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100% found this document useful (1 vote)
163 views17 pagesActivity File Maths
zdfxfgcghvhjm
Uploaded by
zeusujjwal123Copyright
© © All Rights Reserved
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/ 17
Activity 1.3 Algebra of Operations Activity 1
Objective
To verify distributive law for three given non-empty sets A, B and C, that is
AUBOC)=(AUB)N(AUC).
—{P Material Required
Method of Construction *Rirowlng board
1. Take a drawing board and fix a white paper sheet on it with the help —J = A white paper sheet
of board pins. = Compass, pencil, scissors,.¢=
2. Draw a rectangles on the white paper sheet with scale and pencil, Board ploy et
three of them in a horizontal line and remaining two rectangles in
another horizontal line below the first three rectangles as shown in
figure (i), (i), (ii), (iv) and (y).
aes Ax,, .G:
B B
weg (i) AUB)
fiv) AU (BOC) ALB o(AUG
3. These 5 rectangles are universal sets. So write on their left hand side U,
4, Draw three circles and mark them as A, B and C in each of the five rectangles as shown in the five
figures.
5. Shade the portions as shown in these five figures.
Demonstration
1. Each rectangle represents universal set and three sets A, B and C are its subsets.
2. Shaded portion in figure (i) represents (BC).
3. Shaded portion in figure (ii) represents (A U B). Note: In the same way, the other
4, Shaded portion in figure (iii) represents (A U C). distributive law
5. Shaded portion in figure (iv) represents A U (B 0 C). AN BUG =(ANBVAN
6. Shaded portion in figure (v) represents (A UB) 9 (AU C). can also be verified.
Observation
1. Shaded portion of figure (i) represents: (B > C)
2. Shaded portion of figure (ii) represents: (A U B)
3. Shaded portion of figure (iii) represents: (A U C) —sa ~
4, Shaded portion of figure (iv) represents: A U (B/C)
5. Shaded portion of figure (v) represents: (A UB) n (A UC)
6. The common coloured portion in figures (iv) and (v) are equal.
T.. AUBOC=(AUB)N(AUG)
Application
‘This property of set operations (distributive property of set operation) can be used to simplify the problems
involving set operations.
W Viva Voce
‘1. Which property you have verified in this activity.
Ans. I have verified the distributive property of union over intersection.
2. Can we apply this distributive property for intersection over union?
Ans. Yes, this property can be used to distribute union over intersection i.¢.,
An (BUC) =(ANB) LUANG)
Here three sets A, B and C are non-empty sets. Activity 2.2 Identification of Relations and
Functions
Objective Activity 2
‘To identify a relation and a function. 7 25
Method of Construction Bea Ronis | a
1. Take a drawing board. Fix the white paper sheet on it with the half —Jm A drawing board fa
of board pins. & Awhite paper sheet
2. Take two sets A = {a, b, c, d} and B = {1, 2, 3, 4} ‘= Compass, pencil, scale, |¢
3. Draw two circles. In first circle figure (j), write a, b, cand d inside it Board pirsyete 4
Now draw second circle figure (i) and write 1, 2, 3, 4 inside it.
A 8
24
w (ii)
Join ato 1, b to 2 and cto 3 with arrows as shown in figures (i) and (ii).
4, Draw two circles again, figure (iii) and (jv). Write a, b, ¢, d inside circle of figure (iii) and write 1, 2, 3,
4 inside circle of figure (iv).
A B
ay (s)
and write 1, 2, 3, 4 inside circle of figure (iv). Join ato 1, bto2
Write a, b, ¢, d inside circle of figure
and 3, cto 4 and d to 4, with arrows.
5. Draw again two circles as shown in figure (v) and (vi). Write 1, 2, 3, 4 inside circle (v) and a, b, od
inside circle figure (vi).
A 8
\ eT
|
(w) (wi)
Join ato 1, bto 1, cto 2 and d to 3 with arrows as shown in figure (v) and (vi).
fi 6. Draw again two circles as shown in figure (vii) and (vii
=
ii).
8
(vii) (viii)
Write a, b, ¢ dinside left side circle and 1, 2, 3, 4 inside right side circle. Join ato 1, bto 2, cto2 and dto4,
Demonstration an a
1, Set A= {1, 2, 3, 4} shown in left circle represent domain and set B = {a, b, G @, shown in right circle
represents co-domain.
2. Ifeach number a, b, c, dhas been joined to 1, 2, 3, 4 only by one arrow and no element of domain
left without joining it with some element of co-domain, then it is a function otherwise it is a relation]
3. In figures (i), (i), each number a, b, chas been joined to 1, 2, 3, by one arrow but d has not-been joined]
with no element of set B. Therefore figures (i) and (ii) represent a relation and not a function.
4. In figures (iii) and (iv), number b of domain has been joined to co-domain by two arrows to co-domain’
B. Hence this is also a relation and not a function.
5. In figures (v) and (vi), each number a, b, c, dof domain A has been joined with co-domain B with single
arrows and no element of domain A has been left without joining by some number of co-domain B
therefore this is a function. It is also a relation. i
6. In figures (vii) and (viii), each element a, b, c, dof domain A has been joined with co-domain B with single:
arrows and no element of co-domain A has been left without joining to some number of co-domain B.
Therefore this a function. This function is relation also,
Observation
1. In figures (i) and (ii), ordered pairs are: (a, 1), (b, 2), (c, 3). These ordered pairs constitute a relations.
and not a function because element d of domain A has been left without joining.
2. In figures (iii) and (iv), ordered pairs are: (a, 1), (b, 2), (b, 3), (¢, 4), (d, 4). This lati
function because in ordered pairs (b, 2) and (b, 3) first number. pes , rareation end ng,
3. In figures (v) and (vi), ordered pairs are: (a, 1), (b, 1), (c, 2) and (d, 3). This i
relations because no element of domain has been left without joining a como eleitent of cnlenaiitd
4, Rnd ne first number of any ordered pa i repeating therefore itis a function
- In figures (vii) and (vii), ordered pairs are: (a, 1), (b, 2), (c, 2), (a, 4) ‘ i
i » 2 (2 2); (6 2), (a, 4). Here also no element of domain
has been left unpaired, no first ent is ) Iisa aeacee tag
has been left np irst entry in any ordered pair a repeats, therefore this is a function as well
Application
‘This activity can be used to decide whether as given relation is a function or not.
ot.
Viva Voce
ny subset of A x B, g
3,
1. Define a relation from set A to set B,
Ans. A relation R from set A to set B is a
2. Define a function from set A to we
‘Ans. A function F is a relation from set Ato
3. Is every function a relation?
Ans. Yes every function is a relation,
4. Is every relation a function?
Ans. No, every relation is not a function,
cB cl
in which every element of set A has only one image in set 3: 3. Trigonometric Functions
ystems of Measurement of Angles
Activity 3.1
Objective
To verify the relation between degree measure and the radian measure of
an angle.
Material Required
= Drawing board, white
paper sheet, board pins——
= Pencil, compass, thread, .*
scale, protractor
Method of Construction
1, Take a drawing board and fix the white paper sheet on it with the
help of board pins.
2. With the help of compass draw a circle with centre C and any radius.
Draw two diameters PQ and RS intersecting at the centre Cas shown ¢
in figure.
3, Here it is clear that CP = CR = CS = CQ = r, radius of the circle.
P
oe
a Activity 3
Demonstration
1. Draw another circle as shown in figure. Take an angle 0 at the centre 0, radius rand length of the arc
AB = Las shown in figure.
By definition of radian measure
r a
0 = “radians ‘\
r
1 Q t
2. Degree measure of (+) x 360 degree. J
e
Then, | yadians = (4) x 360 degree
r Qnr,
: 180
or 1 radian = —~ degree = 57.27 degree.
Observation
1. With the help of thread, measure the lengths of RP, PS, RQ and QS and record them in the table given
below.
B.No. ‘Are | Lengthof |Radiusofthe| Euipadian theatre!
2a “are (2) cirele(s)
RP Sem
PS Sem 50 5.3.cm Sem
50.
r
GR 106 29.12 radian
4. R 10.6 em Sem ZQCR =~
2. Now with the help of protractor measure the angles in degree and complete the following table.
_ Degree measure =
Angle Degree measure Radian measure Ratio = 2otan meas
2RCP 60° 1.06 radian 56.6
CS 120° 2.12 radian 56.6
ZQCs 60° 1.06 radian 56.6
ZQCR 120° 2.12 radian 56.6
3. The value of one radian is equal to 56.6 degree approximate.
Application
This activity can be used to convert degree measure into radian measures and radian measure to degree
measure.
Vv Viva Voce
1, Define radian measure of an angle.
‘Ans. The angle subtended at the centre of a circle by its arc equal in length to radius of the circle, is equal
to 1 radian.
2. What is the value of one radian in degrees?
Ans. 1 radian = 57.27 degrees.
8. What is the value of 1 degree in radians?
‘Ans. 1 degree = 0.0174 radian. Complex Numbers and Quadratic Equations
ieee ails Act i vity qd
Objective
‘To interpret geometrically the meaning of i= J-1 and its integral powers.
Method of Construction
1. Take the drawing board and place it on a table. Fix the white paper
sheet on it with board pins.
2. Draw X'OX and YOY’ two perpendicular lines intersecting at O. Here
X'OX is x-axis, YOY' is y-axis and O is the origin. The plane is an argand
plane (Plane on which complex number can be plotted) Here x-axis is
real axis and y-axis is imaginary axis. Draw a circle of unit radius and
O as centre. This circle cuts x-axis at A, and A, and y-axis at Ay and —*
As. see the fig.
fs Drawing board, white
paper sheet, board pins,
= Compass, scale, pencil *—
etc.
Demonstration
1. In the argand plane OA = 1, OA, = i, OA) =~1, OAs =i, and
OAy=1
2. OA =i= 1 * i, OA, =-1 =ixi=?, OA, is, and soon.
Here each time OA has been located by 90" and this step is equal
to multiplication by i. Therefore iis multiplying factor for ratio
of 90°.
Observation
1. On rotating OA through 90°, it coincides OA, = 1 x i= i
2. On rotating OA through an angle of 180° (2 right angles)
coincides OA, = 1 * ix i=-1
3. On rotating OA through 270° (3 right angles) it coincides
OAs = 1x ixixiz-i -
4. On rotating OA through 360° (4 right angles) it coincides $i
OA, =1xixixixi=l a ¥
5, On rotating OA through n right-angles of OA, = 1 * ix i* i
xan times = i
Application
‘This activity may be used to evaluate any integral power of i.
¥W Viva Voce
1. What is the value of i?
Ans. i= J1
2. Is ia real number?
Ans. No iis not a real number, it is an imaginary number,
3. Can you define a complex number?
‘Ans. A sum or difference of a real number and an imaginary number is called a complex number. It is
denoted by z= x+ yi. Here xis a real number and yi is an imaginary number. GEE octivitys
Objective
To find the number of subsets of a given set and verify that if a set has n elements, then the total number
of subsets is 2".
Method of Construction
1, Take the empty set Ag it has no elements.
2. Take a set A;, with one clement only
Ao=(h
Ay = {ay}
3. Take a set Ay, which has two elements only
Ag = (a, a}
4. Take a set As, which has 3 elements only
Demonstration
1. Take the set Ag = {}
Ag = {Ary aa, 3}
drawing board
‘A white paper sheet
= Pencil and scale
= Different coloured pencils
a
ss
=
To represent it by venn-diagram, draw a circle and do not write any element inside it. It is an empty
set represented by ¢. (See figure (j)) Here set Ag is subset of itself. Symbolically we represent it
by >.
Hence number of subsets of Ag = 1 = 2
2. Take the set A; = {ay}
Ag = (} has only one subset 4.
plowmber of elements in et A)
Represent it by venn-diagram. Draw two concentric circles. Inner circle has no clement and outer circle
has one element a;. Therefore set A, = {a,} has two subsets 6, (a;} (Sce figure (i))
Number of subsets of Ay = 2 = "umber of elements in set Ai)
2
. Take the set Ay ~ (
Represent it by veni
circles inside it. Inner most circle has no element} so it is
element a, and outer most circle has two elements a,
Hence set A ~ (a}, a,) has subsets >, (ay), (aa) and (a, a
: Number of subsets of set Aj = 4 = 2? = 2 (Number o
aw
liagram. To do so, draw a circle. To divide it into subsets, draw two cencentric
an empty set >, then second circle has one
‘and a, (See figure (ii)).
ements why er ene fr ircle represent Ag = {€,, da, a4}. Now draw
it -diagram, Draw a circle. Outer circle represent Ag = (a1, a,
coveeniie cindceineide inner moe eile eo 20 element, s0 itis ¢. Second circle has fee clement,
and third circle has two elements a, and a,. The fourth circle has 3 elements a,, aa, a. (See figure}
So the set Aj = {a;, a, a3} has 8 subsets as 4, {a,}, {a2}, (a Laan al (aay a3}, {ay, a3} and {ay, a, a.)
en Number of subsets of set Aj = 8 = 2° or 2"
ele} ay 27}}e-— Ay
(iti)
(iw)
5. we continue this process, and take a set A, which as n elements then it will have number of subsets
2 (number ofelements Rec Ag 2
Observation
1. The number of subsets of Ay is 1 = 2°,
2. The number of subsets of A, is 2 = 2),
3. The number of subsets of A, is 4 = 22,
4. The number of subsets of Aj is 8 = 2°,
5. The number of subsets of Ajg = 2".
6. The number of subsets of A, = 2”,
Conclusion
Ifa set A, has n elements, then number of subsets of set An
Application
‘This activity can be used to calculate number of subsets of any given sets.
W Viva Voce
1. Define a set.
Ans. A well defined collection of objects is called a set,
2. Can you give two examples of sets? ;
Ans. (a) A collection of natural numbers is a set,
(+) A collection of real numbers between 1 and 2 is a set.
P= olnumber of element in set Ay)
3. Define subset of a set.
Ans. If all the elements of set A are present in set B, then set Ais called a
7 ‘sub:
4. Which set is subset of all the sets? ‘set of set B. If denoted as A cB.
Ans. Empty set 4 is subset of all the sets,
5. Is a set is subset of itself?
Ans. Yes, every set is subset of itself, ~
Activity 6 Qe
Activity 5.1 Graphical Solutions of Linear
Inequalities in Two Variables
Objective
To verify that the graph of a given inequality by 5x + 4y- 40 < 0 of the form ax + by + c< 0a, b>0,c<9
represents only one of the two half planes.
Method of Construction :
1, Take the drawing board, fix the white paper sheet on it with board pins.
2. Draw two perpendicular lines X'OX and YOY' intersecting at O, on the
white paper sheet.
Here X'OX is x-axis and YOY' is y-axis, Graduate the x-axis and y-axis.
. Take the equation of the line 5x + 4y- 40 = 0 from the inequation
5x+4y-40<0
?
Material Required —
Ju Drawing board, white =
paper sheet, drawing
pins, scale, pencil,
5 5x+ 4y=40 uy .
To draw its graph, make table of points satisfy by the equation 5x + 4y = 40
x [ol] s
y | 10 [ 0
Plot the points A(0, 10) and B(8, 0) and join them with a line.
Line AB is the graph of the linear equation corresponding to the given linear inequation.
4. Make two half planes I and Il as shown in the fig.
5. Mark some points O(0, 0), A(1, 1), B(3, 2), C(4, 3) and D(-1, -1) in half plane I.
Mark some other points E(4, 7), F(8, 4), G(9, 5), H(7, 5) in other half plane Il.
Demonstration
1, Put the coordinates of the point 0(0, 0) in the left hand side of 5x+ 4y- 40 <0.
5x0 +4 0-40 =-40 <0 (True),
The coordinate of the point 0(0, 0) lies in plane I and satisfy the inequation 5x + 4y- 40 < 0.
2. Put the coordinates of the point A(1, 1) in the left hand side of the inequation 5x + 4y-40<0
It gives: 5 x 1 +4 * 1-40 =-31 <0 (True).
Hence the coordinates of the point A(1, 1) lies in plane I satisfy the inequation 5x + 4y-40 <0. 3, Put the coordinates of the poi ‘
It gives: 5x 3+4% 2-40 eipant 2) in the left hand side of the inequation 5x + 4y~ 40 < 0.
7 ~ 40 =-17 <0 (Tru
Hence the coordinates of the poi ; <)
4, put the coordinates of the e point B(3, 2) lying in plane I and satisfy the inequation 5x+ 4y— 40 < 0.
" Teplveat 5444 Point C(4, 3) in the left hand side of the inequation 5x + 4y~ 40 < 0.
It gives: * 3-40 = 20 + 12-40 =-8 <0 (true)
.ce the coordin: i i:
ao the coon einen “ the point C(4, 3) lying in the plane I satisfy the inequation 5x + 4y ~ 40 < 0.
coe '¢ point D(-1, ~1) in left-hand side of the inequation 5x + 4y— 40 < 0.
It gives: 5 * (-1) + 4 (-1) - 40 =-5 4 40 = -49 < 0 (True)
. the coordinates of the point D(-1, 1) lying in the plane I satisfy the inequation 5x + 4y~ 40 <
6, Put the coordinates of the point E(4, 7) in left hand side of the inequation 5x + 4y— 40 < °
Itgives: 5 x 4 +4 * 7-40 = 20 + 25-40 = 48 - 40 = 8 > 0 (False).
Hence the coordinates of the point E (4, 7) lying in the half plane Il does not satisfy the inequation
Sx+ 4y-40 <0.
7, Put the coordinates of the point F(8, 4) in the left hand side of the inequation 5x + 4y—40 <0.
It gives: 5 8 + 4 x 4-40 = 40 + 16 ~ 40 = 16 > 0 (False)
Hence the coordinates of the point F(8, 4) lying in the half plane II does not satisfy the inequation
Sx+ 4y-40 <0.
8. Put the'coordinates of the point G(9, 5) in the left hand side of the inequation 5x + 4y ~ 40 <0.
It gives: 5 x 9 + 4 x 5-40 = 45 + 20 — 40 = 65 - 40 = 25 > 0 (False)
Hence the coordinates of the point G(9,5) lying in the half plane II does not satisfy the inequation
Sx+4y-40 <0.
9. Ina similar way, we can show that the point H(7, 5) also does not satisfy the inequation 5x + 4y~40 <0.
Hence all the points O, A, B, C, and D satisfy the inequation 5x + 4y- 40 <0 and hence lie in the half
plane I and all the points E, F, G, H, which do not satisfy the linear
inequation 5x + 4y - 40 < 0 lie in half plane Il.
Hence graph of the given inequation represents only one of the
two corresponding half planes.
Note: The activity can also be
performed for the inequality of
the type ax+ by +c>0.
Observation
1. The coordinates of the point A(1, 1) satisfy the given inequation and lies in the half plane I.
2. The coordinates of the point 0(0, 0) satisfy the given inequation and lies in the half plane I.
3. The coordinates of the point B(3, 2) satisfy the given inequation and lies in the half plane I.
4. The coordinate of the point C(4, 3) satisfy the given inequation and lies in the half plane I.
5. The coordinates of the point D(-1, -1) satisfy the given inequation and lies in the half plane I.
6. The coordinates of the point E (4, 7) does not satisfy the given inequation and lies in half plane Il.
7. The coordinates of the point F(8, 4) does the satisfy the given inequation and lies in half plane Il.
8. The coordinates of the point G(9,5) does not satisfy the given inequation and lies in half plane Il
9. The coordinates of the point H(7, 5) does not satisfy the given inequation and lies in half plane Il.
Application
This activity can be used to find the half-plane, a given inequality represents.
Ww Viva Voce
1. Write two inequations.
Aas. (a) 4x— 3y- 30 > 0, (b) Sx+ 4y+7>0. Objective
To find the number of ways in which three cards can be selected from given five cards.
Method of Construction a
1. Take the drawing board, place it on the table and cover it with white _*|
paper sheet with board pins.
2, Take the card board sheet and cut out five identical cards of convenient
size from it. Mark these cards as A, B, C, D and E.
?t
Material Required
fa Drawing board, a white |
paper sheet, card board:
sheet
= Cutter, pencil, scale etc,
Demonstration
1. Let us select the first card from the five cards. Let it be card A. Then the other two cards from
the remaining four cards can be BC, BD, BE, CD, CE and DE. Include card A with all the above|
selection.
It gives: ABC, ABD, ABE, ACD, ACE and ADE (6 cases)
2. Now let us assume that first card selected is B. Then the other two cards from the remaining
4 cards are: AC, AD, AE, CD, CE and DE. Now include the first selection of card B with the above|
6 selection.
It gives 3AC, BAD, BAE, BCD, BCE and BDE.
3. Now again we assume that first selected card be C. Then the other two cards from the remaining
4 cards are: AB, AD, AE, BD, BE and DE. Now include the first selection of card C with above
6 selection.
It gives: CAB, CAD, CAE, CBD, CBE, CDE.
4. Again let us assume that first selected card is D. Then the other two cards from the remaining 4 cards
are: AB, AC, AE, BC, BE, CE. Now include the first selected card D with the above 6 selection.
It gives: DAB, DAC, DAE, DBC, DBE and DCE.
5. Now assume that first selected card be E. Then the other selection from the remaining 5 cards are:
AB, AC, AD, BC, BD and CD. Now include the first selected card E in the above 6 selection.
It gives: EAB, EAC, EAD, EBC, EBD and ECD.
6. Record the above selection on the same sheet.
7. Now look at the table where you have record above selection. Therefore total selection are 5 x 6 = 30,
Here each of the selection is repeated thrice. Therefore, the number of distinct selection are
30 +3 = 10, This is same as °C.
Observation
1. ABC, ACB, BAC, BCA, CAB and CBA represent same selection.
2. ABD, ADB, BDA, BAD, DAB, DBA represent same selection.
3. BAE, ABE, EBA BEA and AEB represent the same selection.
4. BAE and ABC represent different selection. pases cer a 1S
5, Among CAE, ADC, ECD, DBE, BDC, ACE, CAE:- CAE, ACE and CAE represent the same selection.
6. ABC, ACB, BAC, BCA, CAB and CBA represent the same selection.
Application
This activity can be used for understanding the general formula for finding out number of selection
when r objects are selected from given n-distinct objects ie. "c, =”!
ri(n=rl
1, Can you define combination?
Ans. Combination is selection of a group of r objects from n objects. It is worked out by the formula.
n nl
*Filw=ai
pherer