Activity Card BIOLOGY Class: 9
Concept area: Breathing process - Inspiration, Expiration.
Activity -12
Observe the illustration and complete the table given below.
The basis of respiratory movements is the expansion and contraction of the thoracic
cavity. The illustruation given below shows the mechanisms that facilitate this re-
peated rhythmic action.
Ribs raise due to Inspiration Ribs lower due to Expiration
the contraction of the relaxation of
intercostal muscles
intercostal
muscles
Diaphragm Diaphragm
contracts relaxes
Volume of the thoracic cavity Volume of the thoracic cavity
increases. The pressure in the decreases. The pressure in
thoracic cavity becomes lower the thoracic cavity becomes
than the atmospheric pressure. greater than the atmospheric
Air enters in. pressure. Air is expelled out.
Inspiration Diaphragm Expiration
Ribs
.......................................... ..........................................
.......................................... Volume in the thoracic cavity ..........................................
.......................................... Pressure in the thoracic cavity ..........................................
.......................................... ..........................................
.......................................... Air ..........................................
GENERAL EDUCATION DEPARTMENT SAMAGRA SHIKSHA, KERALA S C E R T
Activity Card BIOLOGY Class: 9
Concept raea : Smoking and its ill-effects
Activity-13
School health club conducted a seminar in order to develop a positive attitude
in children against smoking. A few slides presented by the pulmonologist in
the seminar is given.
Observe the given slides and answer the questions.
Lung Cancer Lung Cancer
Carcinogens present in tobacco
cause lung cancer.
Slide 1 Normal lungs Lungs affected
by cancer
Slide 2
Emphysema Emphysema
Alveoli rupture due to the loss
of elasticity, by the deposition
of toxic substances contained Normal Alveoli affected
alveoli by emphysema
in tobacco. This reduces the
respiratory surface and reduces Slide 4
Slide 3
vital capacity.
Bronchitis Bronchitis
The tar, carbon monoxide, etc.,
in tobacco leads to the deposi- Normal
tion of mucus and the swelling bronchioles
of lungs due to the proliferation
Affected
of germs in the alveoli. bronchioles
Slide 5 Slide 6
a) What are the respiratory disorders caused by smoking?
............................
GENERAL EDUCATION DEPARTMENT SAMAGRA SHIKSHA, KERALA S C E R T
Activity Card BIOLOGY Class: 9
A person who was addicted to smoking consulted a doctor after experiencing a fre-
quent cough with sputum and difficulty in breathing. After examining the patient, the
doctor diagnosed the disease as bronchitis
What is bronchitis?What causes this disease in smokers?
GENERAL EDUCATION DEPARTMENT SAMAGRA SHIKSHA, KERALA S C E R T
Activity Card BIOLOGY Class: 9
UNIT 5
EXCETION FOR HOMEOSTASIS
Concept Area- Kidneys and related parts
Activity-14
Analyse the diagram and the note given below and complete the illustration
Renal artery
Kidney
Renal vein
Ureter
Urinary bladder
Urethra
Human beings have a pair of kidneys.
They are bean-shaped and are located in the abdominal cavity on either side
of the vertebral column
Renal artery which is a branch of aorta brings blood to the kidneys
The filtered blood reaches the venacava through renal vein
Urine formed in the kidneys reaches the urinary bladder through the uereters
and is expelled out through the urethra.
Position shape Kidneys Blood vessel which
brings blood
a) .......................
b) ......................
tubule which carry
urine to the urinary Blood vessel which carry
blood out side the kidneys
bladder
c) ....................... d) .......................
GENERAL EDUCATION DEPARTMENT SAMAGRA SHIKSHA, KERALA S C E R T
Activity Card BIOLOGY Class: 9
Concept: Internal structure of Kidney
Activity 15
Analyse the description given below and answer the questions.
The light coloured outer part of the kidney is cortex. Ultra filters of the nephrons
are found here.
The dark coloured inner part of the kidney is medulla. Long tubules of nephrons
are found here
The funnel shaped part of the ureter is pelvis. Urine from the filters flows into
pelvis.
a) Identify the parts given below and label them in the diagram indicating their names.
i Part where ultrafilters of nephron are seen
ii Region where urine from the filters flows into.
iii Part where long tubules of the nephrons are found.
iv Part which carry urine to the urinary bladder
Activity-16
Analyse the descripton and answer the questions
Each kidney is comprised of about 12 lakh ultrafilters.These are called nephrons.
Nephrons are the basic structural and functional units of kidneys
Bowmann's capsule- The double walled cup shaped structure at one end of the
nephron. The space between the two walls is called capsular space.
GENERAL EDUCATION DEPARTMENT SAMAGRA SHIKSHA, KERALA S C E R T
Activity Card BIOLOGY Class: 9
Renal tubule- The long tubule which connects the Bowmann’s capsule and the
collecting duct.
Collecting duct- The part where the renal tubules enter.
Afferent vessel- The branch of the renal artery which enters the Bowmann's
capsule.
Glomerulus- The region where afferent vesssel enters the Bowmanns
capsule and splits into minute capillaries.
Efferent vessel- The blood vessel that comes out of the Bowmann’s capsule.
Peritubular capillaries-Blood capillaries seen around the renal tubules as the
continuation of the efferent vessel.
Observe the illustration and answer the questions given below:
A)
B)
C)
D)
a) Identify the figure
b) Identify the parts labelled as A, B, C and D.
A. .................................
B. .................................
C. ..................................
D. ..................................
GENERAL EDUCATION DEPARTMENT SAMAGRA SHIKSHA, KERALA S C E R T
Activity Card BIOLOGY Class: 9
UNIT 6
THE BIOLOGY OF MOVEMENT
Concept area - Importance of exercise
Activity-17
Analyse the description given below and answer the questions
Excercise increases the blood circulation through out the body . At the same time
it makes the cardiac muscles strong. Stored fat is broken down and thereby obesity
is reduced. As the body sweats more , waste is eliminated through sweat.
Exercise makes the exchange of respiratory gases more efective and increases
the vital capacity. More capillaries are formed in the muscles and increases the
efficiency of muscles.
How does exercise become beneficial for the health of the organ systems given below?
a) Respiratory system: .........................................................................................
......................................................................................................................................................
......................................................................................................................................................
......................................................................................................................................................
b) Circulatory system: ................................................................................................
.......................................................................................................................................................
.......................................................................................................................................................
.......................................................................................................................................................
Concept Area- Different Types Of Muscles
Activity-18
Analyse the description and figures and answer the questions
Muscular tissue is one of the basic tissues seen in animals. They have the capacity
to contract and relax. The main function of the muscles is to provide strength and
also to enable body movements. Muscles are classified as Skeletal muscles, Cardiac
muscles and Smooth muscles. Muscles seen attached to bones are called Skeletal
muscles or striated muscles. They have cylindrical cells, have striations and make
voluntary movements possible. Smooth muscles have spindle shaped cells, do not
have striations and make involuntary movements possible.
GENERAL EDUCATION DEPARTMENT SAMAGRA SHIKSHA, KERALA S C E R T
Activity Card BIOLOGY Class: 9
They are seen in internal organs like stomach, small intestine and in blood vessels.
Muscles seen in the walls of the heart are called cardiac muscles. They have
branched cells, striations are seen and make involuntary movements possible.
Skeletal muscles Smooth muscles Cardiac muscles.
Tablulate the following giving suitable heading
Seen in the walls of the heart
Spindle shaped cells
Seen attached to bones
No striations
Enables voluntary movements
Branched cells
GENERAL EDUCATION DEPARTMENT SAMAGRA SHIKSHA, KERALA S C E R T
Activity Card BIOLOGY Class: 9
Concept Area: Different Types of Joints
Activity-19
Analyse the illustration and answer the questions
Pivot joint Hinge joint
Enables Enables movement in
movements one direction only
around the axis
Ball and socket Gliding joint
joint The flattened surface of
Globular tip of the the two bones which
bone is attached glide each other enables
to a cup shaped slipping type movement
socket.
Part where the Name of the joint Peculiarities
joint is present
Elbow
First vertebra of
the vertebral
column
Shoulder bone
GENERAL EDUCATION DEPARTMENT SAMAGRA SHIKSHA, KERALA S C E R T
Activity Card BIOLOGY Class: 9
UNIT 7
DIVISION FOR GROWTH AND REPRODUCTION.
Concept Area: Cell Cycle
Activity-20
Analyse the illustration and description given below, and answer the questions.
Cell Division
Growth begins from the single cell called zygote. A cell attains its complete growth
during interphase (Preparatory phase). A 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 the cell cycle. Growth of the body is brought
about by cell division and cell growth. There are two types of cell division - mitosis and
meiosis.
Mitosis
Type of cell division that helps in the growth of the body.
There is no change in the number of Chromosomes..
Karyokinesis Cytokinesis
Division of Nucleus Division of Cytoplasm
Prophase Chromatin reticulum condensed to form chromosomes.
Metaphase Chromosomes duplicate.
Nucleolus and nuclear membrane disappear.
Spindle fibres arise from the centrioles.
Chromosomes get arranged at the centre of the cell in a single row .
Spindle fibres get attached to the centromere of the chromosome.
Anaphase Chomatids separate.
Chromatids move to each pole of the cell.
Chromatids transform to daughter chromosomes at each pole.
Telophase Nucleous and nuclear membrane reappear.
Daughter nuclei are formed.
Chromosomes become chromatin reticulum.
GENERAL EDUCATION DEPARTMENT SAMAGRA SHIKSHA, KERALA S C E R T
Activity Card BIOLOGY Class: 9
a) List the major events that take place during Interphase.
b) Identify the phase in the division of nucleus in which the following events take place.
i) Daughter chromosomes are formed.
ii) Chromosomes get arranged in a row
at the centre of the cell.
iii) Spindle fibres arise.
iv) Daughter nuclei are formed.
c) Find the correct sequence in connection with division of nucleus.
i) Anaphase Telophase Prophase Metaphase
ii) Prophase Metaphase Anaphase Telophase
iii) Prophase Metaphase Telophase Anaphase
Concept Area: Types of Cell division - Comparison
Activity-21
Compare Mitosis and Meiosis with the help of reading card and complete the table.
Meiosis
• Meiosis occurs in the germinal cells of the reproductive organs.
• It helps in the formation of gametes.
• During meiosis a cell divides twice continuously.
• These divisions are known as Meiosis I and Meiosis II respectively.
• Meiosis II is similar to mitosis.
GENERAL EDUCATION DEPARTMENT SAMAGRA SHIKSHA, KERALA S C E R T
Activity Card BIOLOGY Class: 9
In which kind of cell does it take place? Mitosis Meiosis
Somatic cells Germinal cells
The change in the number of chromosome
Number of daughter cells.
Importance
Concept Area: Division of Cytoplasm.
Activity: 22
Analyse the illustration , compare the division of cytoplasm in animal
cell and plant cell and prepare a short note.
Animal cell Cytokinesis Plant cell
The plasma membrane Small vesicles formed between the
invaginates at the centre daughter nuclei join together to form cell
of the cell. The cell plate. The cell plate extends to both
divides and daughter sides and joins with the plasma
cells are formed. membrane and daughter cells are
formed
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
GENERAL EDUCATION DEPARTMENT SAMAGRA SHIKSHA, KERALA S C E R T
MATHEMATICS
9
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Unit -1
Area
Activity 1
Look at the triangle given below. C
6
A D B
In the figure AB = 8 cm 8
CD = 6 cm
Find the area of the triangle.
Activity 2
A line is drawn through the vertex A parallel to the base of the triangle ABC.
Write the length of PQ (distance between parallel lines)
In this way, how many triangles of same area can be drawn?
PQ = cm
AM = cm
BC = cm
Area of the triangle ABC sq. cm
AP
6
BM CQ
8
Samagra Shiksha, Kerala S C E R T
General Education Department
Activity Card Mathematics Class : 9
ADP
6
B MC Q
8
What is the area of the triangle BCD?
Draw another triangle of the same area.
How many such triangles can be drawn?
What are the general properties of these triangels?
Activity 3
Now find the answers of problems given below.
1. In the figure , the length of sides of rectangle ABCD are 10 cm and 6 cm.
a) What is the area of the rectangle? A M D
6
b) What is the area of the triangle BCM? C
Activity 4 B 10
Draw a triangle with same area of the
triangle given in the figure.
(Hint : Draw a line parallel to QR through P).
P
QR
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Activity 5 C
Look at the figure 6
The heights of the triangle ADC
and triangle BDC are
6 centimeters each.
AM D 5B
5
Area of the triangle ADC 1 5 6 15 sq. cm
2
Area of the triangle BDC 1 5 6 15 sq. cm
2
Ratio of areas of these triangles
15 : 15 = 1 : 1
Ratio in which CD divides AB is 5 : 5 = 1 : 1
That is, a line CD drawn from the vertex C to its opposite side divides the side and the
area of the triangle in the same ratio.
Look at the picture below. C
6
A MD B
64
Area of the triangle ADC
Area of the triangle BDC
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Ratio of areas of these triangles
Ratio in which CD divides AB
Activity 6
See the problem
D
42 sq. cm 14 sq. cm
C
A M
28 sq. cm
B
The ratio of areas of triangles lies above the diagonal AC of the quadrilateral on the
figure.
= 42 : 14
= 3:1
If MC= 7 cm, then
AM = cm
What is the area of triangle AMB?
If DM = 5 cm, then cm
BM =
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Activity 7
Have you observed the given quadrilateral? Shall we draw a triangle with same area of
this quadrilateral?
Step 1
Draw diagonal AC and divide the quadrilateral into two triangles.
Step 2
Draw a line through D, parallel to AC.
Step 3
Draw another triangle with base AC and same area of triangle ADC.
In which context, the newly drawn triangle and triangle ABC joins to get a large triangle?
(Hint : Extend the side BC of the quadrilateral)
D
C
A
B
Draw a quadrilateral on a paper and draw a triangle with same area.
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Unit -2
Decimal Forms
Activity 1
Write the fractions below in decimal form:
Fraction Write the denominator Make the Fraction with Decimal
as the product of prime
denominator denominator as form
factors 2 and 5
as the power of 10 power of 10
13 13 13 5 65 0.65
20 2 2 5 2255 100
22
25 5 5
33
4 22
1 1 55 5
8 222555
77
40 2 2 2 5
21 21 2 2 2
125 5 5 5 2 2 2
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Activity 2
To find the decimal form of 1 1 1 a method is given below:
2 22 23
1 15 5 0.5
2 25 10
1 1 52 25 0.25
22 22 52 100
1 1 53 125 0.125
23 23 53 1000
1 1 1 5 25 125 0.5 0.25 0.125 0.875
2 22 23 10 100 1000
Like this, find the decimal form of 1 1 1
5 52 53
1 1 2 2 0.2
5 5 2 10
1 1 22 .................... .......................
52 52 22
1 .................... ....................... .......................
53
1 1 1 .................... ....................... ....................... .......................
5 52 53
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Activity 3
To find the decimal form of 1 a method is given below:
3
1 1 10 1 3 1 3 1 1 3 1
3 10 3 10 3 10 30 3 10 30
1 1 100 1 33 1 33 1 1 33 1
3 100 3 100 3 100 300 3 100 300
1 1 1000 1 333 1 333 1 1 333 1
3 1000 3 1000 3 1000 3000 3 1000 3000
Summarizing,
the fractions, 3 , 33 , 333 get closer and closer to 1
10 100 1000 3
Hence the fractions with decimal forms 0.3, 0.33, 0.333 ... get closer and closer to 1
3
We write this fact in short as 1 = 0.333 ...
3
Like this, find the decimal form of 1
9
1 1 10 1 1 1 1 1 1 1 1
9 10 9 10 9 10 90 9 10 90
1 1 100 1 11 1 11 1 1 11 1
9 100 9 100 9 100 900 9 100 900
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
1 ............
9
Activity 4
Write the results of the operations below as decimals:
1. 0.111... + 0.222...= 0.111... + 0.222...= 1 2 3 0.333...
9 9 9
2. 0.333... + 0.666...= 0.333... + 0.666...= 36 ...........
99
3. (0.333...)2 = (0.333...)2 = 3 2 ..............
9
4. 0.444..... 0.444..... 4 2 ..................
93
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Unit -3
Pairs of Equations
One day while Raju and Radha were walking, noticed some sheep and shepherds are
there near the meadow. After a while Radha asked to Raju that if there are 56 legs and 16 heads
in total, how many sheep and shepherds are there ? After thinking a while, Raju answered the
question. What is the answer? Can you say? How did Raju fidn the answer? Try yourself.
Activity 1
The price of a book and a pen together is 63 rupees. The price of a book and 5 pens is 75
rupees. What is the price of each?
price of a book + price of a pen = 63
price of a book + price of 5 pens = 75
Here the excess amount is the price of 4 pens.
price of 4 pens = ........ - .......
= .........
Price of a pen = ......
Price of a book = 63 - .....
= ........
We could easily solve this problem. Can we solve the problem of sheep and shepherds
like this ? How ?
How do we understand this problem?
As we know, total number of heads = 16, share minimum 2 legs (common to both sheep
and shephered) to these 16 heads. Now total 16 x 2 = 32 legs are shared.
Total legs – 32 = 56 – 32 = 24 legs.
Do you know, whose legs these remaining 24 legs are ?
Yes. These 24 legs are for the sheep. Already we shared 2 legs to each. So we have to
share 2 more legs to each sheep. In this way we can share this 24 legs to 12 sheep.
Total 16 heads. 12 sheep have 12 heads. So the remaining 4 heads (16 -12 ) are the
heads of shepherds.
No of sheep = 12
No of shepherds = 4
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Activity 2
Sum of two numbers is 15, their difference is 7. What are the numbers ?
Let the numbers be x, y
Sum = 15
x+y = .....
Difference =7
x-y = ......
(x + y) + (x-y) = .....
2x = ......
x = .......
y = .......
Numbers ...... , .......
Activity 3
The price of 3 note books and 5 pens together is 75 rupees. The price of 3 note books
and 10 pens together is 90 rupees. What is the price of a note book ? What is the price of a pen?
Let the price of a note book = x
Price of a pen = y
The price of 3 note books and 5 pens together is 75 rupees ...... + .... = 75
That is ....... x + ...... y = 75
The price of 3 note books and 10 pens together is 90 rupees .... + .... = 90
That is ....... x + ...... y = 90
The increase is due to 5 extra pens , 5y = ......
y = ........
Price of a pen =
Total price of 3 notebooks and 5 pen is 75 rupees
Price of 5 pen is 15 rupees
Price of 3 notebooks =
Price of a note book =
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Activity 5
Sum of digits of a two digit number is 11. If the digits are interchanged the number so got
is 27 more than it. Find the numbers ?
Let the number be xy
Sum of digits = 11
That is ........... + ........... = 11
number = 10x + y
If the digits are interchanged
The number so obtained = ...........
Difference in numbers = 27
............ _ (10x +y) = 27
........... y _ ........... x = 27
On simplification, y _ x = ...........
x = ...........
y = ...........
Number = ...........
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Unit - 4
New Numbers
Activity 1
In the table, fractions closer to 2 and 3 are given. Write fractions closer to other
given new numbers and complete the table.
2 1.4 1.41 1.414
3 1.7 1.73 1.732
22
23
32
33
2+ 3
3_ 2
3 2 3
2 32
3 3 1
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Activity 2
Two familiar triangles are given below.
450 2 300 2
1 3
450 600
11
The sides of right triangles are calculated. Using this, calculate the perimeter of the
triangles given below.
A BAD = ..............................
= ..............................
300 1 300 CAD = .............................. cm
B AB = .............................. cm
BD = .............................. cm
D = .............................. cm
C CD
BC
AC = .............................. cm
Find the perimeter of triangle ABC
R PRS = ..............................
QRS = ..............................
2
= .............................. cm
RS = .............................. cm
= .............................. cm
450 300 PS = .............................. cm
= .............................. cm
PS Q QS
PR
PQ
Find the perimeter of triangle PQR.
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Unit - 5
Circles
Activity 1
We know that the perpendicular from the centre of the circle to the chord bisects the
chord. That is in the figure if the length of chord AB is 8 cm, then AM = 4 cm. If the diameter
of the circle is 10 cm, then OA = ................. cm.
What is the distance between the centre and the chord?
OM = ........... cm
Hint : A O B
M
• OAM is a right triangle. If the hypotenuse is 5 cm
and length of one of the perpendicular sides is 4 cm,
then can you find the length of third side?
Activity 2
See the problem given below. M
O
In the figure the diameter of the circle is 20 cm. Two A B
parallel chords are drawn in the circle. They are N D
AB = 16 cm and CD = 12 cm. Find the distance between
these chords.
CN = cm
BM = cm C
OC = cm
OB = cm
Write the distance from the centre of the circle to the chord AB.
Write the distance from the centre of the circle to the chord CD.
Write the distance between the chords.
If the chords are drawn on the same side of the centre, then
What is the distance between them ?
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Activity 3
We have already studied how to draw the perpendicular bisector of a line segment in the
chapter Equal triangles in standard 8
Draw perpendicular bisectors of all the three sides of the triangle given below.
If you extend these perpendicular bisectors, then they meet at a point . Measure the distance
from this point to the vertices of the triangle .
Draw a circle with this point as the centre and distance to the vertices as the radius. This
circle is called the circumcircle of the triangle.
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
2
Can we draw a circumcircle to a right angled triangle?.Try it.
A triangle with one angle greater than 900 is given below. Draw a circumcircle to this
triangle.
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Observe the circumcentre of these three triangles. Draw circumcircles to more triangles
and write your conclusion.
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Unit - 6
Parallel lines
Activity 1
Draw an equilateral triangle of perimeter 10 centimetres.
Hint
. Draw a line of length 10 cm
. Draw a slanted line AC of length 6 cm from A
. Mark the points M and N such that they divides AC into three equal parts of 2 cms each.
. Join the line BC
. Draw two lines parallel to BC through M and N
. Draw an equilateral triangle with AB as its perimeter.
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Activity 2
In the parallelogram ABCD, the line drawn through a point P on AB, parallel to BC, meets
AC at Q. The line through Q, parallel to AB meets AD at R.
DC
Q
R
A PB
a) From triangle ABC, AQ = AP
QC ............
AQ ............
b) From triangle ACD, QC ............
c) What is the relation between AP and AR ?
PB RD
Activity 3
a) Draw a quadrilateral and mark the midpoints of each side, join these midpoints to form another
quadrilateral. What is the appropriate name for this quadrilateral?
[Rectangle, Square, Trapezium, Parallelogram]
b) Draw a rectangle, mark the midpoints of each side, join these points to form a quadrilateral. Write
the appropriate name for the quadrilateral?
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Unit - 7
Similar Triangles
Activity 1
• In the triangle ABC, AB = 5 cm, A = x0 , B = y0 , BC = 4 cm, AC = 3.5 cm
• In the triangle PQR, P = x0, PQ = 10 cm, PR = 7 cm
• In the triangle DEF, D = x0 , E= y0. EF = 7 cm
R
F
C7 7
3.5 4 x x y
P E
x y 10 QD
A5 B
1. Examine whether these triangles are similar
(Hint : What are the different contexts of similarly of trinagles?)
2. What is the length of QR?
3. Q = ................... R = ................... F = ...................
4. AB = BC
DE .............
5= 4
DE .............
4 DE = .....................
DE = ........................
5. Final the length of FD.
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Activity 2
In the figure, BC = 14 cm. AC = 15 cm, AB= 13 cm. AD = 12 cm. OB=BQ, OC=CR,
OA=AP, D = 900
P
A C
O R
BD
QS
1. OQ = OB .............
2. QR = 2 .............. = .............. cm
3. PR = AC .............. = .............. cm
4. What is the length of PQ?
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
5. Area of triangle 1 × BC ×
2
1 ×14×...........................
2
square centimetres
6. What is the measure of S ?
7. Find the area of triangle PQR.
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Unit - 8
Polynomials
Activity 1
In rectangles with one side is 1 centimetre more 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(1), p(2), p(3), p(4), p(5).
iv) Calculate a(1), a(2), a(3), a(4), a(5).
Length of the shorter side = x cm
Length of the longer side = ___________ cm
Perimeter of the rectangle p(x) = 2(length+breadth) = 2(x+1+x)
= _________________
p(x) = _________________
Area of the rectangle a(x) = length x breadth = x(x+1)
= _________________
a(x) = _________________
p(1) = 6 p(2) =......... p(3) =......... p(4) =......... p(5) = 22
a(1) = 2
a(2) =......... a(3) =......... a(4) =......... a(5) = 30
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Activity 2
If p(x)=3x2_2x+1, find p(_1), p(0), p(1)
p(_1) = 3(_1)2_2(_1)+1 = 3+2+1 = 6
p(0) = 3(0)2_2(0)+1 = ...........................
p(1) = ...........................
Activity 3
Find the first degree polynomial p(x) with p(0) = 1 and p(1) = 2.
General form of a first degree polynomial is p(x) = ax + b.
Here p(0) = 1
Therefore b = _______
Also p(1) = 2
therefore a 1+b = 2
a+b = 2
a = ______
The first degree polynomial p(x) = ___________
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Unit - 9
Circle Measures
Activity 1
• The perimeters of circles are scaled by same factor as their diameters.
• The perimeter of a circle is times its diameter.
• The area of a circle is times the square of the radius.
• Taking radius of the circles as 'r',
Perimeter = 2r
Area = r2
Based on these facts, complete the table.
Diameter Radius Perimeter Area
10 cm 5 cm 10 cm 25 sq. cm
8 cm
3 cm
24 cm
49 Sq.cm.
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Activity 2
The length of arc of a circle is that fraction of the perimeter as the fraction of 3600 that its
central angle is.
The area of a sector is that fraction of the area of the circle as the fraction of 3600 that its
central angle is.
Based on these facts, complete the table.
Perimeter Area of the Central angle What fraction Length of Area of the
of the circle circle of the arc of 3600 arc sector
12 cm 36 sq.cm 600 1 2 cm 6 sq.cm
6
20 cm 100 sq.cm 360
1200 6 cm
64 sq.cm 300
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Unit - 10
Real Numbers
Activity 1 P Q
Consider the number line below.
SR
The distance between any two points on the number line is the smaller of the numbers
denoting them substracted from the larger.
Didstance between the points P, Q is 8_2=6
Didstance between the points R, Q is 8_(-1)=9
Didstance between the points S, P is 2_(-4)=6
Didstance between the points S, Q is 8_(-4)=12
Didstance between the points S, R is -1_(-4)=3
Find the distance between the following points
Distance Points
a) 3,10
b) 1, 12
c) -1, 0
d) -8,-2
e) -10, 10
Activity 2
The mid point of two points on the number line is that point denoted by half the sum of
the numbers denoting these points.
Mid point
Eg: Find the mid point || |
y
x x y
2
| |
10 18
Mid point = 10 18 14
2
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Find the mid point of the points below.
a) 8, 20
b) _4, 2
c) _10, _6
Activity 3
Absolute value of a number is the distance between the number and zero on a number
line. Absolute value of the number 'x' is denoted as |x|.
|4| means distance between 0 and 4 in the number line.
That is |4| = 4
|_4| means the distance between 0 and _4 is the number line
That is |_4| = 4
Like this we can write 1 = 1 and _1 1
2 2 33
Like this absolute value of any number is positive.
Also absolute value of zero is zero itself.
Find the following
a) |100| =
b) |-3| =
c) |-10| =
1 =
d) 5
Activity 4
The distance between two points on the number line is the absolute value of the differ-
ence of the numbers denoting these points.
Distance between the points 'x' and 'y' on a number line can be taken as |x-y|
Distance between 2 and 20 is |2_20| = |_18| =18
Distance between 2 and _3 is |2_(_3)| = |2+3| = 5
Distance between _3, 2 is |_3_2| = |_5| = 5
Find the distance between the points below.
a) 4, 20
b) -10, 5
c) -5, 5
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Activity 5
For any real number x, |x|2 = x2
i.e. |4|2 = 42 =16
|_4|2 = (_4)2 = 16
a) |3|2 =
b) |_5|2 =
Activity 6
What are the numbers x, for which |x-2| = 1
How do we find this? What is the meaning of this? Let's see.
|x-2| =1 means the distance between 2 and a point on the number line is 1. So we can
easily find that numbers by adding and subtracting 1 to the number 2.
x = 2+1 = 3 and x = 2_1 = 1 i.e. x = 3 and 1
Find the following
a) |x_4| = 2 , x = ............
b) |x_5| = 3, x = ............
Activity 7
What are the numbers x, for which |x_3|+|x_9| =6
Let's see
||
39
|x-3| means distance between the point x and 3 on a number line. Similarly |x-9| means
distance between the same point 'x' and 9.
Here we have the sum of these distances is 6 unit. This 6 unit is actually the distance
between the points 3 and 9. So for any point from 3 to 9 will satisfy this equation.
i.e. 3 x 9 | x |
39
Activity 8
What are the numbers x, for which |x_3|+|x_7| =6
Let's see this
4 unit
||
37
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Here the sum of distance 6 is greater than the distance between the points 3 and 7 (7-3=4).
So we can take some values left to the point 3 and some values right to the point 7. Here
do we find this?
Calculate
Total distance =6
Distance between points =7_3=4
their difference =6_4=2
Marks a point half of this differences left to 3 and right to 7 as follows.
That is 3 _ 1 = 2
7+1=8
14 1
The above equation is true for all points from 2 to 8 on the number line.
That is x = 2 or x = 8
a) If |x_4| + |x_10| = 10, then what numbers are x ?
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Unit - 11
Prisms
Activity 1
The base of a prism is in the shape of an equilateral triangle. Its base perimeter
is 6 cm and height 5 cm . Find the volume of the prism.
a) The base edge of the prism (a)
a=
b) Base area of the prism = 3 a2
4
= 3 .........
4
= square centimetre
c) Volume of the prism = Base area x Height
= ........ ........
= ......... cubic centimetre
Activity 2
The length of base edge of a square prism is 8 cm and its height is 10 cm.
Find the total surface area of the prism.
a) Base area of the prism
b) Area of one lateral face of the prism
c) Number of bases
d) Number of lateral faces
e) Total surface area of the prism = 2 x base area + lateral surface area
= ...........+.............= ................................ sq. cm.
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Activity 1 Unit - 12 4
Proportion
2.5 3
1 1.5 2
The measures of circles are given in the table. Complete the table and then write the
answers of questions below.
Radius Perimeter Area r:p r:A p A
r p =2 r A = r2 r r
1
1.5
2
2.5
3
4
1. Which of the following statements are true?
a. Perimeter is 2 times its radius.
b. Perimeter is 2 times its radius.
c. Radius is 1 part of perimeter
2
d. In any circle area is times its radius.
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
e. The ratio of radius to perimeter is always 1 : 2
f. In any circle the ratio of radius to area is 1:
g. The radius and perimeter of a circle are scaled by same factor.
h. The radius and area of a circle area scaled by same factor
i. Area of a circle is proportional to its radius. The constant of proportionality is 3
j. Perimeter is proportional to radius. The constant of proportionality is 2 .
k. If the radius is 1.5 then, the ratio of square of radius to area
(1.5)2 : 2.25
2.25 : 2.25
=1:
l. In any circle the ratio of square of radius to area is 1 :
2. The square of radius of a circle is 3. What is its area?
3. The square of radius of a circle is x. What is its area?
4. The area of a circle is ........... times square of its radius.
5. Is area of a circle proportional to square of its radius? What is the constant of proportionality?
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
6. x is an independently changing measure and y is a dependent measure. If y = kx then,
a. y ...........
x
b. x:y = ............
c. Is y proportional to x? What is the constant of proportionality?
7. Let 'a' be an independent measure and b a dependent measure. If c is the constant of
proportionality, then
b = ....................
b = ....................
a
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Unit - 13
Statistics
Average
The single number which represents a group of measures is called the average of that
group. Arithmetic Mean, simply Mean is one such average.
Mean
To find the Mean of a set of measures, divide their sum by their number.
i.e. Mean = Sum
Number
Find the mean of 12, 10, 8, 11, 9
Mean = Sum = 12 10 8 11 9
Number 5
= 50 10
5
Find the Mean of the following sets.
a) 21, 23, 18, 20, 8
b) 12, 18, 15, 25
Activity 2
The Mean of 8, 12, x, 9. If 10 find the number x.
Mean = 10, 8 12 x 9 10
4
29 x 10,
4
29+ x =40,
x =11
The Mean of 8, 11, , x, 12, 5, 10 is 10 find the number x.
Activity 3
Write 6 different numbers between 20 and 40 so that the arithemetic mean of these numbers
is 30.
30
25, 27, 29, 31, 33, 35
General Education Department Samagra Shiksha, Kerala S C E R T
Activity Card Mathematics Class : 9
Write 10 different numbers between 40 and 60 so that the arithemetic mean of these
numbers is 50.
50
....., ....., ....., ....., ....., .....,
Activity 4
The quantity of rubber sheet produced by a farmer in a month is given in the table. Find
the average production of rubber sheets for the month.
Rubber (Kg) Number of days
10 3
11 5
12 8
13 6
14 5
15 2
16 1
Answer
Rubber (Kg) Number of days Total production
10 3 10 3=30
11 5 11 5=55
12 8
13 6
14 5
15 2
16 1 16 1=16
Total weight of rubbersheet = 30+55+.......+.......+.......+.......+66
Total numbers of days = ..............
Average weight of the rubber sheet = 3+5+8+6+5+2+1
= ..............
Total weight
= Total number of days
= ..............
General Education Department Samagra Shiksha, Kerala S C E R T
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