Answer:
Since quadrilateral ABCD is a trapezium, lines AB and CD are parallel.

So area of ∆ ABD = 20 + 10 = 30 sq.cm
Area of ∆ ABC = 30 sq.cm
(Same base and the third vertex on a line parallel to the base)
Area of ∆ ABC = 30 sq.cm
Area of ∆ ABE = 20 sq.cm
Area of ∆ BEC = 30 – 20 = 10 sq.cm
From ∆ ABC,
AE : EC = area of ∆ ABE : Area of ∆ BEC
AE : EC = 20 : 10 = 2 : 1
From ∆ ACD,
AE : EC = 10 : area of ∆ DEC 2 : 1 = 10: area of ∆ DEC
Area of ∆ DEC = \(\frac{1×10}{2}\) sq.cm
Total area of the trapezium
= 20 + 10 + 10 + 5
= 45 sq.cm

Question 9.
The picture below shows a trapezium divided into four parts by the diagonals.

The area of the blue triangle is 4 square centimetres and the area of the green triangle is 9 square centimetres. What is the total area of the trapezium?
Answer:

Area of ∆ AMD and ABMC are equal.
If area = A sq. cm then
9 : A = BM : MD
A : 4 = BM : MD;
9 : A = A : 4
A² = 36
A = 6 sq. cm
∴ Area of trapezium = 6 + 4 + 6 + 9 = 25 sq. cm

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

Question 1.
a. Where should be the position P in CD to set the isosceles triangle with half the area of the rectangle.

b. Where should be P in CD to get ∠ABP a right angle.
c. If ∠APB is a right angle how can you find the position P?
d. Can such right angled triangle be drawn in any of the rectangle?
Answer:
a. P is the centre of CD
b. It should be in C.
c. The distance from centre of AB to P should be half the length of AB.
d. No.

Question 2.
The lines in the diagram are parallel. Construct ∆ ABC with ∠C right angled

Answer:

AB is the diameter.

Question 3.
Construct ∆ ABC with AB = 8cm, BC = 6cm, AC = 5cm. Draw an isosceles triangle with same area and one side as AB itself.
Answer:

In figure AB ∥ PQ. D is the center point of AB.
∴ AD = BD.
Hence ∆ ABD is an isosceles triangle.

Question 4.
Find the area of the triangles with the given sides.
(a) 21 cm, 17 cm, 10 cm
(b) 12 cm, 16 cm, 20 cm
(c) 34 cm, 30 cm, 16 cm
(d) 20 cm, 21 cm, 29 cm
Answer:

Question 5.
ABCD is a rectangle in the diagram. P is a point on the side CD. Show that the area of ∆ AQB is equal to the sum of the areas of ∆ PQD and ∆ PCB

Answer:
⇒ area of ∆ AQB = area of ∆ APB – area of ∆ BQP
⇒ area of ∆ ADB = area of ∆ DBC area of (∆ ADQ + ∆ AQB) = area of (∆ DQP + ∆ QBP + ∆ BPC ) …….(1)
(∆ DAP = ∆ DBP) ⇒ (∆ ADQ = ∆ QBP) ⇒ area of (∆ ADQ + ∆ AQB)
= area of (∆ DQP + ∆ ADQ + ∆ BPC)
⇒ area of ∆ AQB = area of ∆ DQP + area of ∆ BPC.

Question 6.
In the figure XY, RS are parallel. The area of ∆ ABD is 20 cm²
a) Find the area of the parallelogram ABCD.
b) Find two triangles in the diagram with same area

Answer:
a.
20 + 20 = 40 sq. cm
b.
i. ABD, ABC
ii. DCB, DCA

Question 7.
Find the area of triangle in the dia-gram.

Answer:
\(\frac{1}{2}\) x 6 x 4 = 12 cm

Question 8.
In the figure BP : PQ : QC = 1 : 2 : 1. Area of triangle APQ is 8 sq.cm.
a. Find the area of triangle ABP.
b. Find the area of triangle ABC.

Answer:
a. BP : PQ = 1:2
Area of ∆ ABP is half the area of ∆ APQ.
Area of ∆ ABP = \(\frac{1}{2}\) x 8 = 4 sq.cm

b. Area of ∆ ABC = 4 + 8 + 4 = 16 sq.cm

Question 9.
In the figure line AB is parallel to CD. If AB = 5cm, AC = 4cm and ∠CAB = 90°.
a. Calculate the area of triangle ABC.
b. How much is the area of triangle ∆ ABD? Write reason

Answer:
a. Area of ∆ ABC = \(\frac{1}{2}\) x 5 x 4 = 10 sq.cm
b. Area of ∆ ABD = Area of ∆ ABC
Triangles with the same base and the third vertex on a line parallel to the base, have the same area.
Area of ∆ ABD = 10 sq.cm

Question 10.
Draw triangle ABC with AB = 8cm, BC = 6cm and B = 30°. Draw a right angled triangle of the same area. Measure its perpendicular sides. Calculate the area of the triangle.
Answer:
Draw ∆ ABC with the given measures.

Draw a line through C, parallel to AB.
Mark D on this parallel line such that ∠ BAD = 90°.
∆ ADB is the required triangle.
Measure the required lengths and then find the area.
AB = 8 cm
AD = 2.9 cm
Area of ∆ ADB = \(\frac{1}{2}\) x 8 x 2.9 = 11.6 sq.cm

Question 11.
Draw a quadrilateral with given measures. Draw a triangle of the same area as that of the quadrilateral.

Answer:
‘Draw a line through C parallel to DB. Which cuts AB at E.

∆ AED has the same area as that of quadrilateral ABCD.

Question 12.
If AB = 8cm, AC = 6cm, BC = 9cm,
AD is the bisector of A , then
a. BD : DC = ………….
b. What is the ratio of areas of triangle ABD and triangle ACD?
c. Draw a line of length 9 cm and divide it in the ratio 3 : 4.

Answer:
a. BD : DC = 8 : 6 = 4 : 3
b. Ratio of the areas of ∆ ABD and ∆ ACD = 8 : 6 = 4 : 3
c. AB = 9 cm
AC = 3 cm
CD = 4 cm
Draw CE through C and parallel to DB.

AC : CD = 3 : 4
AE : EB = 3 : 4

You can Download Division for Growth and Reproduction Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Biology Solutions Part 2 Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Biology Solutions Chapter 8 Division for Growth and Reproduction

Division for Growth and Reproduction Textual Questions and Answers

Division For Growth And Reproduction Class 9 Question 1.
The phase at which a cell prepares for division is called
Answer:
Interphase

9th Standard Biology Notes Kerala Syllabus Question 2.
………. takes place after karyokinesis
Answer:
Cytokinesis

Kerala Syllabus 9th Standard Biology Notes Question 3.
………… begins after interphase
Answer:
Cell division

Hss Live Guru 9th Biology Kerala Syllabus Question 4.
What are changes that take place during interphase?
Answer:
a) Division of nucleus (Karyokinesis)
b) Division of Cytoplasm (Cytokinesis)

Hss Live Biology Class 9 Kerala Syllabus Question 5.
State whether true or false
Cytokinesis takes place after karyokinesis
Answer:
True

Kerala Syllabus 9th Standard Biology Notes Pdf Question 6.
Main stages of cell division
Answer:
Interphase, Division of nucleus, Division of cytoplasm

Hss Live Guru Biology 9 Kerala Syllabus Question 7.
What are the important changes that take place during interphase?
Answer:

• Number of cell organelles increase
• Quantity of cytoplasm increases
• Cell size increases
• Genetic material duplicate

Cell cycle

A cell attains its complete growth during interphase. The fully grown cell undergoes division and becomes daughter cells. As the interphase and the division phase get repeated in a cyclic manner, they together constitute cell cycle.

UML full form, stands for, meaning, what is, description, example, explanation, acronym for, abbreviation, definitions, full name.

9th Biology Notes Kerala Syllabus Question 8.
…………..is brought about by cell division and cell growth
Answer:
Growth of the body

Kerala Syllabus Biology 9th Standard Question 9.
What are the two types of cell division?
Answer:
Mitosis and meiosis.

Biology Class 9 Kerala Syllabus Question 10.
What do you mean by mitosis?
Answer:
A parent cell divides to form two daughter cells are called mitosis.

Karyokinesis

Question 11.
Point out the phases taken place in the changes in nucleus?
Answer:
Prophase, metaphase, anaphase and telophase

Question 12.
In which phase does the chromatin reticulum become chromosomes?
Answer:
Prophase

Question 13.
What changes occurs in telophase?
Answer:
In telophase chromosomes that moved to the poles become chromatin reticulum and daughter nuclei are formed.

Question 14.
State whether true or false in prophase chromosomes become chromatin reticulum.
Answer:
False

Question 15.
Complete the table of stages of nuclear division
Answer:

Phases	Changes
Prophase	1. Chromatin reticulum become chromosomes 2. Duplicated chromosomes. 3. Formation of spindle fibers 4. Nucleolus and nuclear membrane get disappeared
Metaphase	Chromosomes have moved to the middle of the cell and, chromo­somes doubled.
Anaphase	1. Chromatids are starting to separate from each other. 2. Formation of two sets of daughter chromosomes
Telophase	1. Formation of daughter nuclei 2. Two daughter nuclei are formed. 3. There will be no change in chromosome number in each daughter nucleus

Cytokinesis (Division Of Cytoplasm)

Question 16.
The division of the cytoplasm is taken place in plant is entirely different. Give reason?
Answer:
Because it is due to the presence of the cell wall in plant cell.

Question 17.
What is the significance of mitosis?
Answer:
The significance of mitosis is that there is no change in the number of chromosomes.

Question 18.
Mitosis helps ………… & ……….
Answer:
For the repair of tissues and growth.

Question 19.
Which condition leads to cancer?
Answer:
Mitosis is a controlled process. A disruption in this controlled process leads to the excessive division of a cell and its proliferation. This condition leads to cancer.

Different Stages Of Growth

Question 20.
List out the different stages in the growth of human beings.
Answer:

• Zygote
• Infancy
• Old age
• Embryo
• Childhood
• Fetus
• Adolescence
• Youth

Question 21.
What are the physical peculiarities of old age?
Answer:
Rate of cell division decreases, Availability of oxygen to the cells decreases, Deterioration of cells increase Muscles shrink, Production of energy decrease

Question 22.
The elders should be cared. Do you agree with this statement? Why?
Answer:
Old age is inevitable in life. The aged who worked for the welfare of their family and society during their younger age deserve special consideration.

Question 23.
What are the differences between the growth in plants and animals? Draw a comparison and complete table
Answer:

Animals	Plants
Animals grow only up to a certain stage Animals do not have localized centers of growth	Plants can grow through­out their lives Growth in plants is localized only at certain parts

Question 24.
Plants grow due to the rapid division and differentiation of cells.
Answer:
Meristematic cells

Question 25.
What do you mean by meristematic cells?
Answer:
Meristematic cells are special types of cells that have the capacity for continuous division.

Question 26.
Plants can grow throughout their life due to the presence of
Answer:
Meristematic cells

Question 27.
……… helps to increase the length of root and stem.
Answer:
Apical meristem

Question 28.
……… helps to increase the girth of the stem.
Answer:
Lateral meristem

Question 29.
Name the meristematic cells which help to increase the length of the stem.
Answer:
Intercalary meristem

Question 30.
Where do you find intercalary meristem?
Answer:
It seen above the nodes of monocot plants.

Question 31.
The stem of monocots increases in length faster than dicots. Why?
Answer;
Because the intercalary meristem is visible only in the monocot plants.

Question 32.
Dicot plants: Lateral meristem
……………..: Intercalary meristem
Answer:
monocot plants

Question 33.
The stem of monocots does not increase its girth beyond an extent. Why?
Answer:
Because lateral meristem is absent in monocot plants.

Growth In Unicellular Organisms

Question 34.
Does cell division in unicellular organisms lead to growth or reproduction?

Answer:
Mitosis leads to reproduction in unicellular organisms.

Meiosis

Question 35.
What do you mean by meiosis? Explain.
Answer:
Meiosis is the mode of cell division in which gametes are formed. Meiosis occurs in the germinal cells of the reproductive organs. Human beings have 46 chromosomes. Germinal cells with 46 chromosomes divide continuously two times. These divisions in meiosis are known as meiosis I and meiosis II. Two daughter cells “with half the number of chromosomes (23 chromosomes) are formed in meiosis I. Each daughter cell again divides in meiosis II. There is no change in the chromosome number in this division. Hence meiosis II is similar to. mitosis. As a result of meiosis, four daughter cells, each with 23 chromosomes, are formed from a germinal cell.

Question 36.
What do you mean by polar body?
Answer:
When meiosis occurs in the female germinal cell, a large ovum and three small cells are formed. The smaller cells are the polar bodies. These sterile cells get destroyed.

Question 37.
Complete the illustration

Answer:
a = 46
b = 23
c= 23
d = 23
e = 23
f = 23
g = 23

Question 38.
What is the number of chromosomes in germinal cells?
Answer:
46

Question 39.
What is the number of chromosomes in the daughter cells formed after meiosis I?
Answer:
23

Question 40.
What is the peculiarity of meiosis II?
Answer:
Each daughter cell again divides in meiosis II. There is no change in the chromosome number in this division. Hence meiosis II is similar to mitosis.

Question 41.
What are the kinds of cell division occur in sexually reproducing organisms?
Answer:
There are two kinds of cell division occur in sexually reproducing organisms that are mitosis and meiosis.

Question 42.
What are the different stages in the growth of human being?
Answer:
Infancy, childhood, adolescence, youth and old age.

Question 43.
Differentiate mitosis and Meiosis

Answer:

Differences	Mitosis	Meiosis
Type of reproduction	Asexual	Sexual
Genetically Chromosome Number	Similar Remains same	Different Reduced by half
Takes place in	Somatic cells	Germ cells
Number of daughters	2 diploid	4 haploid
cells produced	cells	cells

Let Us Assess

Question 1.
The stage of karyokinesis at which daughter nuclei are formed
A. Prophase
B. Metaphase
C. Anaphase
D. Telophase
Answer:
Telophase

Question 2.
List the meristems in various parts of the plant and list their functions
Answer:

Meristem	Function
Apical meristem	Increase the length of root and stem
Lateral meristem	Helps to increases the girth of stem
Intercalary meristem	Helps to increase the length of the stem of monocot plants

Question 3.
In females, only a single ovum is formed from a germinal cell, whereas in males, more than one sperm is formed. Give reason.
Answer:
When meiosis occurs in the female germinal cell, a large ovum and three small cells are formed. The smaller cells are the polar bodies. These sterile cells get destroyed. So only a single ovum is formed from a germinal cell. But in males, after meiosis, four sperms having 23 chromosomes are formed form one germinal cell.

Question 4.
Observe the figures

a) Which stages of mitosis are indicated in the figures?
b) What are the changes that occur in the chromosomes during these stages?
Answer:
a) Metaphase
b) Chromosomes get aligned at the equator of the cell.

You can Download Proportion Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 12 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 12 Proportion

Proportion Textual Questions and Answers

Textbook Page No. 182

Hss Live Guru 9th Maths Kerala Syllabus Question 1.
A person invests 10000 rupees and 15000 rupees in two different schemes. After one year, he got 900 rupees as interest for the first amount and 1200 rupees as interest for the second amount.
i. Are the interests proportional to the investments?
ii. What is the ratio of the interest to the amount invested in the first scheme? What about the second?.
iii. What is the annual rate of interest in the first scheme? And in the second?
Answer:
i. The ratio between the amounts invested = 10000:15000 = 2 : 3
The ratio between the interests = 900 : 1200 = 3 : 4
Since the ratios are different. So, interests will be not proportional to the investments.

ii. The ratio between the amount invested and interest in scheme 1
= 10000 : 900= 100 : 9
The ratio between the amount invested and interest in scheme 2
= 15000 : 1200 = 25 : 2

iii. Rate of interest in the first scheme
= 900/10000 × 100 = 9%
Rate of interest in the second scheme =
1200/15000 × 100 = 8%

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Kerala State Class 9 Maths Solutions Question 2.
The area of A0 paper is one square metre. Calculate the lengths of the sides of A4 paper correct to a millimetre, using a calculator.
Answer:
The area of A1 paper is half of the area of A0 paper. The area of A2 paper is half of the area of A1 paper. Let con-sider the area of the A0 paper be 1 square metre, the area of A1 paper is
1/2 square metre , the area of A2 paper is 1/4 square metre , the area of A3 paper is 1/8 square metre , and the area of A4 paper is 1/16 square metre ,
If the length of A4 paper is √2 x and breadth x
length of A4 paper : breadth = √2 : 1
Area of A4 paper = length x breadth
√2x × x = √2x²
This is 1/16 m², therefore √2 x2 = 1/16

Hsslive Guru 9th Maths Kerala Syllabus Question 3.
In calcium carbonate, the masses of calcium, carbon and oxygen are in the ratio 10 : 3: 12 . When 150 grams of a compound was analysed, it was found to contain 60 grams of calcium, 20 grams of carbon and 70 grams of oxygen. Is it calcium carbonate?
Answer:
The ratio between calcium, carbon and oxygen in calcium carbonate = 10 : 3: 12 The ratio of calcium, carbon and oxygen in the given compound = 60 : 20: 70 = 6 : 2: 7
Since this ratio is not equal to the ratio of calcium, carbon and oxygen in calcium carbonate, it is not calcium carbonate.

Textbook Page No. 185

Hss Live 9 Maths Kerala Syllabus Question 1.
For each pair of quantities given below, check whether the first is proportional to the second. For proportional quantities, calculate the constant of proportionality.
i. Perimeter and radius of circles.
ii. Area and radius of circles.
iii. The distance travelled and the number of rotations of a circular ring moving along a line.
iv. The interest got in a year and the amount deposited in a scheme in which interest is compounded annually.
v. The volume of water poured into a hollow prism and the height of the water level.
Answer:
i. The perimeter of a circle is n times its diameter. That is 2n times the radius.
∴Perimeter = 2πr
∴ The perimeter and radius are proportional.
The constant of proportionality is 2π

ii. The area of a circle is π times the square of the radius.
∴ Area = πr²
∴ Area and radius are not proportional.

iii. When the ring rotates once, the distance travelled is equal to its perimeter. When it rotates twice the distance travelled is twice its perimeter. When it rotates ‘n’ times, the distance travelled is ‘n ‘ times the perimeter of the ring.

iv. If the amount deposited is P and the rate of interest is R.
Annual interest = I = PNR,
I = P × R (N = 1)
The amount and interest are propor-tional.
Constant of proportionality is rate of interest, R.

v. Volume of a hollow prism
= base × height
Volume, of water and height of water level, are proportional. Constant of proportionality is the base area.

Kerala Syllabus 9th Standard Maths Notes Question 2.
During rainfall, the volume of water falling in each square metre may be considered equal.
i. Prove that the volume of water falling in a region is proportional to the area of the region.
ii. Explain why the heights of rainwater collected in different sized hollow prisms kept near one another are equal.
Answer:
i. The volume of water falling in each square metre are equal.
Let the volume of the rain falling on the 1 square metre be k The volume of the rain falling on the 2 square metre = 2k
The volume of the rain falling on the 3 square metre = 3k
The volume of the rain falling on the x square metre y = kx Here x and y are proportional.

ii. Let x be the base area of the vessel and h be the height of the water collected, then the volume of water y = xh.
y/x =h
The volume of water collected is different in hollow prisms having different base area.

But \(\frac { volume }{ Area }\) is always equal to the height of water level. So the height at which rainwater collected is same.

Hsslive Guru Maths 9th Kerala Syllabus Question 3.
When a weight is suspended by a spring, the extension is proportional to the weight. Explain how this can be used to mark weights on a spring balance.

Answer:
Mark the pointer when no weight is hanging on it. Then mark the point when a constant weight is hanging on it. For example when we hang 1 kg on it a 2 cm extension is made. So we mark 1, 2, 3, 4 from the points marked first, i.e.,
2, 4, 6 and 8.
These distances are divided into 10 equal parts, then we can mark the points 1.1 kg and 2.4 kg etc.

Hss Live Guru 9 Maths Kerala Syllabus Question 4.
In the angle shown below, for different points on the slanted line, as the distance from the vertex of the angle changes, the height from the horizontal line also changes.

i. In the picture, perpendicular to the horizontal lines are drawn from points on the slanted line.
Δ ABP, Δ ACQ, Δ ADRand Δ AES are similar triangles. (Because these are a right-angled triangle with is common to all)
∴ Ratio of their sides are equal. That is sides are proportional.

Let k be the constant of proportionality
PB = k × AB, QC = k × AC
RD = k × AD, SE = k × AE ,
i.e., the change in height is proportional to the distance.

An online geometric sequence calculator helps you to find geometric Sequence, first term, common ratio calculator, and the number of terms.

Textbook Page No. 189

Class 9 Maths Chapter 12 Kerala Syllabus Question 1.
i. Prove that for equilateral triangles, area is proportional to the square of the length of aside. What is the constant of proportionality?
ii For squares, is area proportional to square of the length of a side? If so, what is the constant of proportionality?
Answer:
i. Δ ABC is an equilateral triangle
Consider their sides are ‘a’,
Draw a perpendicular line CD to AB
from C. In right-angled triangle ADC, according to the Pythagoras theorem AD² + DC² = AC²
DC² = AC² – AD²

ii. If x be the one side of a square then its area y = x²
Area of the square is proportional to the square of their sides.
Constant of proportionality = 1

Hsslive Guru Class 9 Maths Kerala Syllabus Question 2.
In rectangles of area one square metre, as the length of one side changes, so does the length of the other side. Write the relation between the lengths as an algebraic equation. How do we say this in the language of proportions?
Answer:
Area of the rectangle is the sum product of length and breadth.
Let × be the length and y be the breadth, then
x × y = 1 m²

x = 1/y = m
Example when y = 3 m
x = 1/3 meter
when y = 4 m
x = 1/4 meter
The algebraic equation is x = 1/y
i.e., length of rectangle is proportional to the reciprocal of the breadth, i.e., length of rectangle is proportional to its breadth.

Chapter 12 Maths Class 9 Kerala Syllabus Question 3.
In triangles of the same area, how do we say the relation between the length of the longest side and the length of the perpendicular from the opposite vertex? What if we take the length of the shortest side instead?
Answer:
Let ‘a’ be the large side, h be the length of perpendicular from opposite vertices, ‘A’ be the area, then

Length of larger side is inversely proportional to the length of perpendicular from the opposite vertex.
i.e., The length of small side is inversely proportional to the length of perpendicular line from the vertex of small side.

Hss Live Guru Class 9 Maths Kerala Syllabus Question 4.
In regular polygons, what is the relation between the number of sides and the degree measure of an outer angle? Can it be stated in terms of proportion?
Answer:
The sum of the exterior angles of all polygon is 360°. If ‘n’ is the number of sides.
Measure of an exterior angle = \(\frac{\text { Sum of exterior angles }}{\text { No. of sides }}\)
One outer angle = 360/n
(n = number of sides)
If the measure of an exterior angle is ‘x’.
x= \(\frac{1}{n} \times 360^{0}\)
One outer angle and number of sides are inversely proportional. The constant of proportionality is 1/n.

Hsslive Guru Maths Kerala Syllabus 9th Question 5.
A fixed volume of water is to flow into a rectangular water tank. The rate of flow can be changed by using different pipes. Write the relations between the following quantities as an algebraic equation and in terms of proportions.
i. The rate of water flow and the height of the water level.
ii The rate of water flow and the time taken to fill the tank.
Answer:
i. Let x be the rate of water flowing, y be the height of water in the tank and A be the base area of the tank, then
x = Ay
Height of the water level in the tank is proportional to the rate of the water flowing.

ii. If C is the volume of the tank, V be the volume of water flowing per second, the volume of water in ‘t’ second is given by
C = V × t
\(V=\frac{C}{t}=C \times \frac{1}{t}\)
That is the rate of water flow and the time taken for filling the tank are inversely proportional. C is the constant of proportionality.

Proportion Exam oriented Questions and Answers

Hsslive 9th Maths Kerala Syllabus Question 1.
the weight of 6 spheres of same size made of the same metal is 14 kg. When 9 more spheres are added the weight is 35 kg. Check whether the number of spheres and their weights are proportional.
Answer:
Weight of 6 spheres = 14kg
Ratio of number and weight = 6 : 14 = 3 : 7
Total number of sphers are added = 15
Total weight = 35kg
Ration of number and weight = 15 : 35 = 3 : 7
Since the ratios are equal, the number of sphers and their weights are propor-tional.

Question 2.
Raghu invested Rs. 60000 and Nazar Rs. 100000 and started a business. Within one month a profit of Rs. 4800 was obtained. Raghu took 1800 and Nazar took Rs. 3000 out of the profit obtained. What is the ratio of the investment? Is the investment and the profit divided proportionally?
Answer:
Ratio of investments = 60000 : 100000 = 6 : 10 = 3 : 5
Ratio of profit divided
1800 : 3000 = 18 : 30 = 3 : 5
Ratio of investments and Ratio of profit divided are equal. Hence they are proportional.

Question 3.
The two sides of a triangle having perimeter 10 m are 2\(\frac { 1 }{ 2 }\) m and 3 \(\frac { 1 }{ 2 }\) m.
What is the ratio of the length of the three sides of triangles?
Answer:
Perimeter of triangle = 10 m
First side = 3\(\frac { 1 }{ 2 }\) m = \(\frac { 5 }{ 2 }\) m
Second side = 3\(\frac { 1 }{ 2 }\) m = \(\frac { 7 }{ 2 }\) m
Third side = 4 m Ratio of three sides of a triangle
\(=\frac{5}{2}: \frac{7}{2}: 4 = 5: 7: 8\)

Question 4.
150 litres of water is flowing through a pipe in 6 minutes. If 200 litres of water flows through it in 8 minutes check whether the quantity of water and time of flow are proportional ?
Answer:
Quantity of water flowing in 6 minutes = 150 litres
Ratio between the quantity of water and timeofflow= 150: 6 = 25: 1
Quantity of water flowing in 8 minutes = 200 litres
Ratio between quantity of water and times of flow = 200 : 8 = 25: 1
Since the radios are equal, the amount of water flowing and the time of flow are proportional.

Question 5.
Unniyappam was made using 1 kg rice, 250 g plantain and 750g jaggery. Find the ratio between the ingredients.
Answer:
Rice = 1
kg= 1000g
Plantain = 250 g, Jaggery = 750 g.
Ratio between the ingredients
= 1000 : 250 : 750 = 4 : 1 : 3

Question 6.
Sathyan got Rs. 500 after working for 6 hours. Gopi got Rs. 400 after working for 4\(\frac { 1 }{ 2 }\) hours. Are the wages obtained proportional to the working time ?
Answer:
Ratio of working hours = 6 : 4 \(\frac { 1 }{ 2 }\)
= 12 : 9 = 4 : 3
Ratio of wages = 800 : 600 = 8 : 6 = 4 : 3
Since the ratios are equal, the working hours are proportional to the wages.

Question 7.
Are the length and breadth of a square having same perimeter inversely proportional?
Answer:
for a square the perimeter is 20 cm. so,
length + breadth = 10.
Let x be the length and y be the breadth then possible values of x and y are
x — y
9 — 1
8 — 2
7 — 3
Here xy is not a constant term i.e. the changes in the x and y is not in the firm of xy = kept
hence length and breadth are not in inversely proportional.

Question 8.
The weight of an object having mass 5 kg is 49 Newton. The weight of another object having mass 15 kg is 147 Newton. Check whether the mass and weight are proportional ? What is the constant of proportionality? What is the weight of an object having mass 8 kg ?
Answer:
Weight of 5 kg object = 49 N
Ratio between mass and weight = 5 : 49
Weight of 15 kg massed object = 147N
Ratio between mass and weight = 15:147 = 5:49
Mass and weight are proportional.
Weight of 8 kg massed object = 8 × 9.8 = 78.4N

Constant of proportionality = 9.8

Question 9.
The perimeter of a triangle is 60 cm. Sides are in the ratio 3: 4: 5. Then j find the length of the sides.
Answer:
Ratio of sides = 3 : 4 : 5
Therefore, the sides are the \(\frac { 3 }{ 12 }\), \(\frac { 4 }{ 12 }\) and \(\frac { 5 }{ 12 }\) part of the perimeter .
The perimeter is 60 cm, then the length of sides are

\(60 \times \frac{5}{12}=25 \mathrm{cm}\)

Question 10.
Will the dye which contains 10 L blue colour and 15 L white colour and another dye which contain 12L blue colour and 17 L white colour have j the same colour? Why?
Answer:
In the first dye , blue colour: white colour = 10 : 15 = 2 : 3
In the second dye, blue colour: white colour = 12: 17
The ratios are not same. Hence the both will not have same colour.

Question 11.
The face perimeter of some vessels in the shape of square prisms are equal. 12 litres of water can be filled in the vessel having height 15 cm. 16 litres of water can be filled in the vessel having height 20 cm. Check whether the volumes of the vessel and height are proportional. What is the constant of proportionality?
Answer:
Since the face perimeters are equal,
Volume = face perimeter × height
Volume of the vessel having height 15 cm = 12 litres.
Ratio of height to volume =15 : 12 = 5 : 4 Volume of the vessel having height 20 cm = 16 litres
Ratio of height to volume =20 : 16 = 5 : 4 Since the ratios are equal, height and volume are proportional.
\(\frac { volume }{ height }\) = \(\frac { 4 }{ 5 }\) constant of proportionality
Volume of the vessel having height 35 cm
= 35 × \(\frac { 4 }{ 5 }\) = 28 liters.

Question 12.
The ratio of carbon, sulphur and potassium nitrate to make gun powder is 3 : 2: 1. How much quantity of each is required to make 1.2 kg of gun powder ?
Answer:

Question 13.
Raju got Rs. 400 after working for 8 hours. Damu worked for 6 hours and got Rs. 300. Are the wages obtained proportional to the work time?
Answer:
Ramu working hours = 8 hour
Wages Raju obtained = Rs. 400
Damu working hours = 6 hour
Wages obtained by Damu = Rs. 300

∴ Wages obtained are proportional to the work time.

Question 14.
150 L of water flows through a pipe for 6 minutes. 200 L of water flows for 8 minutes through the same pipe. Are the time and amount of water flowing proportional?
Answer:
Ratio of quantity of water
= 150 : 200 = 15 : 20 = 3 : 4
Ratio oftime =6 : 8 = 3 : 4
The ratios are same , hence they are proportional.

Question 15.
A car with 5L of petrol travels a distance of 75 km. What is the proportionality constant between the distance travelled and the quantity of petrol? How much petrol is needed for travelling 180 km?
Answer:
Taking the distance travelled as x and quantity of petrol as y, then the constant of proportionality

The quantity of petrol needed to travel 180 km= 12 litre

Question 16.
The angles of a triangle are in the ratio 1 : 3: 5. How much is each angle of the triangle?
Answer:
Sum of the angles of a triangle = 180°
The angles of the traingle are \(\frac { 1 }{ 9 }\), \(\frac { 3 }{ 9 }\) and \(\frac { 5 }{ 9 }\)
parts of 180° , hence the angles are

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Kerala State Syllabus 9th Standard Maths Solutions Chapter 8 Polynomials

Polynomials Textual Questions and Answers

Textbook Page No. 123

Polynomial Root Calculator solves the roots of polynomials easily.

Polynomials Class 9 Kerala Syllabus Chapter 8 Question 1.
In rectangles with one side 1 centimetre shorter than the other, take the length of the shorter side as x centimetres.
i. Taking their perimeters as p(x) centimetres, write the relation between p(x) and x as an equation,
ii. Taking their areas as a(x) square centimetres, write the relation between a(x) and x as an equation.
iii Calculate p(l), p(2), p(3), p(4), p(5). Do you see any pattern?
iv. Calculate a(l), a(2), a(3), a(4), a(5). Do you see any pattern?
Answer:
Let x be the shorter side, then the other side will be (x+ 1).
i. Perimeter = 2[x + (x + 1)] = 2(2x + 1) = 4x + 2
That is, p(x) = 4x + 2

ii. Area = x(x + 1)
a(x) = x² + x
Area, a(x) = x² + x

iii. p(x) = 4x + 2
p(1) = 4 × 1 + 2 = 6
p(2) = 4 × 2 + 2 = 10
p(3) = 4 × 3 + 2= 14
p(4) = 4 × 4 + 2= 18
p(5) = 4 × 5 + 2 = 22
Perimeter is a sequence increasing by 4.

iv. a(x) = x² + x
a(1) = 1² + 1 = 2 = 1 × 2 = 2
a(2) = 2² + 2 = 6 = 2 × 3 = 6
a(3) = 3² + 3 = 12 = 3 × 4 = 12
a(4) = 4² + 4 = 20 = 4 × 5 = 20
a(5) = 5² + 5 = 30 = 5 × 6 = 30
Area is the product of x and the number one more than x.

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Polynomials Class 9 State Syllabus Chapter 8 Question 2.
From the four corners of a rect-angle, small squares are cut off and the sides are folded up to make a box, as shown below:

i. Taking a side of the square as x centimetres, write the dimensions of the box in terms of x.
ii. Taking the volume of the box as vfojcubic centimetres, write the relation between v(x) and x as an equation.
iii. Calculate \(\mathrm{V}\left(\frac{1}{2}\right) \quad, \mathrm{V}(1), \quad \mathrm{V}\left(1 \frac{1}{2}\right)\)
Answer:
If 1cm is the length of the small squares they are cut off, then the length of the maked box by folding it up = 7 – 1 – 1 = 5 cm Width = 5 – 1 – 1 = 3 cm, Height = 1 cm
If 2 cm is the length of the small squares that are cut off, then the length of the maked box by folding it up = 7 – 2 – 2 = 3 cm Width = 5 – 2 – 2 = 1 cm, Height = 2cm
i. If x cm is the length of the small squares that are cut off, then the length of the maked box by folding it up = 7 – x – x = (7 – 2x) cm
Width = 5 – x – x = (5 – 2x) cm
Height = x cm

ii. If volume of the box be v(x), then
v(x) = Length × Width × Height v(x) = (7 – 2x) (5 – 2x)x cm3

v (1) = (7 – 2 × 1) (5 – 2 × 1) 1 = (7 – 2) (5 – 2) × 1 = 5 × 3 × 1 = 15

Hss Live Guru 9th Maths Kerala Syllabus Chapter 8 Question 3.
Consider all rectangles that can be made with a 1-metre long rope. Take one of its sides as x centimetres and the area enclosed as a(x) square centimetres.
i. Write the relation between a(x) and x as an equation.
ii. Why are the numbers a(10) and a(40) equal?
iii. To get the same number as a(x), for two different numbers as x, what must be the relation between the numbers?
Answer:
i. 1 m = 100 cm
If one side is x cm, then the other side is 50 – x.
Area of rectangle = a(x) = x(50 – x)
= 50x – x² cm²
a(x) = 50x – x²

ii. a(10) = 50 × 10 – 10² = 500 – 100 = 400
a(40) = 50 × 40 – 40² = 2000 – 1600 = 400
10 and 40 are the sides of rectangle, so a(10) and a(40) are same .

iii. Numbers must be sides of rectangle. If we add the two numbers together we get the sum as half the length of the wire (50 cm) used to make it.

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Hsslive Guru 9th Maths Kerala Syllabus Chapter 8 Question 1.
Write each of the relations below in algebra and see if it gives a polynomial. Also, give reasons for your conclusion.
i. A 1-metre wide path goes around a square ground. The relation between the length of a side of the ground and the area of the path.

ii A liquid contains 7 litres of water and 3 litres of acid. More acid is added to it. The relation between the amount of acid added and the change in the percentage of acid in the liquid.

iii. Two poles of heights 3 metres and 4metres are erected upright on the ground, 5 metres apart. A rope is to be stretched from the top of one pole to some point on the ground and from there to the top of the other pole:

The relation between the distance of the point on the ground from the foot of a pole and the total length of the rope.
Answer:
i. Let the side of the ground be x then the side of the square including the path is x + 2 metres.
Area of path = Area of large square – Area of small square.
a(x) = (x + 2)² – x² = x² + 4x +4 – x²
= 4x + 4
Here a(x) is a polynomial. Here x is multiplied by 4 and 4 is added to it.

ii. Ratio of acid in the first fluid = 3/10 litre that is 30%
If x litres of acid is added to it, then change in the amount of acid is
\(=\frac{3+x}{10+x}\)
Change in the percentage of acid is = \(\frac{3+x}{10+x} \times 100 \%\)
\(\mathrm{b}(\mathrm{x})=\frac{300+100 x}{10+x} \%\)

iii.

In figure, AB = x m, BD = 5 – x
Total length of the wire = BC + BE

It is not a polynomial since it involves square root of variable x.

Sum and difference of cubes calculator, multiplying and dividing rational exponents not possible.

Kerala Syllabus 9th Standard Maths Notes Chapter 8 Question 2.
Write each of the operations below as an algebraic expression, find out which are polynomials and explain why.
i. Sum of number and it’s reciprocal
ii. Sum of a number and its square root.
iii. Product of the sum and difference of a number and its square root.
Answer:
i. Let the number be x, then the reciprocal is 1/x
sum = \(x+\frac{1}{x}\)
This is not a polynomial, because here the operation of reciprocal is involved.

ii. Let the number be x, then the square root is √x
sum = x + √x
This not a polynomial because here the square root is taken.

iii. Let x be the number
(x + √x) (x – √x) = x²– x
This is a polynomial.

Std 9 Maths Notes Kerala Syllabus Chapter 8 Question 3.
Find p(1) and p(10) in the following polynomials,
i. p(x) = 2x + 5
ii. p(x) = 3x² + 6x + 1
iii. p(x) = 4x³ + 2x² + 3x + 7
Answer:
i. p(x) = 2x + 5
p(1) = 2 × 1 + 5 = 2 + 5 = 7
p(10) = 2 × 10 + 5 = 20 + 5 = 25

ii. p(x) = 3x² + 6x + 1
p(1) = 3 × 12 + 6 x 1 + 1
= 3 + 6 + 1 = 10
p(10) = 3 × 10² + 6 × 10 + 1 = 300 + 60 + 1 = 361

iii. p(x) = 4x³ + 2x² + 3x + 7
p(1) = 4 × 13 + 2 × 12 + 3 × 1 +7
=4 + 2 + 3 + 7 = 16
p(10) = 4 × 10³ + 2 × 10² + 3 × 10 + 7
= 4000 + 200 + 30 + 7 = 4237

Hss Live Maths 9th Kerala Syllabus Chapter 8 Question 4.
Find p(0), p(1) and p(-1) in the fol-lowing polynomials,
i. p(x) = 3x + 5
ii. p(x) = 3x² + 6x + 1
iii. p(x) = 2x² – 3x + 4
iv. p(x) = 4x³ + 2×2 + 3x + 7
v. p(x) = 5x³ – x2 + 2x – 3
Answer:
i. p(x) = 3x + 5
p(0) = 3 × 0 + 5 = 5
p(1) = 3 × 1 + 5 = 8
p(-1) = 3 x-1 + 5 = 2

ii. p(x) = 3x² + 6x + 1
p(0) = 3 × 0² + 6 × 0 + 1 = 1
p(1) = 3 × 1² + 6 × 1 + 1 = 10
p(-1) = 3 × (-1)² + 6 × (-1) + 1 = -2

iii. p(x) = 2x² – 3x + 4
p(0) = 2 × 0² – 3 × 0 + 4 = 4
p(1) = 2 × 1² – 3 × 1 + 4 = 3
p(-1) =2 × (-1)² – 3 × (-1) + 4 = 9

iv. p(x) = 4x³ + 2x² + 3x + 7
p(0) = 4 × 0³ + 2 × 02 + 3 × 0 +7 = 7
p(1) = 4 × 1³ + 2 × 12 + 3 × 1 + 7 = 16
p(-1) = 4 × (-1)³ + 2 × (-1)2 + 3 × (-1) +7 =2

v. p(x) = 5x³ – x² + 2x – 3
p(0) = 5 × 0³ – 0² + 2 × 0 – 3 =-3
p(1) = 5 × 1³ – 1² + 2 × 1 – 3 = 3
p(-1) = 5 × (-1)³ – (-1)² + 2 × (-1) – 3 = -11

Scert Class 9 Maths Solutions Kerala Syllabus Chapter 8 Question 5.
Find polynomials p(x) satisfying
each set of conditions below.
i. First degree polynomials with p(1) = 1 and p(2) = 3
ii. First degree polynomials with p(1) = -1 and p(-2) = 3
iii. Second degree polynomials with p(0) = 0, p(1) = 2 and p(2) = 6.
iv. Three different second degree polynomials with p(0) = 0 and p(1) = 2.
Answer:
i. General form of a first degree poly-nomial is
p(x) = ax + b
Let p(1) = 1, then a × 1 + b = l
a + b = 1 ………. (1)
Let p(2) = 3, then a × 2 + b = 3
2a + b = 3 …….. (2)
(1) × 2, 2a + 2b = 2 …… (3)
(3) – (2), b = -1
From (1), a + -1 = 1,
a = 1 + 1 = 2 Polynomial p(x) = 2x – 1

ii. General form of a first degree poly-nomial is p(x) = ax + b
Let p(1) = -1, then a × 1 + b = -l
a + b = -1 ……….. (1)
Let p(-2) =3 , then a × (-2) + b = 3
-2a + b = 3 ………. (2)
(1) × 2, 2a + 2b = -2 ………. (3)
(2) + (3), 3b = 1, b = 1/3
From (1), a = 1/3 = -1

iii. General form of a second degree polynomial is
p(x) = ax² + bx + c
Letp(0) = 0, then a × 0² + b × 0 + c = 0
0a + 0b + c = 0.
c = 0 (1)
Letp(1) = 2 ,then a × 1² + b × 1 + c = 2
a + b + 0 = 2
a + b = 2 ………. (2)
Letp(2)= 6, then a × 2² + b × 2 + c = 6
4a + 2b = 6
2a + b = 3 (3)
(3) – (2), a = 1
From (2), 1 + b = 2, b = 2 – 1 = 1
Polynomial p(x) = x² + x

iv. General form of a second degree polynomial is p(x) = ax² + bx + c
Letp(0) = 0, then a x 0 + b x 0 + c = 0
0 + 0 + c = 0
c = 0
Letp(1) = 2, then a × 1² + b × 1 + c = 2
a + b + c = 2
a + b = 2
Selecting a and b such that a + b = 2 will give different polynomials.
a = 1, b = 1
a= 3, b =-l
a=4, b =-2
Three different second degree poly¬nomials are
p(x) = x² + x
p(x) = 3x² – x
p(x) = 4x² – 2x

Polynomials Exam Oriented Questions and Answers

Hss Live 9th Maths Kerala Syllabus Chapter 8 Question 1.
In a polynomial p(x) = 2x³ + ax² – 7x + b.
p(1) = 3, p(2) = 19. Then find the value of ‘a’ and ‘b’?
Answer:
p(x) = 2x³ + ax² – 7x + b
We have p(1) = 3.
p(1) = 2 x 1³ + a x 1² – 7 x 1 + b = 3
2 + a – 7 + b = 3
a + b – 5 = 3
a + b =8 ………. (1)
We have p(2) = 19.

p(2 ) = 2 x 2³ + a x 2²– 7 x 2 + b = 19
16 + 4a – 14 + b = 19
4a + b + 2 = 19
4a + b = 17…….. (2)
From equation (1), (2)
(2) – (1) 4a + b = 17
\(\frac{a+b=8}{3 a=9}\)
a = 9/3 = 3
From equation (1)
a + b = 8
3+b =8
b =5

Hsslive Class 9 Maths Kerala Syllabus Chapter 8 Question 2.
In a polynomial p(x) = 2x³ + 9x² + kx + 3, p(-2) = p(-3). Find the value of k.
Answer:
p(x) = 2x³ + 9x² + kx + 3
p(-2) = 2(-2)³ + 9(-2)² + k(-2) + 3
= -16 + 36 – 2k + 3 = 23 – 2k
p(-3) = 2(-3)³ + 9(-3)² + k(-3) + 3
= -54 + 81 – 3k + 3 = 30 – 3k
We have p(-2) = p(-3),
23 -2k = 30 -3k
-2k + 3k = 30 – 23
k = 7

Maths Kerala Syllabus Std 9 Notes Chapter 8 Question 3.
From the polynomial p(x) = 2x – 3x +1, find p(0), p(1) and p(-1).
Answer:
p(x) = 2x² – 3x +1
p(0) = 2(0)² – 3(0) + 1 = 1
p(1) = 2(1)² – 3(1) +1 = 2 – 3 + 1 = 0
p(-1) = 2(-1)² – 3(-1) + 1 = 2 + 3 + 1 = 6

Kerala Syllabus 9th Standard Maths Notes Pdf Chapter 8 Question 4.
In a polynomial p(x) = 2x³ – 7x² + kx + 20,
p(2) = p(3).
a. Find the value of k.
b. Using the value of k, write the polynomial.
c. Find p(1).
Answer:
a. p(2) = 2 × 2² – 7 × 2² + k × 2 + 20
= 16 – 28 + 2k + 20 = 8+ 2k
p(3) = 2 × 3³ – 7 × 3²+ k × 3 + 20
= 54 – 63 + 3k + 20 = 11 +3k
P(2) = P(3)
8 + 2k = 11 + 3k
k = -3

b. p(x) = 2x³ – 7x² -3x + 20

c. p(1) = 2 – 7 – 3 + 20 = 12

Chapter 8 Polynomials Answers Kerala Syllabus Question 5.
Simplify the followi ng
i. (2x + 1) (3x + 4) (4x + 3) (3x + 4)
ii. (3x + 4)² – (2x – 1) (3x + 4)
Answer:
i. (2x + 1) (3x + 4) (4x + 3) (3x + 4)
= (3x + 4) [(2x + 1 + 4x + 3)]
= (3x + 4)[6x + 4]
= (3x x (6x) + (3x) x (4)+ 4x(6x) + 4 x 4
= 18x² + 12x + 24x + 16
= 18x² + 36x + 16

ii. (3x + 4)² – (2x – 1) (3x + 4)
= (3x + 4) [3x + 4 – (2x – 1)]
= (3x + 4)[3x + 4 – 2x + 1] = (3x + 4) (x + 5)
= 3x² + 15x + 4x + 20
= 3x² + 19x + 20

Hsslive Guru Maths 9th Kerala Syllabus Chapter 8 Question 6.
7x³ – 4x² – x + 4 is a polynomial.
a. Write the terms of the polynomial.
b. Write the coefficient of x².
c. Write the constant terms of the polynomial.
d. What is the degree of the polynomial ?
Answer:
a. Terms = 7x², -4x², -x, 4
b. Coefficient of x² = -4
c. Constant term = 4
d. Degree of the polynomial = 3

Hss Live Guru Class 9 Maths Kerala Syllabus Chapter 8 Question 7.
In the polynomial p(x)=3x² – ax + 1,
Find ‘a’ satisfying p(1) = 2.
Answer:
p(x)=3x² – ax + 1
p(1) = 3(1)² – (a × 1) + 1 = 3 – a + 1 =4 – a
Given p(1) = 2
That is, 4 – a = 2 , a = 4 – 2 = 2

Hsslive Maths Class 9 Kerala Syllabus Chapter 8 Question 8.
If p(x)= x³ + 2x² – 3x + 1 and q(x) = x³ – 2x² + 3x + 5
a) Find p(x) + q(x). What is its de-gree?
b) Find p(x) – q(x). What is its de-gree?
Answer:
a) p(x) = x³ + 2x² – 3x + 1,
q(x) = x³ – 2x² + 3x + 5
p(x) + q(x)
= x³ + 2x² – 3x + 1 + x³ – 2x² + 3x + 5
= 2x³ + 6
Degree = 3

b) p(x) – q(x)
= x³ + 2x² – 3x + 1 – (x³ – 2x² + 3x + 5)
= x³ + 2x² – 3x + 1 -x³ + 2x² – 3x – 5
= 4x² – 6x – 4
Degree = 2

9th Std Maths Notes Kerala Syllabus Chapter 8 Question 9.
A right-angled triangle of perpendicular sides 3 cm and 4 cm are ex-tended equally then get another large right-angled triangle. Write the algebraic form of the hypotenuse of the large right-angled triangle.
Answer:
Hypotenuse 2 = base 2 + height 2
(According to pythagorus theorem)
Let the perpendicular side be x cm Length of other sides = 3 + x cm, 4 + x cm

Question 10.
In the polynomial p(x)= 3x² – 4x + 7, check whether p(1) + p(2) = p(3) and P(2) × p(3) = p(6).
Answer:
p(x) = 3x² – 4x + 7
p(1) = 3(1)² – 4(1) + 7 = 3 – 4 + 7 = 6
p(2) = 3(2)² -4(2) + 7 = 12 – 8 + 7 = 11
p(3) = 3(3)² – 4(3) +7 = 27 – 12 + 7 = 22
p(l) + p(2) ≠ p(3)
p(6) = 3(6)² -4(6) + 7 = 108 – 24 + 7 = 91
P(2) × p(3) ≠ p(6)

Question 11.
In the polynomial p(x)= 2x² + ax² – 7x + b,
p(1) = 3 and p(2) = 19. Find a and b.
Answer:
p(x)=2x² + ax² – 7x + b
p(l)=2 x 13+ a x 12 – 7 x 1 +b
=2 + a – 7 + b =a + b + 2 – 7 = a + b – 5
p(1) = 3, a + b – 5 = 3
a + b = 8……… (1)
p(2) = 2 x 2³ + a x 2² – 7 x 2 + b
= 16 + 4a – 14 + b = 4a + b + 2
p(x) = 19
4a + b + 2 = 19
4a + b = 17 ……….. (2)
(2) – (1)
3a = 9, a = 3 a + b – 5 = 3
From equation (1),
a + b = 8
3 + b = 8
b = 5

Question 12.
If p(x) = x² + 3x + 1 and q(x) = 2x – 4, then
i. Find the degree of the polyno-mial p(x) q(x).
ii. If the degree of p(x) × r(x) is 5, then find the degree of the polynomial r(x) .
iii. If p(x) is a third degree poly-nomial and q(x) is a fourth de-gree polynomial then find the degree of p(x) × q(x).
Answer:
i. p(x) = x² + 3x + 1 , q(x) = 2x – 4
p(x) × q(x) = (x² + 3x + 1) (2x – 4)
The degree of p(x) x q(x) is 3.

ii. p(x) = x² + 3x + 1 is a second degree polynomial.
If the degree of the polynomial p(x) × r(x) is 5, then p(x) × r(x) must have a term of x⁵.
x² × x³ = x⁵
So, r(x) is a third-degree polynomial.

iii. If p(x) is a third-degree polynomial, then p(x) must have a term of x³.
q(x) is a fourth-degree polynomial, then q(x) must have a term of x⁴
Then in p(x) × q(x), must have a term of x³ × x⁴= x⁷
So p(x) × q(x) is a seventh-degree polynomial. In general,
If p(x) is an m^th degree polynomial and q(x) is an n,h degree polynomial then p(x) × q(x) is an (m + n),h degree polynomial.

Question 13.
If p(x) = 4x² + 3x + 5 and q(x) = 3x² – x – 7, then find p(x) + q(x) and p(x) – q(x).
Answer:
p(x) + q(x) = (4x² + 3x + 5) + (3x² – x – 7)
= 4x² + 3x² + 3x – x + 5 – 7
= 7x² + 2x – 2
p(x) – q(x) = (4x² + 3x + 5) – (3x² – x – 7)
= 4x² + 3x + 5 – 3x² + x + 7
= x² + 4x + 12

Question 14.
In polynomial p(x) = ax³+ bx² + cx + d, p(l) = p(-l). Prove that a + c = 0.
Answer:
p(x) = ax³ + bx² + cx + d
p(l) = a(1)³ + b(1)² + c(1) + d
= a+ b + c + d
p(-1) = a(-1)³ + b(-1)² +c(-1) + d
= -a + b – c + d
p(1) = p(-1) is given ie, a + b + c + d = -a + b – c + d
2a + 2c = 0; 2(a + c ) = 0;
ie, a + c = 0

Question 15.
If we divide a polynomial by (x-2) we get quotient as x² + 1 and remainder as 5. Find the polynomial.
Answer:
Polynomial = (x² + 1) (x – 2) + 5
= x³ + x – 2x² – 2 + 5 ,
= x³ – 2x² + x + 3

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Kerala State Syllabus 9th Standard Maths Solutions Chapter 13 Statistics in Malayalam Medium

Statistics Questions and Answers in Malayalam

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फूलों का शो Textual Questions and Answers

9th Hindi Notes Kerala Syllabus प्रश्ना 1.
किसी एक त्योहार की ऐतिहासिक या सांस्कृतिक पृष्ठभूमि पर लेख लिखें।

उत्तर:
ओणम : फूलों का त्योहार
ओणम केरल का त्योहार है। केरल के लोग बड़े उत्साह से ओणम’ मनाते हैं। ‘ओणम’ के साथ राजा महाबली और वामन-अवतार की पौराणिक कथा जुड़ी हुई है। राजा महाबली बहुत ही न्यायप्रिय और परम दानवीर था। उसके शासन में प्रजा बहुत सुखी थी। ऐसा विश्वास है कि ‘ओणम’ के दिन महाबली अपनी प्रजा को देखने आते हैं। प्रजा उनका स्वागत-सत्कार करती है। वैसे भी ‘ओणम’ को ‘फूलों का पर्व’ कहा जाता है। क्योंकि ओणम के दस दिन पहले से ही घरों के आँगन में फूलों से रंगोली बनाते हैं।

 

फूलों का शो Summary in Malayalam and Translation

 

फूलों का शो शब्दार्थ

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Kerala State Syllabus 9th Standard Social Science Solutions Part 2 Chapter 9 Economic Systems and Economic Policies

Economic Systems and Economic Policies Textual Questions and Answers

9 Class Social Science Notes Kerala Syllabus Question 1.
Name the different economic systems
Answer:

• Capital economy
• Socialist economy
• Mixed economy

9th Standard Social Science Notes Pdf In English Question 2.
What do you mean by capitalist economy? Point out its important features.
Answer:
Capitalist economy is the economy in which the ownership of means of production is with private individuals who work with the motive of making profits. Other features of capitalist economy are as follows:

• Freedom for the entrepreneurs to produce any commodity
• Right to private property
• Profit motive
• Transfer of wealth to legal heir
• Free market with no control over price
• Consumers sovereignty
• Competition among entrepreneurs to sell products.

9th Standard Social Science Guide Kerala Syllabus Question 3.
In a capitalist economy, major economic decisions are taken by
Answer:
Price Mechanism

Scert Class 9 Social Science Notes Malayalam Medium Question 4.
Define socialist economy. What are the features of a socialist economy?
Answer:
Socialist economy is an economic system in which the means of production are owned by the public sector. This economic system works on centralised planning. Let us analyse other features of a socialist economy:

• Activities aimed at social welfare
• Absence of private entrepreneur
• Absence of private ownership of wealth and transfer of wealth to legal heir
• Economic equality

Scert Class 9 Social Science Notes Pdf Kerala Syllabus Question 5.
Point out demerits of socialism
Answer:

• Less investment.
• Absence of private ownership of wealth
• Limited choice of products.

Social Science Class 9 Notes State Syllabus  Question 6.
What is mixed economy? What are its features?
Answer;
Mixed economy is the economy that has certain features of both the capitalist economy and socialist economy. India has adopted mixed economy. Let us analyse some of the features of a mixed economy.

• Existence of both private and public sectors.
• Economy works on the principle of planning
• Importance to welfare activities
• Existence of both freedom of private ownership of wealth and economic control.

9th Standard Social Science Notes Pdf Kerala Syllabus Question 7.
Categorise the following countries into different economic systems.
1. USA
2. China
3. Cuba
4. India
5. Sri lanka
6. U.K.
Answer:
Capitalist economy:
1. USA
6 U.K.
Socialist economy
2. China
3. Cuba
Mixed Economy
4. India
5. Sri lanka

Class 9 Social Science Notes Kerala Syllabus Question 8.
Relaxation of government control and influence over the economic activities in a country is:
a) Privatisation
b) Liberalisation
c) Globalisation
d) Marketisation
Answer:
b) Liberalisation

Social Notes For Class 9 State Syllabus Question 9.
India adopted new economic policy in
Answer:
1991

9 Class Social Science Notes Malayalam Medium Question 10.
Mention the changes that were brought about as a result of liberalisation.
Answer:

• Relaxation of control in setting up industries.
• Reduction of import tariff and tax.
• Changes in foreign exchange rules.
• Abolition of market control
• Permission of foreign investment in many sectors.
• Reduced the role of government in the basic industries and basic infrastructure development.

Social Guide For Class 9 State Syllabus Question 11.
Prepare a note on globalization.
Answer
Globalisation is the economic integration and interdependence of nations as a result of free flow of capital, labour, goods and services, and exchange of technology irrespective of boundaries.

Towards the end of the twentieth century, developments in computer, mobile phones, internet, etc helped in improved communication. Container ships, flights, bullet trains, etc. contributed to the fast means of transportation these changes helped globalization.

Social Science Textbook Class 9 Kerala Syllabus Pdf Download Question 12.
WTO was formed on
Answer;
1^st January 1995

Social Science Class 9 Kerala Syllabus Question 13.
Name Bretton woods twins
Answer;
IMF and World Bank

Let Us Assess

Social Science Class 9 Malayalam Medium Question 14.
Why is the capitalist economy known as Market economy?
Answer;
Capital economy functions aiming profit, complete each other and there is no control over prices. So capital economy is also called market economy.

State Syllabus Class 9 Social Science  Question 15.
Elucidate the concept of police state.
Answer;
In capital economy there is very little government intervention in the economic activities in capitalist economy. The main functions of the nation are to maintain law and order and to defend the country from foreign invasion. Such nations are known as police state.

9th Class Social Science Notes Kerala Syllabus Question 16.
Planning is the main feature of the socialist economy. Explain.
Answer;
It is through planning that the basic problems of an economy are solved by the socialist economy. Regarding what to produce, how to produce and for whom to produce is done to ensure social welfare. Goods and services are produced and distributed accordingly. The amount of goods and services required for the society is calculated and production is done accordingly.

Std 9 Geography Kerala Syllabus Question 17.
Today, purely capitalist and socialist economies can not be found in the world. Substantiate.
Answer;
Pure form of capitalist or socialist economies can not be seen anywhere today. There is government intervention in capitalist economies like those in the United Kingdom and the United States of America. Private ownership of wealth and freedom of market have been permitted in socialist countries like China and Cuba.

Class 9 Social Science Notes State Syllabus Question 18.
List the actions taken by the government of India as a part of liberalisation.
Answer:

• Relaxation of control in setting up industries.
• Reduction of import tariff and tax
• Changes in foreign exchange rules
• Abolition of market control
• Permission of foreign investment in many sectors.
• Reduced the role of government in the basic industries and basic infrastructure development.

Question 19.
Make a note on the working of multinational companies.
Answer:
Multinational companies are those companies registered in the home country but operating in many countries. These companies with high technology and huge capital viewed neoliberalisation as an opportunity. Instead of producing goods in a country and exporting it to other countries, the multinational companies have invested their capital in developing countries so that raw materials, labour and market available there can be used in their favour.

Question 20.
Do you support globalisation policies? Why?
Answer:
Globalisation have its own merits and demerits. As part of globalisation a wide variety of products, ability to use the most advanced technology, increased competition in the market, more employment opportunities and increase in national income are possible and at the same time increase economic disparity, excess exploitation of natural resources, fall in the price of domestic products lose of government control over economic system happens. Even though globalisation is good for developed nations but it has adverse effect for developing nations.

Question 21.
What are the features of mixed economy?
Answer:

• Existence of both public and private sectors.
• Economy works on the principle of planning.
• Importance to welfare activities.
• Existence of both freedom of private ownership of wealth and economic control.

Question 22.
What is known as globalisation?
Answer:
Globalisation is the economic integration and inter-dependence of nations as a result of free flow of capital, labour, goods and services and exchange of technology irrespective of boundaries.