GOVERNMENT OF MAHARASHTRA...2 Bridge Course for Std – 10th Maths (Part 1& 2) • Promotor : Department of School Education, Government of Maharashtra. • Publisher : State Council

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GOVERNMENT OF MAHARASHTRA

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Bridge Course for Std – 10th Maths (Part 1& 2)

• Promotor : Department of School Education, Government of Maharashtra.

• Publisher : State Council of Educational Research and Training, Maharashtra, Pune

• Motivation : Hon. Smt. Vandana Krishna, (I.A.S.)

Additional Chief Secretary, Department of School Education Gov.of Maharashtra.

• Guider : Hon. Shri.Vishal Solanki, (I.A.S.)

Commissioner (Education), Maharashtra, Pune

Hon. Shri.Rahul Dwivedi (I.A.S.)

State Project Director, Maharashtra Prathmik Shikshan Parishad, Mumbai

• Editor : Shri.Dinkar Temkar

Director, SCERT, Maharashtra, Pune

• Co-Editor : Dr.Vilas Patil

Join-Director, SCERT, Maharashtra, Pune

• Executive Editor : Shri.Vikas Garad,

I/C Principal, SCERT, Maharashtra, Pune

Dr. Prabhakar Kshirsagar

Senior Lecturer, Department of Mathematics, SCERT, Pune

Smt.Vrushali Gaikwad

Lecturer, Department of Mathematics, SCERT, Pune

• Editing Assistancet : Smt.Vaishali Gadhave and Smt. Bhakti Joshi

Subject Assistant, Department of Mathematics, SCERT, Pune

• Creative Team : 1. Dr.Nitu Gavande Senior Lecturer, DIET, Nagpur

2 . Smt.Nilofer Patel Lecturer ,RAA ,Aurangabad

3. Shri.Atul Patva Co-Teacher , Bhausaheb Firodiya Highschool,Ahmednagar

4. Shri. Sanjiv Bhor Co-Teacher,S.i R. M.Vidya.,Rase.Tal Khed Dis.Pune

5. Shri. Pramod Bendre Co-Teacher,R.D.M.V.Ambawade Bk, T/D Satara

` 6 . Shri.Shivprasad Mehtre Co-Teacher,S. T. Vidy.Khadaki, Ahmednagar

7. Smt.Manisha Nahar Co-Teacher, H.H.C.P.H. for Girls, Hujurpaga, Pune

8. Shri.NageshChalpelwar Co-Teacher, Z.P.P.,Adhul,T.Paithan, Aurangabad

• Translation Support: 1.Dr.Bhavana Joshi H.M. Arnyeshwar Madhyamik Vidyalaya,Pune

2. Shri. Pramod Bendre Co-Teacher,R.D.M.V.Ambawade Bk, T/D Satara

3. Smt.Manisha Nahar Co-Teacher, H.H.C.P.H. for Girls, Hujurpaga,Pune

4. Shri. Sandip Panchbhai, Co-Teacher, Nagpur

5. Smt. Madhuri Mone Co-Teacher Gurunanak High School Jr.College,Pune

6. Shri. Rajendra Aphale Co-Teacher, New English School, Satara

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Instructions for Students

Dear students, due to pandemic situation in the last academic year you

continued your learning and education through online and in various digital modes.

This Bridge Course has been prepared for you with the objective of reviewing the

previous year's syllabus at the beginning of the present academic year and helping you

to prepare for this year's syllabus.

1. The bridge course lasts for a total of 45 days and consists of three tests after a

certain period oftime.

2. The bridge course will help you to understand exactly what you have learned in the

previous academic year and to understand the curriculum for the next class.

3. This bridge course should be studied on a day-to-day basis.

4. It consists of day-to-day worksheets. You are expected to solve the worksheet on

your own as per the given plan.

5. Seek the help of a teacher, parent or siblings if you have difficulty solving the

worksheet.

6. The video links are provided to better understand the text and activities given in

each worksheet for reference, try to understand the concept using them.

7. Solve the tests provided along with as planned.

8. Get it checked with the teacher after completing the test.

9. Seek the help of teachers, parents or siblings to understand the part that is not

understood or seems difficult.

Best wishes to you all for the successful completion of this Bridge Course!

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Instructions for Teachers, Parents and Facilitators

As we all are very well aware about the fact that due to pandemic situation, the

schools were formally closed during the last academic year and the actual classroom

teaching and learning could not take place. There is uncertainty even today as to when

schools will restart in the coming academic year. On this background various efforts

have been made by the government in the last academic year to impart education to

the students through online mode. Accordingly, the Bridge Course has been

prepared with the dual objective of reviewing the studies done by the students in the

previous academic year and helping them to learn the curriculum of the present

class in this academic year.

1. The bridge course lasts for a total of 45 days and consists of three tests after a

certain period oftime.

2. The bridge course is based on the syllabus of previous class and is a link between

the syllabi of previous and the current class.

3. This bridge course has been prepared class wise and subject wise. It is related to

the learning outcomes and basic competencies of the previous class’ textbook and is

based on its components.

4. The bridge course includes component and sub-component wise worksheets. These

worksheetsare generally based on learning outcomes and basic competencies.

5. The structure of the worksheet is generally as follows.

Part One - Learning Outcomes/Competency Statements.

Part Two - Instructions for teachers / parents and facilitators

Part Three - Instructions for Students

Part Four - Learning Activity

Part Five - Solved Activity/ Demo

Part Six - Practice

Part Seven - Extension Activity/Parallel Activity/Reinforcement

Part Eight – Evaluation

Part Nine - DIKSHA Video Link/E-Content/QR Code

Part Ten - My Take Away/ Today I Learnt

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6. This bridge course will be very important from the point of view to revise and

reinforce the learning of the students from the previous class and pave the way to

make their learning happen in the next class.

7. Teachers/parents and facilitators should help their children to complete this bridge

course as perday wise plan.

8. Teachers/parents and facilitators should pay attention to the fact that the student will

solve each worksheet on his/her own, help them where necessary.

9. The teacher should conduct the tests from the students after the stipulated time

period, assess the test papers and keep a record of the same.

10. Having checked the test papers, teachers should provide additional supplementary

help to thestudents who are lagged behind.

Best wishes to all the children for the successful completion of this Bridge Course!

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State Council of Educational Research and Training, Maharashtra

Standard 10th: Mathematics Part 1 and part 2

Bridge Course

INDEX

Sr.

No. Day Name of the Topic

Page

No.

1 1 Set- Concept, methods of writing set and its elements 8

2 2 Subset 10

3 3 Point, line and plane 12

4 4 Coordinates of the point, distance and betweenness . 14

5 5 Segments, rays, and pairs of angles. 16

6 6 Conditional statements, converse and proof of the

theorems. 19

7 7 Properties of rational numbers 21

8 8 Operations on like surds: addition and subtraction 23

9 9 Operations on like surds: multiplication, division and

rationalization of surds 25

10 10 Properties of parallel lines 27

11 11 Tests of parallel lines 30

12 12 Degree of Polynomial, Standard form of Polynomial,

Coefficient form of the Polynomial 32

13 13 Operations on polynomials 35

14 14 To find value of the polynomial when value of the variable

is given. 37

15 15 Factorization of the polynomial 39

16 16 Unit Test No.1 41

17 17 Types of triangles 42

18 18 Congruence of triangle 44

19 19 Properties of triangles measuring °30-°60-°90 and

°45-°45-°90 47

20 20 Similar triangles 49

21 21 Some properties of triangle 51

22 22 Basic Constructions 53

23 23 Basic Construction- Triangle 55

24 24 Equations in two variables-Concept 57

25 25 Simultaneous Equations – Concept and method (1) of

finding its solutions 59

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26 26 Method 2 - Elimination method of solving simultaneous

equations by substitution 62

27 27 Solution of the simultaneous equations of type ax + by = p

and bx + ay = q 64

28 28 Word problems based on simultaneous equations Part 1 66

29 29 Word problems based on simultaneous equations Part 2 69

30 30 Unit Test No.2 71

31 31 Quadrilateral and its types 73

32 32 Some properties of a Quadrilateral and Triangle 75

33 33 Various parts of a circle 77

34 34 Properties of arcs of a circle 79

35 35 Properties of chords of a circle 81

36 36 Operations on equal ratios 83

37 37 Proportion and continued proportion 85

38 38 Word problems based on ratios Part 1 87

39 39 Word problems based on ratios Part 2 89

40 40 Point – Coordinates and Position 91

41 41 Equations of lines 95

42 42 Graphs of linear equations in general form 97

43 43 Introduction to trigonometry and trigonometric ratios 99

44 44 Trigonometric ratios of angles having particular measure. 102

45 45 Final Test No. 3 104

46 Answer Key 106

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DAY: 1st

Standard:10th Subject:Mathematics 1

Topic: Set

Sub Topic: Set –Concept, methods of writing set and its elements

Competency Statements:1) To identify Set.

2) To write given set using the listing method

3) To write number of elements in a set

Let’s recall:

1) Classify the following numbers as Natural numbers, Whole numbers

and Integers.12, -7, 0, 32, -1, 9

2) Classify the following numbers as Odd numbers, Even numbers and Prime

numbers. 11, 14, 27, 31, 52, 97, 2

Important Points:

Set:

• If we can definitely and clearly decide the objects of a given collection then that

collection is called a set.

Generally, the name of the set is given using capital letters A, B, C....., Z.

• The members or elements of the set are shown by using small letters a, b, c.

• Now let us observe the set of numbers.

N = {1, 2, 3, . . .} is a set of natural numbers.

W = {0, 1, 2, 3, . . .} is a set of whole numbers.

P = Set of colours in the rainbow.

Listing method or roster form

In this method, (i) We write all the elements of a set in the curly brackets.

(ii) Each of the elements is written only once and separated by commas.

(iii) The order of elements is not important but it is necessary to write all the

elements of the set.

Natural Numbers Whole Numbers Integers

Even Numbers Odd Numbers Prime Numbers

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For Example: 1) The set of even numbers between 1 and 10, can be written as,

A = {2, 4, 6, 8} or A = {8, 6, 2, 4}

2) The set of letters in the word ‘mathematics’is written as

B = {m, a, t, h, e, i, c, s}

Number of elements in a set:

Suppose D = {1, 4, 9, 16, 25, 36} is given set. There are 6 elements in the set D.

Number of elements in set D is denoted as n(D). Thus n (D) = 6

Exercise

Q. 1 Choose correct option.

(i) Out of following collections which is a set?

(A) Days of week (B) Happy people in the town

(C) Easy examples in the text book (D) Clever students of the school

(ii) If P is set of letters in the word ‘college’ then which of the following set is set P

by listing method?

(A) {c, o, l, l, e, g, e} (B) {c, l, e } (C) {c, o, l, e, g } (D) {c, o,l, g}

(iii) If M = { 3, 6, 9, 12, 15, 18 } then n(M) =?

(A) 5 (B) 6 (C) 4 (D) 7

Q. 2 Complete the following table

SET By listing method

1) A is a set of prime numbers between 1 to 20

2) B is a set of letters in the word ‘cricket’

3) C is a set of vowels in English.

4) D is a set of multiples of 4 less than 50

5) E is a set of perfect squares between 1 to 50

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DAY :2nd

Sub Topic: Subset

Competency Statements:1) To write subset of a given set.

2) To decide which set is subset of the other set from the

given different sets.

Important points

Subset: If A and B are two given sets and every element of set B is also an element of

set A then B is a subset of A which is symbolically written as B ⊆ A. It is read as 'B is

a subset of A' or 'B subset A'.

For Example: (1) A = {11, 12, 13, 14, 15, 16, 17, 18}

B = {11, 13, 17}

Every element of set B is also element

of set A. Thus B ⊆ A

Properties: (i) Every set is a subset of itself.A ⊆ A

(ii) Empty set is a subset of every set. ⊆ A

(iii) If A = B then A ⊆ B and B ⊆ A

(iv) If A ⊆ B and B ⊆ A then A = B

Exercise

Q.1 Choose correct option.

(i) If A = { 1, 2, 3, 4, 5, 6 }, B = { 4, 6, 8}, C = { 1, 4}, D = { 2, 4, 6}

Then which of the following statements is false?

(A) A⊆A (B) B⊆A (C) D⊆A (D) C⊆A

(ii) If P = { a, e, i, o ,u } then which of the following set is subset of P.

(A) {a, b, c } (B) {e, f, g } (C) { } (D) {o, p, u}

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15

11

13

12 14

A

B

16

17

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Q.2:Solve the following sub questions.

1) If A= { 2, 3, 5, 7, 11} then write any four subsets of the A.

2) From the figure given below state which set is subset of which set.

3) From the figure given alongside state the subset relation between the sets..

i) X = set of points on the line.

ii) Y = set of points on the segment AB.

iii) Z = set of points on the segment BC

iv) T = set of points on the ray BA.

Link:

https://diksha.gov.in/play/collection/do_312528209300701184153324?referrer=utm_source

%3Dmobile%26utm_campaign%3Dshare_content&contentId=do_313018565848219648178

• • •

l A B C

https://diksha.gov.in/play/collection/do_312528209300701184153324?referrer=utm_source%3Dmobile%26utm_campaign%3Dshare_content&contentId=do_313018565848219648178


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DAY : 3rd

Topic: Basic Concepts in geometry

Sub Topic: Point, line and plane.

Competency Statements:To understand the concepts: Point, Line and Plane.

Let’s recall :

1) Why the word ‘coplanar’ is used in the definition of parallel lines?

2) How did you experience that the light travels in a straight line?

Just recall the scientific experiments performed in previous standards.

Important points:

Point, line and plane: -

We do not define a point, line and plane also. These are basic concepts in Geometry.

Lines and planes are sets of points.

Point: In simple language point is a dot made by pencil on the paper. Point does not

have length, breadth or height. Points are denoted by capital letters.

For example, point A , point B, point C.

Collinear and non collinear points: If there is a single line passing through all the

given points, the points are said to be collinear, otherwise points are said to be

non collinear.

Line 1) 2)

Name of this line is line q Name of this line is line PQ or line QP

Plane:

A flat surface which extends infinitely on

all directions.

If there is one and only one plane containing all the lines then the lines are said

to be Coplanar lines otherwise, they are said to be Non coplanar lines.

q

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Do it yourself

Solve the following by doing activity.

1) How many lines can pass through a given one point?

2) How many lines can pass through two distinct points?

3) How many planes will be there containing a line and a point outside it?

4) Is it possible that three points are coplanar?

5) Is it possible that two intersecting lines are coplanar?

6) What is the intersection of a plane and a line intersecting it?

7) What is the intersection of two planes?

Link :



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DAY :4th

Topic: Euclidean geometry

Sub Topic: Co-ordinates of a point, distance, and betweenness .

Competency Statements: 1.To find the distance between the points when the

Co-ordinates of the points are given.

2 .To determine the betweenness when the distances between points are known .

Let’s think

There are three cities Pune, Ahmednagar and Shirur on the straight highway.

The distance between Pune and Ahmednagar is 120 Km. The distance between

Shirur and Ahmednagaris 50 Km. whereas the distance between Pune and

Shirur is 70 Km. then find which city is between other two cities?

Important points :

Co-ordinates of points and distance) -

Observe the number line given below,

Co-ordinate : Here, the point D on the number line denotes the number 1. So, it is

said that 1 is the co-ordinate of point D.

To find the distance between two points, consider their co-ordinates

and subtract the smaller co-ordinate from the larger.

The distance between Points E and D is denoted by d(E, D).

This is the same as ℓ (ED), i.e. the length of the segment ED.

Let’s find d (A, B):

The co-ordinate of A is - 5 and that of B is −3 as − 3 >− 5

∴ d (A, B) = −3 − (−5)

= −3+5

= 2

The distance between two points is obtained by subtracting smaller

co-ordinate from the larger co-ordinate.

The distance between any two points is a non-negative real number.

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Betweenness –

If P, Q, R are three distinct collinear points, there are three possibilities.

1) Point Q is between 2) Point R is between 3) Point P is between

the points P and R the points P and Q the point Rand Q

If d (P, Q) + d(Q, R) = d (P, R) then point Q is said to be between point P

And point R. This betweenness is shown as P − Q – R

Midpoint of a segment -

If A-M-B and seg AM ≅seg MB, then M is called the midpoint of seg AB.

Every segment has one and only one midpoint.

Exercise

1) On the number line points A, B and C are such that, d (A, C) = 10, d (C, B) = 8

then find d (A, B). Consider all possibilities.

2) On the number line the co-ordinate of point P is -3. Then find the co-ordinate of

the point which is at a distance of 5 units from P.

3) Point M is mid-point of seg AB and AB = 18 then find AM.

Link:https://diksha.gov.in/play/collection/do_312528194785001472250597?referrer=utm_s

ource%3Dmobile%26utm_campaign%3Dshare_content&contentId=do_3130185625388892

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DAY :5th

Topic: Line and angle Sub Topic:Segment, ray and pairs of angles

Competency Statements:1 .To identify pairs of angles.

Let us think :

1 .let us think about pairs of angles formed by lines intersecting each other.

Important points :

Pairs of angles:

1. Adjacent angles –

Two angles are said to be adjacent

if they have

a) common vertex

b) a common side .

c) disjoint interiors

2. Angles in linear pair

When the sum of measures of pair of adjacent angles is 180o then they

form a linear pair of angles.

Which means that angles in the linear pair are always adjacent but

adjacent angles may not form a linear pair.

3. Supplementary angles –

If the sum of the measures of two angles is 180°, they are known as supplementary

angles.

It means angles in a linear pair are supplementary but it is not necessary that angles

which are supplementary form a linear pair of angles.

In the figure (a), BOD and AOD, similarly in the figure (b) AOB and

COB are angles in linear pair.

येथे 40o + 140o = 180o

Figure (a) Figure (b)

B

O

A

C

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4. Complementary angles –

If the sum of the measures of two angles is 90°, they are known as complementary

angles.

5. Opposite angles –

If sides of two angles form two pairs of opposite rays then angles are said to be

opposite angles.

In the figure, ∠AOC and ∠BOD is a pair of vertically opposite angle.

Similarly, ∠AOD and BOC is also a pair of vertically opposite angles.

Vertically opposite angles are always congruent.

40o

50

1300

A C

B

500

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Exercise

Qu: Read the given question carefully, draw the figure for the same and write

appropriate answers.

1) Is it possible that two acute angles are complementary angles

of each other?

2) Is it possible that two obtuse angles are supplementary angles

of each other?

3) Is it possible that two acute angles are supplementary angles

of each other?

4) When will be two adjacent angles are complement of each other?

5) Is it possible that two obtuse angles form a linear pair of angles?

Link :



1




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DAY : 6th

Topic :Euclidean geometry

Sub Topic: Conditional statement, converse of a statement and proofs of the theorem

Competency Statements:

1.From the given statement write ‘What is given’ and ‘What is to be proved’.

2 .Write the proof of given statements giving logical reasons.

Conditional statements: The statement which is written in the ‘If-then’ form is

called a conditional statement.

Antecedent: The part of the statement following ‘If’ is called the antecedent.

Consequent: The part following ‘then’ is called the consequent.

For example, consider the statement: The diagonals of a rhombus are perpendicular

bisectors of each other.

The statement can be written in the conditional form as, ‘If the given quadrilateral

is a rhombus, then its diagonals are perpendicular bisectors of each other.’

Converse: If the antecedent and consequent in a given conditional statement are

interchanged, the resulting statement is called the converse of the given statement.

Proof: Some self-evident geometrical statements which are accepted directly

to be true are called Postulates. On the basis of the postulates some more

properties can be proved logically. Such properties are called theorems.

The logical argument made to prove a theorem is called its proof.

Theorem: Properties proved logically are called Theorems.

Some of Euclid’s postulates are given below.

(1) There are infinite lines passing through a point.

(2) There is one and only one line passing through two points.

(3) A circle of given radius can be drawn taking any point as its centre.

(4) All right angles are congruent with each other.

(5) If two interior angles formed on one side of a transversal of two lines add

up to less than two right angles then the lines produced in that direction

intersect each other.

When we are going to prove that a conditional statement is true, its antecedent is

called ‘Given part’ and the consequent is called ‘the part to be proved’.

There are two types of proofs, Direct and Indirect.

logical argument made to prove a theorem is called its proof. There are two types of proofs, Direct and Indirect. right angles then the lines produced in that direction

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Exercise

1) Write following statement in ‘If’ and ‘Then’ form: Opposite angles of cyclic

quadrilateral are supplementary.

2) Write converse of following statement: Diagonals of rectangle are congruent.

3) For the following statement write ‘Given’ and ‘To prove’

If all three sides of a triangle are congruent then all three angles also are congruent.

4) For the statement given below draw suitable figure with names and write “given’

and ‘to prove’ for the same.

(i) Two equilateral triangles are similar.

(ii) If angles in the linear pair are congruent then each angle is right angle.

Useful Link :





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DAY :7th

Topic:Real Numbers Sub Topic: Properties of rational numbers

Competency Statements:1)To use addition property of rational numbers.

2) To use property of multiplication of rational numbers.

Let’s Recall:Carry out operations on fractions given below and fill proper numbers in

the boxes.

1) 13

8−

7

8= 2)

10

3−

15

3=

3) 6

7×

14

3= 4)

20

9÷

10

3=

20

9 × =

Important Points:

Properties of rational numbers:If a, b and c are rational numbers then,

Property Addition Multiplication

Commutative a + b = b + a a × b = b × a

Example

5

7+

3

7=

8

7 and

3

7+

5

7=

8

7

∴5

7+

3

7=

3

7+

5

7

5

7×

3

7=

15

49and

3

7×

5

7=

15

49

∴5

7×

3

7=

3

7×

5

7

Associative (a + b) + c = a + (b + c)

a × (b× c) = (a × b)× c

Example

(14 + 9 ) + (-8) = 23 – 8 =15

14 + [ 9 +(-8)] = 14 +1 = 15

∴(14 + 9) + (-8) = 14 + [ 9+(-8)]

14 [ 9 (-8)] =14 (-72)= -1008

(14 9) -8 =126 (-8)= -1008

∴14 [9(-8)] = (14 9) -8

Identity a + 0 = 0 + a = a a × 1 = a = a ×1

Example −11

8+ 0 = 0 +

−11

8= −

11

8 −

11

8× 1 = 1 ×

−11

8= −

11

8

Inverse a + (-a) = 0 𝟏

𝐚 × a = 1

Example 7

3+

−7

3= 0

1

9× 9 = 1

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Exercise

Q. 1 Choose correct alternative for the following questions.

(i) 8

7−

3

7=?

(A) 11

7 (B)

5

7 (C)

−11

7 (D)

−5

7

(ii) 10

9

5

7=?

(A) 15

9 (B)

15

7 (C)

50

63 (D)

50

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Q. 2 Complete the following activity.

(i)To find the value of 10

7+ (

15

7+

50

14) complete the following activity.

Activity:10

7+(15

7+

50

14)

= 10

7+(−−−

14 +

50

14)

= 10

7 + −−−

14

= 10

7+

−−−

7

=

(ii) Complete the following activity to find the value of 9

7× ( −14

15×

2

3 ).

Activity:9

7 ( −14

15×

2

3 )

= 9

7 −−−

45

= −−−

5

23

DAY:8th

Sub Topic: Operations on similar surds: Addition and Subtraction

Competency Statements:1) To carry out addition of similar surds

2) To carry out subtraction of similar surds.

Let’s Recall : 1) 42 = Square root of = 4 √16 = 4 2) 92 = Square root of 81 = √81 = 3) If √36

□ = 6 then the number in the box = ? Important Points:

Surds: If √𝑎𝒏

is a surd then√ symbol is called radical sign. ‘n’is the order

of the surd and ‘a’ is called radicand.

The surd of order 2 is called Quadratic surd.

Example: (i) In the surd √8𝟓

, order of the surd is 5 and radicand is 8.

(ii) In the surd, √𝟕 Order of this surd is 2, hence it is a quadratic surd.

Similar or Like surds:Two surds are said to be like surds if their order is same and

radicands are equal.

For Example: 5√3,2

7√3, −6√3 are similar surds.

Operations on like surds:

1) Addition and Subtraction: Mathematical operations like addition and subtraction,

can be done on like surds

Example: 1) Simplify. 7√5 + 12√5

Solution: 7√5 + 12√5

= (7 + 12)√5

= 19√5

2) Simplify: 14√7 − 9√7

Solution:14√7 − 9√7

= (14 − 9)√7

= 5√7

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3) Simplify: 3√8 + √50 − 4√2

Solution: 3√8 + √50 − 4√2

= 3√42 + √252 − 4√2

= 3 2√2 + 5√2 − 4√2

= 6√2 + 5√2 − 4√2

= (6 + 5 − 4)√2

= 7√2

Exercise

Q.1 Choose correct option.

(i) 2√3 + √3 = ?

(A) 2√3 (B) 2√9 (C) 2√6 (D) 3√3

(ii) 10√5 − 4√5= ?

(A) 14√5 (B) 6 (C) 6√5 (D) −6√5

Q.2 Solve the following.

1) Simplify:3√7 + 7√63 − √7

2) Simplify:7√5 − 4√5 + √80

3) Simplify:4√12 − √75 − 7√48

25

DAY:9th

Sub Topic: Operations on surds: Multiplication, division and rationalization

of the surds.

Competency Statements:1) To carry out multiplication of surds.

2) To carry out division of surds.

3) To carry out rationalization of denominator.

Important points:

1)Multiplication and Division: Multiplication and division can be done on surds.

Example: 1) Simplify,4√12 × 2√3

Solution: 4√12 × 2√3

= (4 × 2)√12 × 3

= 8√36

= 8 6

= 48

2) Simplify,8√35 ÷ 4√7

Solution:8√35 ÷ 4√7

= 8√35

4√7

= 8

4√

35

7

= 2√5

2) Rationalization of surd: If the product of two surds is a rational number, each

surd is called a rationalizing factor of the other surd.

Example:1) If surd√3 is multiplied by surd√3 we get√9.

√9 = 3 which is a rational number. Here rationalization of √3 is done

2) If the surd√3is multiplied by surd √12 then we get √36.

√36 = 6 is a rational number.

Here rationalization of √3 or √12 is done.

26

Exercise

Qu. 1 Choose the correct option.

(i) 3√6 × 2√2= ?

(A) 5√8 (B) 6√8 (C) 12√3 (D) 5√12

(ii) 6√12 ÷ 3√6

(A) 2√2 (B) 18√2 (C) 9√6 (D) 3√6

Qu. 2 Solve the following sub questions.

1) Simplify: 4√15 × 3√3

2) Simplify: 8√28 ÷ 2√7

3) Simplify: ( √5 + √2 ) ( √5 + 3√2 )

4) Rationalize the denominator.

(i) 1

√11 (ii)

3

5√7

Link:

https://diksha.gov.in/play/collection/do_312528209300701184153324?referrer=utm_s

ource%3Dmobile%26utm_campaign%3Dshare_content&contentId=do_31301856577

9193856123




27

DAY :10th

Topic: Parallel lines Sub Topic: Properties of Parallel lines


1. To Identify pairs of angles made by parallel lines and their transversal .

2. To understand various properties of pairs of angles and their application .

Let’s Recall :

1) What do you mean by parallel lines?

2) Write the names of all angles formed at

point R and Point S.

Important points :

Some important properties :

1) When two lines intersect, the pairs of opposite angles formed are congruent.

2) The angles in a linear pair are supplementary.i.e.the sum of measures of

angles in the linear pair is 180°

3) When two lines are intersected by a transversal, eight angles in all are formed

near their points of intersection.

4) Pairs of Corresponding angles

i) ∠a and ∠p ii) ∠b and ∠q

iii) ∠c and ∠r iv) ∠d and ∠s

Pairs of interior angles

i) ∠d and ∠p ii) ∠c and ∠q

Pairs of alternate exterior angles

i) ∠a and ∠r ii) ∠b and ∠s

Pairs of alternate interior angles

i) ∠d and ∠q ii) ∠c and ∠p

5) When one pair of corresponding angles is congruent, then all the remaining

pairs of corresponding angles are congruent.

6) When one pair of alternate angles is congruent, then all the remaining pairs

of alternate angles are congruent.

7) When one pair of interior angles on one side of the transversal is

supplementary, then the other pair of interior angles is also supplementary.

A B

C D

P

Q

R

S

L a

m

b

s

p

d

n

c

r q

28

l a

m

b

h

e

d

f

n

c

g

8) Interior angles theorem :If two parallel lines are intersected by a

transversal, the interior angles on either side of the transversal are

supplementary.

9) Corresponding angles theorem :The corresponding angles formed by a

transversal of two parallel lines are of equal measure.

10) Alternate angles theorem :The alternate angles formed by a transversal

of two parallel lines are of equal measures.

Exercise

Qu1: Choose correct option.

1) A transversal intersects two parallel lines. If the measure of one of the

interior angles is 45° then the measure of another angle is ............

(A) 350 (B) 450 (C) 1350 (D) 1250

2) The number of pairs of corresponding angles formed by a transversal of two

lines is ............

(A) 2 (B) 4 (C) 6 (D) 8

3) Two parallel lines are intersected by a transversal. If measure of one of the

alternate interior angles is 85° then the measure of the other angle is .............

(A) 150 (B) 1050 (C) 950 (D) 850

Qu2: Complete the following activity.

1) In the figure line AB ∥ line CD and line PQ is

their transversal which intersects line AB and line

CD in points R and S respectively.

Complete the following activity

Activity: (i) mARS = ……. ( Corresponding angles)

(ii) mPRB = ....... ( alternate angles )

(iii) If mBRS = 600 then mRSD = ..(Using property of interior angles)

(iv) mCSR + mRSD = ..... (Linear pairs of angles )

2) In the figure, line l ∥ line m, line n is their transversal.

If d = 700 Then find measure of g

Activity: line l || line m ..... (Given)

d = f ...........

But d = 700 ......... (Given)

∴ ∠ f =

∠ f + ∠ g = ………( Linear pair of angles )

A B

C D

P

Q

R

S

29

∴ 70° + ∠g = 180°

∴ ∠ g =

3)The corresponding angles formed by a transversal

of two parallel lines are ofequal measure.

To prove this complete following activity .

Given : line l ∥ line m

To prove : ∠ a =

Proof : ∠ a + = 180° …… (I) ( angles in linear pair)

+ ∠ c = 180° ……….(II) (property of interior angles of parallel lines)

∠ a + ∠ c = ∠ b + ∠ c …………… ( from (I) and (II))

∴ ∠ a =

(4) PQRS is a trapezoid in the figure.

Side PQ ∥ side SR and S = 650

So find the measure of P.

Activity : □PQRS is a Trapezium.

Side PQ ∥ ………. ( given )

Side PS is their transversal

∠ P + = 180° ……….(property of interior angles of parallel lines)

∴ ∠ P + = 180°

∴ ∠ P = 180° - 65°

∴ ∠ P =

Qu.3Solve following subquestions . )i ( In figure, if line l|| line m,

line nis their transversal and if

∠c= 50° find the values of ∠ a, ∠ q, ∠ r, ∠s.

(ii)The line p ∥ line q and the line r is their

transversal, From given information in the

figure find the value of x .

Link :https://diksha.gov.in/play/collection/do_312528194785001472250597?referrer=utm_source%3Dmobile%26

utm_campaign%3Dshare_content&contentId=do_313006482740953088130

l a

m

b

s

p

d

n

c

r q

q

120° r

p

l a

m b

n

c

R S

Q P

65°





30

:DAY :11th

Topic: Parallel lines Subtopic :Tests for parallel lines

Competency Statement: Application of tests for parallel lines.

Let’s recall:

From Figure write down all pairs of

Corresponding angles, alternate angles,

and interior angles.

Some Important Points :

(1)Interior angles test : If the interior angles formed by a transversal of two distinct

lines are supplementary, then the two lines are parallel.

(2)Alternate angles test: If a pair of alternate angles formed by a transversal of two

lines is congruent then the two lines are parallel.

(3) Corresponding angles test :If a pair of corresponding angles formed by a

transversal of two lines is congruent then the two lines are parallel.

Corollary I :If a line is perpendicular to two lines in a plane, then the two lines are

parallel to each other.

Corollary II :If two lines in a plane are parallel to a third line in the plane then those

two lines are parallel to each other.

Theorem :The sum of measures of all angles of a triangle is 180°.

Assignment

Q.1 :Select the correct alternative and fill in the blanks in the

following statements.

1)In ∆PQRQ = 700,R = 450 then P = .............. .

(A) 650 (B) 750 (C) 850 (D) 950

2) A transversal intersects two lines. If the measures of pair of

interior angles are 79° and 990then two lines are ............

(A) parallel (B)intersecting (C) equal (D) perpendicular

x a

y

b

s

p

d

z

c

r q

31

Q 2 : Complete the activity.

(1)Theorem : If a pair of alternate angles formed

by a transversal of two lines is congruent then

the two lines are parallel.

Given :line n is a transversal of line l and linem.

∠ a and ∠b is a congruent pair of alternate angles.

i.e. ∠a = ∠ b

To Prove :

+ ∠ c = 180° ……….. (I) (angles in linear pair) Proof :

∠ a = ∠ b ……… (II) (Given)

∴ + ∠ c = 180° ………. ( From (I) and (II))

But ∠ c and ∠ b are interior angles on one side of transversal line

line l ∥ …………. ( Interior angle test )

(2) In figure measures of some angles

are shown. Using the measures find the

measures of x and y and hence

show that line l || line m.

Solutin : ∠x = ……. Properties of opposite angles

∠y = 50° ……..

∠x + 50° = + 50°

∴ ∠x + 50° =180°

∴ line l ∥ line m ………..

Q 3. Solve

)i( In figure line ris a transversal of line p and line q.

If∠m = 58०and∠n = 121०then

are the lines p and q parallel ? Justify

(ii) Prove that, if a line is perpendicular to one of the two

parallel lines, then it is perpendicular to the other line also.

(iii) Prove that, if one angle of a parallelogram is right angle then that quadrilateral

is a rectangle.

Link : Use of properties of parallel lines and tests for parallel lines

:https://diksha.gov.in/play/collection/do_312528194785001472250597?referrer=utm_source


l 130°

m y

n

x

50°

l a

m b

n

c

p

q

r

m

n





32

DAY : 12th

Topic :Polynomials

Subtopic: Degree of the polynomial, standard form, coefficient form of a

polynomial


1) To identify the degree of polynomial.

2) To write down the polynomial in standard form. 3) To write down the polynomial in coefficient form.

Important Points and Revision :

Degree of a polynomial :The highest power of the variable is called the

degree of the polynomial.

For example : In the polynomial 2x + 5x2-3, variable is ‘x’ and highest power

is 2. So degree of this polynomial is 2. When the degree is 2 the polynomial is

called as Quadratic Poliynomial.

Standard form of the polynomial: When the polynomial is written in

descending or ascending powers of its variable, then it is called the standard form of

the polynomial.

Ex. -7+6x2+4x. This polynomial can be written in standard form as 6x2+4x-7

Coefficient form of the polynomial: One can write the coefficients of the variables

by considering all the missing terms in the standard form of the polynomial. This

form of the polynomial is called Coefficient form.

Ex:(i) Coefficient form of polynomial 6x2+4x-7 is ( 6,4,-7)

(ii) Coefficient form of polynomial 5x2-2x-3 is ( 5,-2,-3)

In the same way coefficient form of the polynomial can be written in index form.

Degree of the polynomial is less than 1 by total number of coefficients.

Ex. Write the polynomial in coefficient form (5,-4,0,3) into index form.

Coefficient form of the polynomial is (5,-4,0,3)

Total terms are 4, therefore degree = 4-1 = 3

By using x as a variable index form of the polynomial is 5x3+4x2+0x+3

Dear friends try to solve the examples based on above concepts.

33

Assignment

Qu.1:Select the correct alternative and fill in the blanks in the following statements.

(1) Degree of the polynomial 3x3+ 5x2+5x– 5 si ....

A) 1 B) 2 C) 3 D) 4

(2) Standard form of 5x3 – 2x + 6x4 + 1 is . .............

A) 6x4 + 5x3 – 2x + 1 B) 5x3 – 2x + 1 + 6x4

C) 6x4 + 5x3 + 1 – 2x D) 6x4– 2x + 1+ 5x3

(3)Polynomial 4y2 + 5y – 3 is written as …………….in coefficient form.

A) (4, 5, 3) B) (4, -5, -3) C) (4,-5,3) D)(4, 5, -3)

Q 2: Complete the activity

1) Complete the table .

Sr.

No. Polynomial Degree

1 4p5 + 7p3 – 8p + 1

2 6a5 + 7a8 – 9

3 C9 – 1

2) Complete the table


Sr.No. Polynomial Standard form

1 4d3 + 5d4 – 3d2 + 1

2 6y2 – 3y5 + 7y3 – y4 + 10

3 9p2+ 8p3 – 9 + 5p

Sr.

No. Polynomial Coefficient form

1 7p2 – 8p + 3

2 9t3 + 7t2 – 5t + 9

3 10h – 4

34


Link

:https://diksha.gov.in/play/collection/do_312528209300701184153324?referrer=utm_source%3Dmobile%26utm_campaign%3Dshare_content&contentId=do_31243867414684467211791

Sr.No. Polynomial in

coefficient form

Polynomial in index form

( use x as variable)

1 (1,-2, 3)

2 (1, 0 , 0 , -3)

3 ( -4, 0, 1, 2 , 9 )



35

DAY:13th

Topic :Polynomials

Sub topic : Operation on Polynomials

Competency Statements :

1) Can do addition and subtraction of polynomials:

2) Can do multiplication of polynomials:

Important points and revision:

The methods of addition, subtraction, multiplication and division of

polynomials is similar to the operations of algebraic expressions

Addition

Ex :(3x2 – 5x + 3) + (8x + 4x2 – 4)

= 3x2+ 4x2– 5x + 8x +3 – 4 ……(like terms taken together )

= 7x2 + 3x – 1

Subtraction

Ex : (4P3 + 3P2 – 5) – (2P – 5)

= 4P3 + 3P2 – 5 – 2P + 5……......(solving bracket)

= 4P3 + 3P2– 2P– 5 + 5..........(like terms are arranged )

= 4P3 + 3P2– 2P

Multiplication :

Ex.1)3a ( 4a – 5)

= (3a x 4a) – (3a x 5)

= 12a2 – 15a

2) (2a + 3 ) ( 5a2 + 4a)

= 2a (5a2 + 4a) +3 (5a2 + 4a)

= (2a x 5a2) + (2a x 4a) + (3 x 5a2) + (3 x 4a)

= 10a3+ 8a2 + 15a2 + 12a

= 10a3 + 23a2 + 12a

Dear friends, try to solve the examples based on above concepts.

36

Assignment

Q 1 :Write ( ) in appropriate box

Sr.

No Question Yes No

1 Are 2a2 and 3a like terms?

2 Can we add 2a2 and 3a?

3 2a2 3a = 6a3, Is it true?

4 Is it correct?

2a2- 3a = a (2a – 3)

Qu 2 : Solve the examples :

1)Add the given polynomial.

a) (3b2 + 4b – 5) + (5 + 4b2 – 8b)

b) (15w3 + 12w2 – 5w) + ( 13w4 – 12w2 + 7w -18)

2) Subtract :

a) (2a4 – 23a2 – 8) – (3a4 – 3a2 -9)

b) (3a2 – 5a + 4a3) – (4a3 – 10a +11)

3) Multiply :

a) 3p (3p + 9)

b) 5a (-9a2 + 3a)

c) (3q – 2 ) (5q + 3)

d) (-2b + 3) (3b2 + 8b)

37

DAY :14th

Topic: Polynomials

Subtopic: Value of the polynomial.

Competency statement: 1) To find the value of polynomial when value of variable

is given.

Important points and revision :

Value of a polynomial: In a polynomial if variable is replaced by a given number

then we get the value of that polynomial for that number.

Ex. P(x) = x2 + 5x – 10 , Find the value of polynomial if x = 3.

In above example ,x2 + 5x – 10 , x is variable and we want to find out

the value of polynomial at x=3.

Steps are as follows :

P(x) = x2 + 5x – 10

P(3) = 32 + 5(3) – 10

= 9 + 15 – 10

= 14

Thus the value of polynomial P(x) = x2 + 5x – 10, at x=3 is 14.

Friends, solve the questions based on above concept.

Assignment

Qu.1 Select the correct alternative and fill in the blanks in the following statements.

1) The value of the polynomial x+5 for x = 2 is ……………

A) 2 B) 4 C) 5 D) 7

2) The value of the polynomial z2 – 5z + 2 for z=0 , is …….

A) 2 B) 7 C) -7 D) -2

3) The value of the polynomial y-12 for y = -1 is ……………

A) -12 B) -13 C) 11 D) -11

38

Q 2 : Complete the activity

1) If p=3, find the value of the polynomial 3p3+ 2p2 – 4p -3 Complete the activity:

Solution : p(p) = 3p3 + 2p2 – 4p -3

p(..) = 3(3)3 + 2(3)2 – …….(3) -3

= 81 + 18 – (…..) – 3

= (…)

2) If the value of the polynomial ax2 – 2x - 8 for x = - 2 is 4, then find the value

of a.

Solution : p(x) = ax2 – 2x - 8

p(..) = a(-2)2 – 2(-2) – (….)

= 4a + 4 – 8

= 4a – (……)

But x = -2 …. (given)

∴ 4a – 8 = 4

∴ 4a = (…)

∴ a = (…)

Q 3 : Solve the examples :

1) If p(x) = x2 – 5x + 9, x = 2 then find p (x )

2) If s = 3 , find the value of the polynomial s3 – 4s + 8 .

3) Value of the polynomial 2y2 – ky + 3 for y = -1 is 6 .Find the value of k.

39

DAY :15th

Unit: Polynomials

Subunit: Factors of the Polynomial

Competency Statement : Students factorize the polynomial.

Important points and Revision

Factors of the polynomials:

• ax2 + bx + c This type of polynomial is called as quadratic polynomial. In

std.8th we have learnt the method of finding out the factors of this type of

polynomial

• Solve some examples.

Solved Examples:

Find the factors of the polynomials given below.

1) 4x2 – 8x

= 4x (x – 2)…………..(4xis common )

2) y2 – 25

= ( y – 5 ) (y + 5) ………[By the formula (a2– b2) = (a-b)(a+b) ]

3) 3x2 + 7x + 2

For factorization , Here we find the product of coefficient of first term and third

term

3 x 2 = 6

Now consider the factors of 6 such that their sum is 7.

Thus is 7=6+1

Now, 3x2 + 7x + 2 = 3x2 + 6x+ 1x + 2

= 3x ( x + 2 ) + 1 ( x + 2 )

= (x + 2 ) ( 3x + 1 )

Dear friends try to solve the examples based on above concepts .

40

Assignment

Qu.1 Select the correct alternative and fill in the blanks in the following statements.

1) Factors of the polynomial x2 + 3x + 2 are .................

A) (x + 2 ) (x + 1) B) (x + 3 ) (x + 1)

C) (x + 3) (x + 2) D) (x+3) (x – 2)

2) Factors of the polynomial 4p2– 9 is …………..

A) (2p – 3) ( 2p – 3) B) (2p – 3) ( 2p + 3)

C) (2p +3) ( 2p +3) D) (4p – 3) ( 4p –3)

3) Factors of the polynomial 5x2 - 125 is ………….. .

A) ( x – 5 ) (x - 5) B) 5 ( x + 5 ) (x + 5)

C) 5 ( x – 5 ) (x + 5) D) ( x – 5 ) (x + 5)

Qu .2. Complete the activity .

1) For factorization of the polynomial 5x2 – 12x – 9, complete the activity.

= 5x2 – (……..) + (……..) − 9

= 5x(.........) + 3 (……….)

= (x – 3) (………..)

2) For factorization of the polynomial n2 – 5n + 6, complete the activity.

= n2 – (……..) - (……..) + 6

= n (.........) - 2 (……….)

= (n – 3) (………..)

3) For factorization of the polynomial p2 + 7p + 10 complete the activity.

= (…….) + 5p + 2p + 10

= p (………) + (…..) (p + 5)

= (……) ( p + 5)

Qu.3 Factorize the following polynomials.

1) 6x2 – 54x

2) 7p2 – 63

3) a2 + 11a – 42

4) 2h2 + 12h + 18

5) 3k2 – 18k + 15

41

1

TEST - 1 DAY:16th

Std :10th Subject : Mathematics

Time :1 Hour Marks :15

Q.1. Select the correct alternative and fill in the blanks in the following

statements. [4 marks]

i) If A = {a ,b ,c ,d ,e ,f, g, h {= B , } a ,c ,e { =C, } a ,e, i {= D , } g ,h ,i}

Then which statement is true from the following.

(A) CA ) B) BA (C) DA (D) BD

ii) A transversal intersects two parallel lines. If the measure of one of

the angles is 40° then the measure of its corresponding angle is .............

(A) 140° (B) 80° (C) 50° (D) 40°

iii) Coefficient form of the polynomial 2x2-5x+12 is ………………….

(A) (2, -5, 12) (B) (2, 5, -12 ) (C) (2, 5, 12 ) (D) ( 2, -5, -12 )

iv)Two parallel lines are intersected by a transversal. If measure of one of

the alternate interior angles is 50° then the measure of the other angle is .......

(A) 50° (B) 130° (C) 40° (D)1400

Q.2. Solve [3marks]

i)When two distinct lines intersect each other, how many points are there in their

intersection?

ii) Point Q is a midpoint of seg PR. If PR=10 cm find PQ=?

iii) Coordinates of points A and B are -5 and 3respectively. Find (A,B).

Q.3. Solve the following examples [8 marks]

i) In figure line RP || line MS and line DK is their

transversal. DHP = 85° Find the measures of

following angles.

(a) ∠RHD (b) ∠PHG (c) ∠HGS (d)∠MGK

ii) (a) write down the set of even prime numbers

from 1 to 25 in list form.

(b) B ={ } then n(B) =?

iii ) Simplify : ( 4x2+7x-1)-(x2+8x-7)+(x2+10)

iv) Simplify :3√2 + √8 − √50

42

DAY :17th

Topic :Triangle Subtopic:Types of triangle

Competency Statement:1)To Indentify types of triangles based on sides and angles.

2) In isosceles triangle to understand the relation between congruent

sides and angles.

Let’sRecall:Draw any triangle. Measure all sides of a triangle using scale.

Measure of all angles of a triangle by using protractor.

Important points :

Triangle :A closed figure obtained by joining three non-collinear points is called

a triangle. Triangle has three vertices, three sides and three angles.

Types of triangles (On the basis of sides) -

1) Equilateral triangle

An equilateral triangle is a triangle in which all

three sides have the same length.

In △ABC,AB = BC = AC.∴△ABC is an

equilateral triangle.

Each angle of an equilateral triangle is of 60°.

2) Isosceles triangle

An isosceles triangle is a triangle in which two

sides have same length.

In △UVW,UV = UW . ∴△UVW is isosceles triangle.

3) Scalene triangle -

Scalene triangle is a triangle in which all sides have different lengths.

Types of triangles (On the basis angles)

1) Acute angled triangle– An acute angled triangle is a triangle with three

acute angles. In figure △DEF is acute angled triangle.

2) Right angled triangle–Right angled triangle is a triangle with one right angle.

.In figure PQR is a right angled triangle.

3) Obstuse angled triangle – Obtuse angled triangle is a triangle in which one

angle is obstuse. In figure △LMN is an obstuse angled triangle.

43

Isosceles triangle Theorem and its converse

1) Theorem : If two sides of a triangle are congruent then

the angles opposite to them are congruent.

∴ In △ABC if side AB ≅ side AC, then∠B ≅∠C.

2)Converse :If two angles of a triangle are congruent then

the sides opposite to them are congruent.

∴ In △ABC if∠B ≅∠C then side AB ≅ side AC.

Assignment

Qu1) Choose the correct alternative answer for the following questions.

1) If sides of a triangle are 3.4 cm, 3.4 cm, and 5 cm.What is the type of triangle?

(A) Equilateral triangle (B) Scalene triangle

(C) Scalene triangle (D) Right angled triangle

2) In △XYZ ,∠Z ≅∠X then which sides are congruent of△XYZ ?

(A) Any two sides (B) XY and XZ (C) XY and YZ (D) XZ and YZ

Qu 2) Complete the following activity;

From figure, find the value of x and y

(a) In △ABC AB ≅ AC ……( given)

∴ ∠ABC ≅∠ (angles opposite to

congruent sides )

∴ x =

(b) In △BDC BD ≅ DC ……( given)

∠∴ DCB ≅∠ …… ( angles opposite to congruent sides )

∴ y =

Qu 3) 300 and 450 are the measures of two angles of triangle. What is the type of a

triangle?

Qu 4) Perimeter of an equilateral triangle is 16.5 cm. Find the length of its side.

Link :Isosceles triangle Theorem and its converse

https://diksha.gov.in/play/collection/do_312528194785001472250597?contentId=do_313

018562688786432174



44

DAY :18th

Topic:Triangle Subtopic : Congruence of triangles.

Competency statement :To apply the tests of congruence of triangles .

Let’s Recall :

1) If lengths of two segments are equal then they are called as …….segments..

2) If measures of two angles are equal then they are called as ………. Angles.

3) Which sign is used for congruence?

Important points :

Congruence triangles :

If all corresponding sides and corresponding

angles are congruent of two triangles then

those triangles are called as congruent

triangles .

In △ABC and △DEF

(a) seg AB ≅ seg DE, seg BC ≅ seg EF, seg AC ≅ seg DF

…….. ( corresponding sides are congruesnt )

(b) ∠A ≅∠D, ∠B ≅∠E, ∠C ≅∠F (corresponding angles are congruent)

∴△ABC ≅△DEF

Tests of Congruence :–

1) S-S-S Test : In a correspondence, if three sides of

ABC are congruent to three sides of PQR , then

the two triangles are congruent.

.

2) S-A-S Test :In a correspondence, if two sides of

XYZ are congruent to two sides of LMN and the

angles included by the respective pairs of sides are also

congruent, then the two triangles are congruent by S-A-S

test.

3) A-S-A Test :In a correspondence, if two angles of

ABC are congruent to two angles of PQR and the

sides included by the respective pairs of angles are

also congruent, then the two triangles are congruent by A-S-A test.

45

4( ) S-A-A Test :In a correspondence, if two

sides of LMN are congruent to two sides of

DEF and the angles included by the respective

pairs of sides are also congruent, then the two

triangles are congruent by S-A-A Test.

(5)Hypotenuses side Test : In a correspondence, if

hypotenuses of two right angled triangle are

congruent and one pair of corresponding sides are

congruent, then the two triangles are congruent by

hypotenuse-side test.

Ex. (1) RST≅PMA write down all pairs of corresponding sides and angles.

Solution: Pairs of corresponding sides

(1) seg RS ≅ seg PM (2) seg ST ≅ seg MA (3) seg RT ≅ seg PA

Pairs of Corresponding angles :

(1) ∠R ≅∠P (2) ∠S ≅∠M (3) ∠T ≅∠A

Ex. )2 ( As shown in the figure ,write the test which

assures the congruence of the two triangles.

Solution: In △HEG and △FGE

seg HE≅ seg GF ……. ( given)

seg HG ≅ seg EF ……. (given)

seg EG ≅ seg EG ……. ( common side)

∴△HEG ≅ ΔGFE (S-S-S test )

Assignment

Q 1)Select the correct alternative and fill in the blanks in thefollowing statements

1) By one-one correspondence of DEF ↔ RST,seg DE ≅ seg RS, seg DF≅seg RT

and ∠D≅∠R then by which test two triangles are congruent?

(A) S-S-S Test (B) S-A-S Test (C) A-S-ATest (D) S-A-A Test

2) In △NTS and △PQR, seg NT ≅ seg QR, seg TS ≅ seg PR and seg SN ≅ seg PQ

. Which of the following statement is appropriate for congruent triangles?

(A)△NTS ≅△PQR (B) △STN ≅△PRQ (C) △TNS ≅△RPQ (D) △PQR ≅△TSN

3) In △ABC and △DEF seg AB ≅ seg FD and ∠A ≅∠D .Which additional

information is to be given in △ABC व△DEF, so that they will be congruent

by S-A-S test.?

(A) Seg AC≅ Seg DE (B) Seg BC≅Seg EF (C) Seg AC≅Seg EF (D) Seg BC≅Seg DE

46

Qu 2) △TUV ≅△ LMN, complete the boxes.

(a) side TU ≅ side (b) side ≅ side MN (c) side ≅ side

(d) ∠ ≅∠L e) ∠ U ≅∠ (f) ∠ ≅∠

Qu 3) State the test by which the triangles in each pair are congruent.

Qu 4)If in △ABC segAB ≅ segAC and ray AD is an angle

bisector of ∠BAC, then complete the activity to prove

∠B ≅∠C

Activity :

In △ABD and △ACD

segAB ≅ segAC ……… (given)

∠BAD ≅ …….. (angle bisector)

segAD ≅ ……… ( common side)

∴ △ABD ≅△ACD ………

∴∠B ≅ ..… ( corresponding angles of congruent triangle)

Link :

1) Congruent Triangles -


0140222755061761158

2) Congruent Triangles –


30185626663813121656

...…………………test ...…………………test ...…………………test

1) 2) 3)





47

DAY :19th

Topic :Triangle Subtopic : Property of triangle 30°-60°-90°and 45°-45°-90°

Competency statement: To understand the properties of 300-600-900 and 450-450-900

theorem and use of it.

Let’s Recall :

1) In △ABC write down the relation between sides of a

triangle by Pythagoras theorem.

2) Draw a square. Draw its diagonal. Measure the angles of any

triangle formed by this diagonal.

3) What is the measure of angles formed by altitude in an equilateral triangle ?

Important points :

300-600-900 triangle theorem:

If the acute angles of a right angled triangle have

measures 300 and 600 then the length of the side

opposite to 300 is half the length of the

hypotenuse, and side opposite to √3

2is °60

hypotenuse.

a) Opposite side of angle 300= 1

2 × hypotenuse

∴ AB =1

2 × AC

(b) Opposite side of angle 600= √3

2 × hypotenuse

∴ BC = √3

2 × AC

450 –450–900 triangle theorem:

If measures of angles of a triangle are 450,45 0and 900

then the length of each side containing the right angle

is hypotenuse

√2

Opposite side of angle 450 = 1

√2×hypotenuse

∴ AB = BC = 1

√2 ×AC

Ex )1: ( From adjacent figure if PR=12cm then find PQ

and QR.

Solution :𝒎∠P=60°, 𝒎∠Q=90°, 𝒎∠R=30°

∴ By 30°-60°-90° triangle theorem,

(a) opposite side of angle 300 = 1

2 × hypotenuse

48

∴ PQ = 1

2 × PR =

1

2 × 12 = 6cm

(b) opposite side of angle 600 = √3

2 × hypotenuse

∴ QR = √3

2 × PR =

√3

2 × 12 = 6√3 cm

Ex. )2: ( If diagonal of a square is cm, find its side.√210

Solution :◻ABCD is a square

AC = 10√2cm

△ABC is 45°-45°-90° triangle.

Opposite side of angle 450 = 1

√2×hypotenuse

∴ AB = BC = 1

√2 × AC =

1

√2 ×10√2

= 10

∴ Side of a square = 10

Assignment :

Qu1) Choose the correct alternative answer for the following questions.

)1 ( In an isosceles right angled triangle what is the measure of each acute angle.

(A) 30° (B) 45° (C) 60° (D) 90°

(2) Side of an equilateral triangle is 12cm .Find the height of that triangle?

(A) 4cm (B) 4√3 cm (C) 6 cm (D) 6√3cm

Qu 2. From figure for finding out AB and BC, Complete the activity.

Activity :AB ≅ BC …. (given)

∴∠A ≅∠C …. (isosceles triangle theorem)

∴𝒎∠A = 𝒎∠C = and𝒎∠B = 90°

∴ By45°-45°-90° theorem

AB = BC = × AC

= × √8

=

Q3 In right angled triangle △LMN, LM is hypotenuse∠L = 30° and MN = 4cm .

Find LM and LN.

Link :1) Property of 30°-60°-90° triangle https://diksha.gov.in/play/collection/do_312528194785001472250597?contentId=do_3130185627115520001168 2) Property of 45°-45°-90° triangle https://diksha.gov.in/play/collection/do_312528194785001472250597?contentId=do_313018562733211648172





49

DAY :20th

Topic:Triangle Sub Topic: Similar triangles

Learning Outcomes:To identify similar triangles and to write the ratio of their

corresponding sides.

Let’s Recall:

1) How many different way we can show one to one correspondence of

the vertices of a triangle.

2) Write corresponding sides and corresponding angles for the

correspondence, GST↔ KLM. Important points:

Similar triangle:

For two triangles, if corresponding angles of two triangles are congruent and the

corresponding sides are in proportion, then the triangles are said to

be similar triangles.

In △ABC and △PQR

1) ∠A ≅ ∠P, ∠B ≅ ∠Q, ∠C ≅ ∠R

… (Corresponding angles are congruent)

2) AB

PQ=

BC

QR=

AC

PR

… (Ratios of corresponding sides are equal)

Then △ABC△PQR

Ex. (1): In the figure given alongside decide

whether the triangles are similar or not? If

they are similar then write correspondence

of the vertices.

Solution :In △DEF and △GHI,

∠D ≅ ∠G, ∠E ≅ ∠H, ∠F ≅ ∠I …………. (1) and

DE

GH=

9

6=

3

2,

EF

HI=

12

8=

3

2,

DF

GI=

15

10=

3

2…….(2)

Form )1 ( and )2 ( corresponding angles are congruent and

corresponding sides are in proportion.△∴DEF ~△GHI

△DEF and △GHI are similar by the correspondence DEF ↔ GHI of vertices.

Ex2):If △ABC ~△PMN, AB = 6, BC = 10, AC = 8 and MN = 4 then find PM

and PN .

Solution : △ ABC ~ △PMN

∴ AB

PM=

BC

MN=

AC

PN ..…… …(Corresponding sides are in proportion)

∴ 6

PM=

10

4=

8

PN

50

(a) 6

PM=

10

4 ∴ 10 × PM = 6 × 4 ∴ PM =

24

10 ∴ PM = 2.4

(b) 10

4=

8

PN ∴ 10 × PN = 8 × 4 ∴ PN =

32

10 ∴ PN = 3.2

Assignment

Qu.1) Choose correct options from the given options.

(1) If △ABC ~△PQR, 𝒎∠A = 40° and 𝒎∠B=35° then𝒎∠R =?

(A) 75° (B) 15° (C) 95° (D) 105°

(2) If △LMN ~ △QTP, 4LM = 3QT andLN = 6 cm then find the length

of PQ

(A) 5.4 cm (B) 4.5 cm (C) 8 cm (D) 13 cm

Qu.2) RST ∿XYZ then complete the following statements.

(a) ∠R ≅ …, ∠S ≅ …, ∠T ≅ …

(b) RT

XZ=

.….

YZ,

RS

XY=

ST

.….,

XY

…..=

YZ

ST

Qu.3) If DEF~UVW, 𝒎∠D=70° and 𝒎∠ E=50°then complete the activity to find

measures of angles UVW

(a) △DEF~△UVW

∴ ∠D ≅ ……..(Corresponding angles are congruent)

but 𝒎∠D=70° ∴ 𝒎∠U=

(b) Similarly∠E ≅ …….( Corresponding angles are congruent)

but 𝒎∠E=50°∴𝒎∠V=

(c)Now 𝒎∠U + 𝒎∠V + 𝒎∠W = (Sum of measures of angles of

triangle)

∴ 70 + 50 + 𝒎∠W = 180

∴𝒎∠W =

Q 4) Draw rough figure of similar triangles and name it. Show their corresponding

angles with same marks and write their corresponding sides using proper numbers.

Link :1) Similar figures

https://diksha.gov.in/play/collection/do_312528209258020864153216?contentId=do_

313000776267538432114

2) Similar triangle

https://diksha.gov.in/play/collection/do_312528194785001472250597?contentId=do_

3130185628145827841657

https://diksha.gov.in/play/collection/do_312528209258020864153216?contentId=do_%20%20313000776267538432114

https://diksha.gov.in/play/collection/do_312528209258020864153216?contentId=do_%20%20313000776267538432114

https://diksha.gov.in/play/collection/do_312528194785001472250597?contentId=do_%203130185628145827841657

https://diksha.gov.in/play/collection/do_312528194785001472250597?contentId=do_%203130185628145827841657

51

DAY :21st

Topic:Triangle Sub Topic: Some properties of triangle

Competency Statements:To understand some properties of triangle and to apply

them to solve examples.

Let’s Recall: Draw any one triangle. Show all its exterior angles. How many exterior

angles of triangles can be drawn?

Important Points:

Exterior angles of a triangle:

If any side of a triangle is extended then the angle which forms

linear pair with corresponding interior angle of a triangle, that

angle is called exterior angle of a triangle. In the figure given

alongside ∠PRS is an exterior angle of △PQR. ∠P and ∠Q are

remote interior angles w.r.t. ∠ PRS

Theorem of remote interior angle:

In a triangle, the measure of an exterior angle is equal to the sum of remote

interior angles.

∠PRS=∠P+∠Q

Theorem of perpendicular bisector :

(1) Every point on the perpendicular bisector of a

segment is equidistant from the end points of the

segment. In the figure given alongside,

line 𝑙 is perpendicular bisector of seg AB.

∴𝑙(AP) = 𝑙(PB)

(2) Any point equidistant from the end points of a

segment lies on the perpendicular bisector of the

segment

If (AP) = (PB) then point P is on the perpendicular bisector of seg AB

Median:

i) Definition-A segment joining the vertex and the

midpoint of the opposite side is called the median

of a triangle.

ii) Property-All three medians of a triangle are

concurrent. Their point of concurrence (i.e.centroid)

divides each median in the ratio 2:1.

In △ABC, seg AD is median. Point G is the point of concurrence of median.

∴ AG:GD = 2:1

52

Property of sides of a triangle:

The sum of any two sides of a triangle is greater than the third side .

Similarly, the difference between any two sides is always smaller than the third

side of the triangle.

In △ABC (a) AB + BC > AC (b) AB + AC > BC (c) BC + AC > AB

Also (a) AB – BC < AC (b) AB – AC < BC (c) BC – AC < AB

Assignment

Qu.1) Choose the correct option.

(i) In △DEF, ∠D = 50°, ∠E=80° then find the measure of exterior angle ∠DFG

of △DEF.

(A) 130° (B) 100° (C) 70° (D) 50°

(ii) Two sides of a triangle are of length 5 cm and 1.5 cm then out of following

which cannot be the length of third side.

(A) 3.7 cm (B) 4.1 cm (C) 3.8 cm (D) 3.4 cm

Qu.2) Observe the figure and answer the flowing questions.

(a) Name the exterior angle of the triangle.

(b) With respect to this exterior angle, name the pair

of remote interior angles.

(c) Which property can be used to find the measure

of ∠ACD?

(d) If ∠B = 50°and ∠A=75° then find m∠ACD.

Qu.3) In △PQR, PT is median and point G is centroid. If GT= 2.5 cm then find PG

and PT .

Link :

1)Theorem of remote interior angle

https://diksha.gov.in/play/collection/do_312528194785001472250597?contentId

=do_3130185626182451201620

2) Median of a triangle and its

propertieshttps://diksha.gov.in/play/collection/do_312528194785001472250597?content

Id =do_3130185627472117761508

3) Theorem of perpendicular and angle

bisectorhttps://diksha.gov.in/play/collection/do_312528194785001472250597?contentI

d =do_313018562765660160175

https://diksha.gov.in/play/collection/do_312528194785001472250597?contentId%20%20%20=do_3130185626182451201620

https://diksha.gov.in/play/collection/do_312528194785001472250597?contentId%20%20%20=do_3130185626182451201620

https://diksha.gov.in/play/collection/do_312528194785001472250597?contentId%20=do_3130185627472117761508




53

DAY :22nd

Topic : Geometric Construction SubTopic : Basic Constructions

Competency Statements :To carry out basic geometric constructions using various

geometric instruments.

Let’s recall:1) Draw seg AB of length 4.5 cm.

2) Draw a circle of radius 3.5 cm.

Important Points:

Essential Basic Constructions:

1) To draw perpendicular bisector of a given segment.

Example: Draw seg AB of length 4.3cm and bisect it.

Steps of construction:

1) Draw seg AB of length 4.3 cm

2) Keep metal point of compass on point A, take more than half distance of distance

AB and draw arcs on both the sides of the segment AB.

3) Keep same distance in the compass and keep metal point of the compass on point

B and draw arcs which will intersect the previously drawn arcs.

4) Mark the points of intersection of the arcs and

draw a line passing through them.

2) To draw perpendicular to a line from a point on the line.

Example: Draw line l. Take any point P on it. Draw line m perpendicular to line l

andpassing through point P.


1) Draw a line and name it as l.

2) Take any point P on it.

3) Keep pointed end of the compass on point P and draw two arcs on the line l on

both sides of point P at a same distance and name the points as point A and

point B.

54

4) With A as centre and radius equal to more than

half of the distance AB draw arcs on

both the sides of line l.

5) Keeping same distance in the rounder, keep

pointed end of the rounder at point B draw two

arcs which intersect previous arcs.

6) Draw line m passing through the point of

intersection of these two arcs.

7) Line m is perpendicular to line l.

3) To draw a circle of given radius or diameter. In that circle to draw chord of

given length.

Example:Draw a circle of diameter AB of length 6.3 cm. Draw a chord MN of

length 4cm.


1) Draw seg AB of length 6.3 cm.

2) To find midpoint of seg AB draw perpendicular

bisector of seg AB and name it as line l. Point O is the

midpoint of seg AB which is the center of the circle.

3) Now keep pointed end of the compass on point ‘O’

and taking distance OA or OB in the compass draw a

circle passing through points A and B.

4) Taking distance 4 cm in the compass decide position

of chord MN.

Exercise

1) Draw seg AB of length 4.7 cm and bisect it.

2) Draw line 𝑥. Take any point A on it. Draw line 𝑦 from point A perpendicular to

line 𝑥.

3) Draw segment PQ of length 5.7cm. Take any point R on it such that 𝑙 (QR) = 3cm

and P-R-Q. Draw perpendicular to segment PQ from point R.

4) Draw a circle of diameter 7.3 cm. Draw chord XY of length 5.7cm.

Link :

1) To draw perpendicular bisector of a segment.

https://diksha.gov.in/play/content/do_3130384857885818881157


55

DAY: 23rd

Topic:Geometric Constructions Subtopic:Basic Constructions

Competency Statements: Will be able to construct a triangle.

Let’s remember:Draw∠ABC of measure 75°

Important points:

Important basic construction –

1) To construct a triangle when length of two sides and measure of angle in

between them is given.

Ex.:In∆PQR 𝑙(PQ) = 5.5 cm,m∠P = 50°, 𝑙(PR) = 5cm then

construct △PQR.

Steps of construction :

(1) Draw base PQ of length 5.5cm .

(2) Draw a ray making an angle of 50o at point P .

) 3) With P as centre and radius equal 5 cm draw an arc to

get point R.

(4) Join points R & Q

2) To draw congruent angle to the given angle:

Ex.:Draw ∠PQR which is congruent to angle ∠ABC

without using protractor.


1) Draw ray QR.

2) Take sufficient distance in between

two points in compass. Keep metal

point of compass on point B and draw

an arc intersecting two sides of

∠ABC.

3) Now with the same distance and by keeping compass on point Q draw an arc

as shown in figure.

4) With the help of compass take distance equal to an arc drawn from vertex B of

angle ABC.

5) Now keep the metal point of compass on point where arc drawn from point Q has

been intersected ray QR. Draw another arc intersecting the arc drawn before.

Name the point of intersection as O and then draw ray QP passing through point O.

Congruent Angle Given Angle

56

3) To divide a line segment in a given ratio.

Ex.:Draw seg AB of length 7.6 cm.Divide it in

the ratio 5:8.


1) Draw seg AB of length 7.6 cm

2) Draw any ray AX , making an acute angle

with AB

3) With the help of compass mark 5 + 8 = 13

points at equal distance on ray AB. Name them as A1,

A2, A3, …. , A13

4) Join point B with A13.

5) Draw a parallel segment to segment BA13 from A5.

6) Parallel line drawn from A5 will intersect seg AB. Name the point as C.

C is a point on seg AB which divides it in the ratio 5:8

Exercise

1) In △STU, 𝑙(ST) = 6cm, 𝑙(TU) = 4.5cm &𝑙(SU) = 5 cm then construct△STU.

2) In △DEF, 𝑙(DE) = 5.5cm, 𝒎∠E = 70° & 𝑙(EF) = 6.3 cm then construct△DEF.

3) In△PQR 𝑙(PR) = 7cm, 𝒎∠P = 40° and 𝒎∠R = 75° then construct△PQR.

4) Draw any angle ABC. Construct another angle PQR congruent to angle ABC

without using protractor.

5) Draw seg MN of length 6.5cm. Divide in the ratio 3:2.

Link :

1) To construct a triangle –


2) To divide a line segment in a given ratio


30007769159270401210




57

Day :24th

Topic:Equation in two variables.

SubTopic: Equation in two variables-concept

Competency Statements:1) Will be able to explain concept of equation in two

variables.

2) Will be able to find solution of equations in two variables.

Lets revise :

Qu.1: Observe the equations in the following table and identify variables in them and

degree of the equation.

Equation Variable Degree

x + 2 = 5

m – 3 = 7

y + 5 = 10

Qu.2: Solve following equations.

1) m + 7 = 3 2) 2x + 5 = 13 3) y + 3 = 2

In all the above equations there is only one variable and its degree is one hence this

type of equations are called as equation with one variable.

Important points:

I) Equation in two variables:

Ex.:1) Find two numbers whose sum is 12.

Equation form : x+ y = 12

Above equation is equation in two variables.In this equation two different

variables are used. Degree of both the variables is 1. We can get more than 1

solution for above equation and those are as follows,

9+3 =12, 7+5=12, 8+4 =12, 6+6 =12, (-1)+13 =12, 10+2=12, ...

Hence solutions for above equation are (9, 3), (7, 5), (8, 4), (6, 6), (-1, 13), (10, 2 ), ...

Ex.:2) Find two numbers with difference of 8.

Equation form : x - y = 8

Above equation is equation in two variables.

We can get more than 1 solution for above equation and those are as follows,

9 –1 = 8, 10 - 2 = 8, 11 - 3 = 8, 12 - 4 = 8,13 - 5 = 8, 15 - 7 = 8,…

Hence solutions for above equation are (9,1), (10, 2), (11, 3), (12, 4), (13, 5),

(15, 7) ….

58

II) General form of equation in two variables:

ax + by + c = 0 in this equation a, b, c are real numbers. a and b cannot be 0 at the

same time.

Ex.: i) 3x + 2y + 2 = 0 ii) 3x +5y - 4 = 0

iii) 5x + y + 7 = 0 iv) x +3y - 6 = 0

These are equations in two variables.

Exercise

Qu.1: Write following statements in equation form using two variables.Write 5

solutions for each equation.

1) Find two numbers whose sum is 12.

2) Find two numbers whose sum is 16.

3) Find two numbers whose difference is 25.

4) Find two numbers whose difference is 14.

Qu.2: Find two numbers whose difference is 2. (Complete the activity.)

Activity: Let greater number be x and smaller be y.

x − y = 2 is an equation.

We get infinite solutions for this equation.

10 – 8 = 2 , 9 – 7 = 2

– = 2, – = 2

Let us write the solution of the above equation as follows.

(10, 8) (9, 7) ( , ) ( , ) ………..

Qu.3: Find at least 5 solutions of following equations.

i) m + n = 12

ii) m – n = 4

Qu.4: Write at least 5 equations in two variables using variables x and y.

59

DAY :25th

SubTopic: Concept of simultaneous equations and to find their solutions.


1) Able to explain concept of simultaneous equations.

2) Will be able to give various examples of simultaneous equations.

3) Will be able to find solutions of simultaneous equations by elimination method.

Let’s revise:

Find solutions of following equations in two variables.

i) x + y = 14

Solutions of above equations are (9, 5) (7, 7) (8, 6) (4, 10) (-1, 15) …..

ii) x - y = 2

Solutions of above equations are (7, 5) (0, -2) (8, 6) (-2, -4) (5.2, 3.2) ….

If we think of above two equations then they have (8, 6) as a common solution.

Now we will see new concept related to this relation of two equations.

Important points:

I) Simultaneous equations:Two linear equations in two variables having same

variables are together called as simultaneous equations.

Now let’s enjoy following activity based on above concept.

Activity 1) Find two numbers whose sum is 34 and difference is 12.

i) x + y = 34 ii) x – y = 12


i) x + y = 45 ii) x – y = 17


i) x + y = 60 ii) x – y = 22

29 23 11

29

29

60

II)Examples of simultaneous equations :

I) x + 2y = 5 ; 3x + 2y = 7 ii) 2x +y = 8 ; 5x + 2y = 12

iii) 3x + y = 5; 2x + 3y = 1 iv) 3x - 4y = 15; x + y = 2

III) To find solutions of simultaneous equations by elimination method:

We have learnt what simultaneous equations are, now we will see how to find

solution of these simultaneous equations by elimination method.

Ex.1) Solve . 2x + y = 5 ; 3x − y = 5

Solution :2x + y = 5 ………….. (I)

3x − y = 5 ………….. (II)

By adding equation (I)and(II)

2x + y = 5

3x − y = 5

5x + 0 = 10

x = 10

5

x = 2

Substitute x = 2 in equation.(I)

2x + y = 5

4 + y = 5

y = 5 −4

y = 1

Here (2,1) is a common solution of two given equations.

Ex.2) Solve: 3x + y – 5 = 0; 3y + 2x = 1

Solution: 3x + y – 5 = 0

We will transfer constant term -5 from left to right side.

3x + y = 5…………… (I)

3y + 2x =1

Write above equation by rearranging the terms.

2x + 3y = 1………….. (II)

Since coefficient of none of the terms are same, to eliminate one of the terms from

both the equations we will equalize the co-efficients.

+

61

Multiply equation (I) by 3.

(3x 3) + (3 y) = (5 3)

9x + 3y = 15 ………….. (III)

Now subtract equation (II) from equation (III).

x = 14

7

x =2

Substitute x = 2 in equation (I).

3x + y = 5

6 + y = 5

y = 5 − 6

y = -1

(2, -1) is the solution of both the equations.

Excercise

Q.1: Write following statements in equation form.

1) Sum of two numbers is 58 and their difference is 4.



Q.2: Solve following simultaneous equations.

I) x + y = 4 ; 2x – 5y - 1 = 0

ii) 4x −3y−17 = 0; 5x – y = 13

iii) 2y – x = 0; 10x +15 y = 105

link

:https://diksha.gov.in/play/collection/do_312528209300701184153324?contentId=do_3130185661956833281629

− − − − −2x + 3y = 1

9x + 3y = 15

7x + 0 = 14




62

DAY :26th

SubTopic:To solve simultaneous equations by substitution method.

Competency Statements:Will be able to solve simultaneous equations by

substitution method.

Important points:

To solve simultaneous equations by substitution method.

There is one more method of solving simultaneous equations and that is by

substituting the value of one of the variables in terms of the other variable.

Let’s understand with help of following example.

Sample example :

Ex. (1) Solve. 2x + 3y = 11 ; 2x – y = -1

Solution: 2x + 3y = 11 ……………….(I)

2x – y = -1 ………………(II)

It will be easier to represent the value of y in terms of x using equation (II).

Hence 2x – y = -1

y = 2x + 1 ……………… (III)

Substitute the value of y in equation (I).

2x + 3y = 11

2x + 3(2x + 1) = 11

2x + 6x +3 = 11

8x +3 = 11

8x = 11 −3

8x = 8

x = 1

Substitute x = 1 in equation (III)

y = (21) + 1

y = 2+1

y = 3

(1, 3) is the solution of given simultaneous equations.

63

Ex. (2) Solve. 5y−3x =14 ; 3y− 2x =1

solution: 5y−3x = 14 …………. (I)

3y− 2x = 1…………..(II)

It will be easier to represent the value of y in terms of x using equation (II).

3y = 2x + 1

y = 2𝑥+1

3…………..(III)

Substitute the value of y in equation (I)

5( 2𝑥+1

3 ) −3x =14

10𝑥+5

3 − 3x =14

10𝑥+5−9𝑥

3 = 14

x + 5 = 42 ........ (multiplying both the sides by 3)

x = 42 − 5

x = 37

Substitute x = 37 in equation(III)

y = 2 ×37 +1

3

y = 74 +1

3

y = 75

3

y = 25

(37, 25) is the solution of given equations.

Excercise

Q 1: Solve following simultaneous equations.

1) 3x + 2y = 8; x + 2y = 4

2) x + y = 11; 2x − 3y = 7

3) 2y + 5x = 13; x − 2y = 5

4) 5m + 8n = 9; m + n = 3

Links:


8566225076224180



64

DAY :27th

SubTopic:To find solutions of ax + by = p and bx + ay = q types of simultaneous

equations.

Competency Statements:Will be able to find solutions of simultaneous equations.

Important points:

To find solutions of ax + by = p and bx + ay = q types of simultaneous equations

We have already seen two methods of solving simultaneous equations.

Now we will see how to solve simultaneous equations in ax + by = p;bx + ay = q

form.

Observe following simultaneous equations.

7x + 13y = 47; 15x− 14y = 117;

13x + 7y = 53 14x− 15y = 115

If you observe both the equations in box 1 and 2 , you will be able to see that co-

efficient of x and y are exchanged. To solve such type of equations we need to add

and subtract both the equations to get two more equation in the form like x + y and

x – y. Then by simply adding those two equations we can get value of x or y.

Ex.1) Solve: 7x + 13y = 47 ; 13x + 7y = 53

Solution: 7x + 13y = 47..................... (I)

13x + 7y = 53 .....................(II)

By adding equations (I) and (II)

− − −

7x + 13y =47 13x + 7y = 53

20x + 20y = 100 x + y = 5 ....(dividing by 20)

............(III)...................(III)

+

Now subtract equation (II) from equation (I).

7x + 13y =47 −

13x + 7y = 53

−6x + 6y = − 6 x − y = 1 .... (dividing by 6 )...........(IV)

65

By adding equations (III) and (IV)

Substitute x =3 equation in (III)

3 + y = 5

y = 5 − 3

y = 2

( 3, 2 ) is a solution of given simultaneous equations .

Exercise

Qu.1) Without finding the value of x and y find the value of (x + y) and (x−y)for the

following equations.

i) 12x + 13y = 51; 13x + 12y = 49

ii) 5x − 3y = 14; 3x − 5y = 2

Qu.2) Solve following simultaneous equations.

1) 33m + 32n = 34; 32m + 33n = 31

2) 15x − 17y − 28 = 0; 15y − 17x + 36 = 0

x−y = 1

x+y = 5 +

2x = 6

x = 3

66

DAY :28th

SubTopic: Word problems based on simultaneous equations part - I


1) To form an equation with the help of given equation.

2) To solve simultaneous equations using suitable method .

3) To verify solutions of simultaneous equations satisfy given conditions or not .

Important points:

Steps to solve word problems using simultaneous equation method:

Read and understand the word problem carefully.

As per the information given in the word problem allot variables to unknown quantity.

Form two equations with help of conditions given in the word problem.

Solve simultaneous equations using appropriate method.

Get solutions.

verify your answer.

Write your answer.

If we follow above steps then we can find answer of any word problem very easily.

We are going to solve word problems based on following topics,

(1) Age

(2) Numbers

(3) Fractions

Ex.1) A two digit number is 3 more than the four times the sum of its digits.

If we add 18 to the number then the places of digits get interchanged. Find the two

digit number.

67

Solution: Let the digit at unit place be ‘x’and tens place be‘y’.

By the given first condition,

10y + x = 4(y + x) + 3

10y + x = 4y + 4x + 3

x – 4x + 10y – 4y = 3

-3x + 6y = 3

−x+ 2y = 1 (dividing by 3 ) ......(I)

By the given second condition,

10x + y = 10y + x + 18

10x− x + y− 10y = 18

9x−9y = 18

x − y = 2 (dividing by 9)

x = y + 2 .......... (II)

Substitute x = y + 2 in equation (I)

−x+ 2y = 1

−(y + 2) + 2y = 1

− y − 2 + 2y = 1

y = 2 + 1

y = 3

Substitute y =3 in equation (II)

x = 3 + 2

x = 5

Original two digit number: 10 y + x = 10 (3) + 5

= 35

Digit at

tens place

Digit at

unit place

Number Sum of

digits

Original number y x 10y + x y + x

Number obtained after

interchanging the digits x y 10x + y x + y

68

Exercise

Qu1: Choose the correct alternative answer.

1) Seema is younger than Reena by 5 years. Sum of their ages is 25.

Find the age of Seema.

A) 50 B) 15 C) 10 D) 5

2) Sum of two numbers is 125 and difference between them is 25 then find

the numbers.

A) 75 and 50 B) 73 and 52 C) 72 and 47 D) 65 and 60

Qu.2 :Solve following examples .

1) Age of Yusuf is 24 years more than half the age of Ajay. 5 years before sum

of their ages was 41. Find their present ages.

2) Sum of numerator and denominator of a fraction is 15. Denominator of a

fraction is 3 more than twice the numerator then find the fraction.

3) A two digit number is 3 more than six times the sum of its digits. If 18 is

added to the number obtained by interchanging the digits then we get original

number. Find the two digit number.

69

DAY :29th

SubTopic: Word problems based on simultaneous equations part - 2

Important points:

Now let’s see next type of word problems.

(1) Examples based on finance/money.

(2) Examples based on geometric figures.

(3) Examples based on speed, distance and time.

Example:

1) Length of a rectangular mobile is 3 cm more than its breadth. Its perimeter is

34cm.Then find length and breadth of the rectangle.

Solution: Step 1: To understand word problem.

Step 2: let the length and breadth of rectangular mobile be x and y respectively.

Step 3: length of rectangular mobile is 3 cm more than its breadth

hence x = y + 3.

Perimeter of rectangle is 34cm.

Perimeter of rectangle = 2(l + b)

Hence 34 = 2 (x + y) is equation (II).

Step4 :Now we will find solution.

x = y + 3................ (I)

34 = 2 ( x + y )

x + y = 17.......... (dividing by 2) .......... (II)

Substitute the value of x in equation x + y = 17.

y + 3 + y = 17

2y + 3 = 17

2y = 14

y = 7

Step 5: Substitute y = 7 in equation (I)

x = y + 3

x = 7 + 3

x = 10

Step 6: Hence length of mobile is 10 cm and breadth is 7 cm.

70

Exercise

Qu1: Choose the correct alternative.

(i) Choose correct mathematical equation for given statement.

‘If 5 is subtracted from length and breadth of a rectangle then its perimeter

becomes 26’.

A) x – y = 8 B) x + y = 8 C) x + y = 23 D) 2x + y = 21

Q 2 : Solve following sub questions.

1) Cost of 5 books and 7 pens is Rs 79 and cost of 7 books and 5 pens is

Rs 77 .Then find the cost of 2 books and 3 pens.

2) Difference between opposite angles of a cyclic quadrilateral is 12.Then find

measures of those angles.

3) In the figure given below length and breadth of rectangle is given by using

variables. With the help of that information find actual length and breadth of

the rectangle.

Link:


2x +

y +

2

4x + 2y

2x + 3y +4

3x − y + 3




71

1

TEST - 2 DAY:30th

Std :10th Subject : Mathematics

Time :1 Hour Marks :15

Qu .1Choose the correct alternative option for the question given below.

[4 Marks]

i) Write two numbers whose sum is 17 .

(A) (-1, -18) (B) (18, 1) (C) (-1, 18) (D) (18, -1)

ii)…………… is a solution of equation 12 = 0 –x + y :

(A) (10, 2) (B) (10, -2) (C) (14, -2) (D) (-2, -10)

iii) In a right angled triangle length of hypotenuse is 15 cm then find the length of

median drawn on hypotenuse.

(A) 7.5 cm (B) 30 cm (C) 15 cm (D) 10 cm

iv) In ∆ABCA = 400, B = 700 thenC = ..........

(A) 1100 (B)700 (C) 600 (D)900

Qu.2.Solve following sub questions.[ 3 Marks]

i) If = 65 x + y and = 35 y – x then find the value of x and y .

ii) Write the following statement in equation form using variables x and y.

‘Sum of two numbers is 15 and difference is 11’ .

iii) Identify the test by which given two triangles are congruent.

72

Qu.3. Solve following subquestions [8 Marks]

i) Divide seg AB of length 7 cm into a ratio 3:2

ii) If ∆ABC ∆PQR then

a) AP, B......, C......,

b) AB

PQ=

BC

……=

……

PR

iii) Solve: 14 = 3y + x; 20 = 2y -3x

iv) Total cost of 6 pens and 4 pencils is Rs 60. Total cost of 4 pencils is Rs 40.

Find the cost of one pen and one pencil.

73

DAY: 31st

Topic – Quadrilateral SubTopic–Quadrilateral and types of quadrilateral

Competency statements –To identify and study properties of types of quadrilaterals

and their applications.

Let’s revise:

1) Draw◻ABCD. Write pairs of adjacent sides, opposite sides, adjacent angles

and opposite angles. Also write the names of diagonals.

2) How many types of quadrilaterals are there? Write their names.

Important points:

Quadrilateral:

A closed figure bounded by four segments is called as quadrilateral.

A quadrilateral has four vertices, four sides, four angles and 2 diagonals.

Types of quadrilateral –

1) Parallelogram – If opposite sides of a

quadrilateral are parallel to each other then it is called

as Parallelogram.

◻ABCD is a parallelogram.

∴ sideAB ।। sideDC, side AD ।। side BC

Properties of parallelogram–i) Opposite sides are congruent.

ii) Opposite angles are congruent.iii) Diagonals bisect each other.

2) Rectangle – If all the four angles of a quadrilateral are right

angles then it is called as rectangle.

◻ABCD is a rectangle.

∴𝒎∠A = 𝒎∠B = 𝒎∠C = 𝒎∠D = 90°

Properties of rectangle -i) Opposite sides are congruent. ii) Diagonals are congruent

and bisect each other.

3) Square -If all the sides of a quadrilateral are congruent and

all the angles are right angles then it is called as square.

◻ABCD is a square.

∴side AB = side BC = side CD = side AD

Also 𝒎 ∠A = 𝒎∠B = 𝒎∠C = 𝒎∠D = 90°

Properties of Square-i) Diagonals are congruent and perpendicular bisectors of each

other.

74

4) Rhombus – If all the four sides of a quadrilateral are

congruent then it is called as rhombus.

◻ABCD is a rhombus .

∴ side AB = side BC = side CD = side AD

Properties of Rhombus –i) Opposite angles are congruent ii) A diagonal bisects

opposite angles iii) Diagonals are perpendicular bisectors of each other.

5)Trapezium –If only one pair of opposite sides of a

quadrilateral is parallel to each other then it is called

as trapezium.

◻ABCD is a trapezium.

sideAB || side CD .

Exercise

Qu1) Choose the correct alternative option for the following questions.

1) fI only one pair of opposite sides of aquadrilateral is parallel

then it is called ...............

A) parallelogram B) rectangle C) rhombus D) trapezium

2) Diagonals of ……………. are congruent and perpendicular bisectors of each

other.

A) rectangle B) rhombus C) square D) parallelogram

Qu 2) Identify who am I?

1) All my angles are right angles.

2) All my sides are congruent.

3) All my sides and angles are congruent.

4) My diagonals are congruent but are not perpendicular to each other.

Qu 3)◻PQRS is a parallelogram then fill in the blanks .

(a) side PQ II side

(b) side QR ≅ side

(c) ∠P ≅∠

(d) ∠P + ∠S =

Qu 4) If length and breadth of rectangle are 10 cm and 6 cm respectively then find

its perimeter.

Link : https://diksha.gov.in/play/collection/do_31259888027666841621857?contentId=do_31301401952796

67201261






75

DAY :32nd

Topic:Quadrilateral SubTopic:Properties of quadrilaterals and triangle

Competency statements :To study and apply few properties of a triangle and

quadrilateral

Let’s remember :

1) Draw any quadrilateral. Measure all its angles and add them.

2) Draw any triangle. Locate midpoint of each of its sides.

Important points:

Theorem of sum of angles of quadrilaterals –

Sum of all the four angles of a quadrilateral is 360o

∴ ◻ABCD , 𝒎∠ A +𝒎∠B +𝒎∠C +𝒎∠D = 360°

Tests of parallelogram –

1) If opposite sides of a quadrilateral are

congruent then it is a parallelogram.

In ◻PQRS if side PS ≅ side QR and

side PQ ≅ side SR then ◻PQRS is a

parallelogram.

2) If opposite angles of a quadrilateral are congruent

then it is a parallelogram.

In ◻PQRS, if ∠ PQR ≅∠ PSR and ∠QPS ≅ ∠ QRS then ◻PQRS is a

parallelogram.

3) If diagonals of a quadrilateral bisect each other then it is a parallelogram.

In ◻PQRS, if seg PT ≅ seg TR and seg QT ≅ seg TS then ◻PQRS is a

parallelogram.

4) If a pair of opposite sides of a quadrilateral is parallel as well as congruent then

it is a parallelogram.

In ◻PQRS, if seg PS ≅ seg QR andseg PS || seg QR then ◻PQRS is a

parallelogram.

Theorem of midpoints of two sides of a triangle:

The segment joining midpoints of any two sides of

a triangle is parallel to third side and half of it.

In △ ABC, point P & point Q are midpoints of side AB

and side AC respectively .

seg PQ || seg BC and PQ = 1

2 BC

76

Converse of midpoint theorem:

If a line drawn through the midpoint of one side of a triangle is parallel to the other

side then it bisects the third side.

∴ In △ ABC, point P is midpoint of segAB and seg PQ || seg BC then AQ = QC

Exercise

Qu1) Choose the correct option for the following questions.

1)If adjacent sides of a rectangle are congruent then it is called as ………….

(A) trapezium (B) square (C) rhombus (D) parallelogram

Qu 2) State whether the given statements are true or false.

i) Every parallelogram is a rhombus.

ii) Every rhombus is a rectangle.

iii) Every rectangle is a parallelogram.

iv) Every square is a rectangle.

Q3) In △ABC, point E and F are the midpoints of

sides AB and AC respectively. If EF = 5.6 then complete the following activity

to find the length of side BC.

Activity :In △ ABC, point E and F are the midpoints of sides AB andAC

respectively .

∴EF =1

2× BC ….........

∴ = 1

2× BC

∴ BC = × 2 =

Q 4) If measures of 3 angles of a quadrilateral are 65o, 95o and 45o then find the

measure of the fourth angle.

Q 5) If breadth of a rectangle is less than 4 cm of its length and its perimeter is 32

then find its length and breadth.

link -


0192177



77

DAY :33rd

Topic:Circle Subtopic :Various parts of circle

Competency statements : To understand various parts of circle.

Let’s remember :Draw a circle and measure its radius with the help of scale.

Important points:

Circle and its parts:

Circle – The set of points in a plane which are equidistant from a fixed point in the

plane is called a circle.

Centre of a Circle: The fixed point is called the centre of the circle.

Radius(r)– The segment joining the centre of the circle and a point on the circle is

called a radius of circle.

The distance of a point on the circle from the center of a circle is also called

the radius of the circle.

Chord -The segment joining any two points on the circle is called as chord of a circle .

Diameter – A chord passing through the centre of a circle is called a diameter of the

circle.

Diameter is the largest chord of a circle.

Diameter is twice/double of radius of the same circle.

In the figure point O is centre of the circle and

OD is its radius.

PQ is a chord and AB is a diameter.

Important Formulae-

1) Diameter = 2 x Radius or Radius = Diameter

2

2) Circumference of a circle (c) = 2𝜋r

Also circumferecnce (c) = 𝜋𝑑

3) Area of a circle = 𝜋𝑟2

is an irrational number and approximate value of is 22

7 or 3.14

78

Circles in a plane:

Congruent circles–Circles with equal radii are called as congruent circles.

Concentric circles –Circles with common center are called as concentric circles.

Excercise

Qu1) Choose the correct option for the following questions.

i) Which of the following formula is used to find area of a circle?

(A) 2πr (B) πr2 (C) πd (D) 1

2πr2

ii)What is the length of longest chord of a circle with radius 2.7 cm?

(A) 2.7cm (B) 5cm (C) 7.2cm (D) 5.4cm

iii) In a circle with centre O and radius 3.5 cm l(OA) = 3.7 cm then the

point A will lie …………

(A) on the circle (B) inside the circle (C) outside of the circle (D) on the centre.

Qu 2) Solve the following sub questions.

1) Radius of a circle is 8cm in length then find the length of its longest chord .

2) Radius of circular garden is 70m. Find its area.

3) A circular wire with radius 14cm is cut and made straight then what will be its

length?

Link:




79

DAY :34th

Topic:Circle SubTopic:Properties of arc of a circle

Competency Statements:To understand properties and applications of arc of a circle

Let’s remember:

A Horse is tied at the corner of square shaped meadow with a string is 10m long then

find the area of sector in which it can graze.

Important points:

Central Angle –

An angle in a circle whose vertex is center of a circle is called as central angle.

Arc of a circle –

A chord of a circle divides a circle into two parts. Each of the parts is called as

arc of a circle.

There are 3 types of arcs

1) minor arc 2) major arc 3) semicircular arc.

Minor and major arc:

If due to a chord circle doesn’t get divided into two equal

parts then the smaller part is called as minor arc and

greater part is called as major arc.

An arc which has centre of a circle is called as major arc

and one without centre is called as minor arc.

In adjoining figure seg AB is a chord of a circle. ∠ACB is a central angle.

Arc AXB is a minor arc and arc AYB is a major arc.

Semicircular arc: Due to a diameter, circle gets divided into two equal parts. Each

partis called as semicircular arc.

Measures of an arc:

1) Measure of a circle = 360°

2) Measure of semi circle = 180°

3) Measure of minor arc = measure of corresponding central angle

4) Measure of major arc = 360°- measure of corresponding minor arc

Congruent arcs:

If two arcs of a same circle (or congruent circles) have equal

measures then those arcs are said to be congruent.

If m(arc AXB) = m(arc CYD)

then arc AXB ≅arc CYD .

80

Properties of arcs of a circle and their corresponding chords:

1) Chords corresponding to congruent arcs also congruent.

In a circle with center O

if arc AXB≅ arc CYD

Then chord AB ≅ chord CD.

2) If two chords of a circle are congruent then their

corresponding minor and major arcs also congruent.

If chord AB ≅ chordCD

then arcAXB ≅ arcCYD.

Exercise

Qu.1) Choose the correct alternative option for the following questions.

i) Measure of an arc formed by diameter of a circle is ...........

(A) 360° (B) 90° (C) 180° (D) 45°

ii) If measure of minor arc of a circle is 1100 then find the measure of its

corresponding major arc is …

(A) 70° (B) 250° (C) 110° (D) 180°

Q 2) solve the following sub questions.

1 ) With the help of given figure,

a) Write the names of minor and major arcs.

b) If m∠AOQ = 70° then find m(arcAYQ) and

m(arcAXQ).

2) In a circle with centre C, points G, D, E and F are on the

circle .

If 𝒎∠ECF =50° and (arcDGF) = 200°, then find

m(arcDE) and m(arcDEF)

Link:




81

DAY: 35th

Topic: Circle Subtopic: Properties of chord of a circle

Competency Statements: Understand the properties of chord of circle and make use

of them.

Let’s remember:

Take any circular object like bangle and measure its circumference with the help of

thread. Find its radius by using circumference.

Important points:

Properties of chords of a circle

1) Theorem – A Perpendicular drawn from the center of a circle

on its chord bisects the chord.

In a fig. O is a center of a circle and seg AB is a chord

If seg OP ⊥ chord AB then seg AP ≅ seg BP.

2) Converse – The segment joining the center of a circle and the

midpoint of its chord is perpendicular to the chord.

If point P is a midpoint of seg AB then seg OP ⊥ chord AB .

Properties of congruent chords:

Theorem: Congruent chords are equidistant from the center of

a circle.

In the adjoining fig. O is a center of a circle.

seg OP ⊥ chord AB and seg OQ ⊥ chord CD.

∴ If chord AB ≅ chord CD

Then OP = OQ .

Converse –The chords equidistant from the center of a circle are congruent.

∴ If OP = OQ then chord AB ≅ chord CD.

Sample example:

1) In a circle with radius 20cm chord CD is at a distance of 12 cm from the center

then find the length of chord CD.

Solution: In a circle with center O segOP ⊥ chord CD.

∴ OD = 20 cm and OP = 12 cm

In right angled △OPD by Pythagoras Theroem

∴ OD2 = OP2 + PD2

∴ 202 = 122 + PD2

∴ 400 = 144 + PD2

82

∴ 144 + PD2 = 400

∴ PD2 = 400 − 144

∴ PD2 = 256

∴ PD = 16 ….. (by taking square root)

but CP = PD ...... perpendicular from center to the chord bisect the chord.

∴ CP = 16

∴ CD = CP + PD = 16 + 16 = 32

∴ length of chord CD is 32 cm.

Excercise

Qu 1) Choose the correct alternative option for the following questions

i) In a circle with radius 5cm, a chord is at a distance of 3cm from the center then

finds the length of a chord.

)A( 5 cm (B) 10 cm (C) 6 cm (D) 8 cm

ii) Perpendicular drawn from the center to the chord bisect the……………...

(A) radius (B) chord (C) diameter (D) circumference

Qu 2) solve the following sub questions.

i) In a circle with center P chord AB is of length 15cm. If seg PQ ⊥ chord AB

then find l(QB)

ii) In a circle with center P and radius 10cm. If one of the chord is of length 12 cm.

Then find the distance of the chord from the center.

iii) In the adjoining fig. In a circle with center C and

chord DE, seg CF ⊥ chord DE. If diameter of a circle

is 20 cm and DE= 16 cm, then CF = ? Give reason for

your answer.

Link:







83

DAY : 36th

Topic : Ratio and Proportion

Subtopic : Operations on equal ratios

Competency Statements: Students are able to perform operation at the equal ratio.

Important points and Review:

Operations on equal ratios :

Invertendo:

𝑎

b =

𝑐

𝑑 using invertendo

b

a =

d

c

Alternando:

𝑎

b =

𝑐

𝑑 using alternando

a

c =

b

d

Componendo:

𝑎

b =

𝑐

𝑑 using componendo

a+b

b =

c+d

d

Dividendo:

𝑎

b =

𝑐

𝑑 using dividendo

a−b

b =

c−d

d

Componendo-Dividendo:

𝑎

b =

𝑐

𝑑 using componendo-dividendo

a+b

a−b =

c+d

c−d

We have already learned the properties of equal ratios in 9th standard. Now try

to solve the examples on it.

84

Exercise

Qu : Solve the following questions.

1) If 𝑎

𝑏=

3

2 then find the value of

a) 𝑏

𝑎 b)

𝑎+𝑏

𝑏

2) If 𝑎

𝑏=

3


a) 𝑎−𝑏

𝑏 b)

𝑎+𝑏

𝑎−𝑏

3) If 𝑎

𝑏=

5


3𝑎+2𝑏

3𝑎−2𝑏

4) If 𝑥

𝑦=

7


3𝑥2+2𝑦2

3𝑥2−2𝑦2

Link :

https://www.youtube.com/watch?v=0VklDDDQ-lg

https://www.youtube.com/watch?v=KYkkAMzVOAA

https://www.youtube.com/watch?v=Xp4tUwZkHBs




85

DAY : 37th


Subtopic : Proportion and Continued Proportion

Competency Statement:

1) Students are able to solve examples based on proportion and continued proportion.

Important Points:

Proportion:

If 𝑎

𝑏=

𝑐

𝑑 then a, b, c, d are said to be in proportion.

If 𝑎

𝑏=

𝑐

𝑑 is equivalent to 𝑎 × 𝑑 = 𝑏 × 𝑐

Simple Example (Solve problem)

1) If a, b, c, d are in proportion and a = 5, b = 10, c = 3 then find the value of d.

Solution :-

a, b, c, d are in proportion

∴𝑎

𝑏 =

𝑐

𝑑

∴ 𝑎 × 𝑑 = 𝑏 × 𝑐

5 × 𝑑 = 10 × 3

𝑑 =10 ×3

5

𝑑 = 6

Continued Proportion :

When two ratios are equal, like 𝑎

𝑏 =

𝑏

𝑐

then we can say that a, b, c are in continued proportion

If ac = b2, then dividing both sides by bc we get 𝑎

𝑏 =

𝑏

𝑐

If ac = b2, then a, b, c are in continued proportion

When a, b, c are in continued proportion then b is known as Geometric mean

of a and c or Mean proportional of a and c.

86

Hence all the following statements convey the same meaning.

1) 𝑎

𝑏 =

𝑏

𝑐 2) 𝑏2 = 𝑎𝑐 3) a, b, c are in Continued Proportion

4) b is the Geometric mean of a and c 5) b is the Mean Proportion of a and c

Ex : If x is the geometric mean of 16 and 4, then find the value of x.

Solution : x is the geometric mean of 16 and 4

x2 = 16 4

x2 = 64

x = 8

We learned the operations on equal ratios in std 9th. Now try to solve the

problem based on them.

Exercise

1) If a, b, c, d are in proportion and a =4, b = 8, c = 5 then find the value of d.

2) If a = 2, b = 3, c = 4, d = 6 then verify a, b, c, d are in proportion or not?

3) If ‘m’ is the geometric mean of 12 and 3 then find the value of ‘m’.

Link :







87

DAY : 38th


Sub topic : Word problems based on ratio & proportion part I


Students can solve word problem based on ratio & proportion.

Revision of Important Points:

We learned the word problems based on direct proportion or inverse

proportion in std 8th.

Solve Example

1) The ratio of two numbers is 5 : 7 and their sum is 96. Find those number.

Solution :-

In the above example we have given the ratio of two numbers.

Let, The common factor of two numbers be x.

Now, We can say that two numbers can be denoted as,

The first number = 5x and the second number = 7x

Now the sum of two numbers is 96

∴ 5𝑥 + 7𝑥 = 96

12𝑥 = 96

𝑥 = 96

12

𝑥 = 8

∴ 8 is the common factor of those two numbers.

∴ The first number = 5𝑥 = 5 × 8 = 40

and the second number = 7𝑥 = 7 × 8 = 56

88

Solve Example

2) The ratio of ages of Abha and her mother is 2:5. At the time of Abha’s birth

her mother’s age was 27 yrs. Find the present ages of Abha and her mother.

Solution :-

The common factor of ages of Abha and her mother be x

∴ The present age of Abha = 2x yrs

and the present age of her mother = 5x yrs

At the time of Abha’s birth

The age of Abha = 0 yrs

So, the difference between their ages is 27 yrs

( Note: this difference is always the same )

∴ 5𝑥 − 2𝑥 = 27

3𝑥 = 27

𝑥 = 27

3

𝑥 = 9

∴ The present age of Abha = 2x = 2 × 9 = 18 yrs

and the present age of her mother = 5𝑥 = 5 × 9 = 45 𝑦𝑟𝑠

Exercise

1) The ratio of two numbers is 9:5 and their difference is 96 then find the number.

2) The ratio of two numbers is 3:4 and their sum is 112 then find the numbers.

3) The ratio of ages of Meenu and her brother Sudhir is 3:4. The difference between

their ages is 5 yrs then find their present ages.

89

DAY : 39th

Subtopic : Word problems based on Ratio & Proportion part II


Students should be able to solve word problems on ratio.

Some of the problem in geometry can be solved by using concept of ratio.

For that we also need to know the properties of geometric figure.

Simple Example

1) The ratio of measures of adjacent angles of a parallelogram is 2:3 then find the

measures of the angles.

Solution :-

If the above example we know that the sum of measure of adjacent angles of a

parallelogram is 1800

Common factor of ratio is x

∴ m∠A = 2x0 and m∠B = 3x0

∴ 2x + 3x = 1800

5x = 180

𝑥 = 180

5

𝑥 = 36

∴ 𝑚∠𝐴 = 2𝑥 = 2 × 36 = 720

∴ 𝑚∠𝐵 = 3𝑥 = 3 × 36 = 1080

90

Exercise

1) The ratio of measures of three angles of a triangle are 1:2:3. Then find the

measures of the angles of a triangle.

2) The ratio of length of adjacent sides of a rectangle is 5:2 and its perimeter is 52cm

then find the length of rectangle.

3) The perimeter of an isosceles triangle is 35 cm. The ratio of the length of congruent

side to the base is 2:3. Find the length of all sides of a triangle.

Link :

https://www.youtube.com/watch?v=bvX8NcOcu30

https://www.youtube.com/watch?v=k-x7FX63u7U

https://www.youtube.com/watch?v=j4gf9ve4yFU

https://www.youtube.com/watch?v=bvX8NcOcu30

https://www.youtube.com/watch?v=k-x7FX63u7U

https://www.youtube.com/watch?v=j4gf9ve4yFU

91

DAY : 40th

Topic : Co-ordinate geometry Subtopic : Point, Co-ordinates and Plotting Points


1) To enable to explain the meaning of coordinates and plotting the points

in the plane.

2) Be able to describe the position of the point with help of co-ordinate.

3) Being able to plot the point by using the given co-ordinate.

4) To be able to tell the co-ordinate of the point in the plane.

Important Points:

Axis : A horizontal line drawn at convenient place in a plane to indicate the

position of a point on the plane is called the X-axis. Another number line

intersecting the X-axis at convenient point marked as ‘O’ for origin and

Perpendicular to the X-axis is called Y-axis.

Origin : The point ‘O’ (origin) is represented zero on both the axes. This point

is called the origin.

On the X-axis, positive numbers are shown on the right of ‘O’ and negative

numbers on the left. In the Y-axis positive numbers are shown above and

negative numbers below it.

Quadrant: The X and Y axes divide the plane into four parts, each of which is

called a quadrant. As shown in the figure, the quadrants are numbered in

the anticlockwise direction. The points on the axes are not included in the

quadrant.

Quadrant II

(-X , +Y)

(-X , -Y) (+X , -Y)

(+X , +Y)

Quadrant I

Quadrant IV Quadrant III

92

The Co-ordinates of Points:

i) x co-ordinate of a point means the

perpendicular distance of the

point from Y-axis.

ii) y co-ordinate of a point means the

perpendicular distance of the

point from x -axis.

iii) The convention for describing the

position of a point is to mention x-co-ordinate

first then y-co-ordinate.

Position of Points & their co-ordinate:

i) If the point lies on I quadrant both the co-ordinates are positive (+, +)

ii) If the point lies on II quadrant then x co-ordinate is negative and y co-ordinate is

positive (- ,+)

iii) If the point lies on III quadrant both the co-ordinates are negative (-, -)

iv) If the point lies on IV quadrant then x co-ordinate is posetive and y co-ordinate is

negative (+, -)

v) If a point is on X- axis it’s y co-ordinate is zero (a, 0)

vi) If a point is on Y- axis it’s x co-ordinate is zero 0, b)

vii)The co-ordinate of origin are (0, 0)

Exercise

Qu 1: Choose the correct alternative answer for the following questions.

i) What is the form of co-ordinates of a point on the Y-axis

A) (a,a) B) (o, b) C) (a, o) D) (b, b)

ii) In which quadrant does the point (-5, 7) lie?

A) Fourth B) Third C) Second D) First

Qu 2: Complete the following activity

i) State in which quadrant or on which axis do the following points lie.

Sr.No. Co-ordinate Quadrant/Axis

1 ( - 4 , - 3 )

2 ( 0 , - 5 )

3 ( 3.5 , 4.2 )

4 ( 7 , 0 )

5 ( - 5 , 7.5 )

6 ( 0 , 3.5 )

7 ( - 2.1 , 0 )

8 ( 2 , - 3 )

93

(ii) Observe the following graph and write the co-ordinate of the points.

Sr.No. Points Co-ordinate

1 P

2 J

3 E

4 A

5 H

6 C

7 L

8 Q

(iii) Plot the following points on the same co-ordinate system.

A ( 4 , - 5 ), B ( 5 , 0 ), C ( 4 , 5 ), D ( 0 , 3.5 ), E ( - 3 , - 5 ),

F ( 0 , - 2) , G ( 3.5 , 2.5 ), H ( - 7 , 2 ), I ( - 7 , 0 ), O ( 0 , 0 )

Que 3 : On the graph paper plot the points (3, 4), (3, -4), (-3, -4), (-3, 4). Join the

points pairwise and observed the figure formed. If it is quadrilatreral

name which type of quadrilateral .

94

Link :

Cartesian system

Part 1.

https://diksha.gov.in/play/collection/do_312528194785001472250597?referrer=ut

m_source%3Dmobile%26utm_campaign%3Dshare_content&contentId=do_31300031

42421544961125

Part 2.

https://diksha.gov.in/play/collection/do_312528194785001472250597?referrer=utm_sou

rce%3Dmobile%26utm_campaign%3Dshare_content&contentId=do_31300031444973

9776193

Part 3.



67681118

Ploting Points

Part 1.



43681119

Part 2.



60001138

Part 3.



72481114

Part 4.



6672171






















95

DAY : 41st

Topic : Co-ordinate geometry Subtopic : Linear equation

Competency Statement: To draw the graph of Linear equation in one variable.

Important Points:

i) The equation of line parallel to the X-axis is in the form of y = a (a 𝜖 R)

ii) The equation of line parallel to the Y-axis is in the form of x = b (b 𝜖 R)

iii) The equation of X-axis is y = 0

iv) The equation of Y-axis is x = 0

Exercise

Qu 1 : Choose the correct alternative answer for the following questions.

(i) Write the equation of the line parallel to the Y-axis at a distance of 3 units from

it to its left.

(A) y = 3 (B) y = - 3 (C) x = 3 (D) x = - 3

(ii) Write the equation of the line passing through the points ( 4, 3 ), ( - 5, 3 ), ( 7, 3 )

(A) y = - 4 (B) x = 4 (C) y = 3 (D) x = - 5

Qu 2 : Complete the following activity.

(i) State the graphs of the following equations will be parallel to which axis.

Sr.No. Equation Parallel to which axis

1 y – 2 = 0 X – axis

2 y = - 1.5

3 x = - 5

4 2x – 6 = 0

5 y = 5

6 x + 3 = 0

7 2x + 5 = 0

8 y + 7 = 2

96

(ii) Draw the graph of following equations.

x = 3 and y = - 5

Solution :

Note : i) Draw the X-axis and Y-axis on the graph paper

ii) Since it is given that x = 3, draw a line on the right of the Y-axis at a

distance of 3 units from it and parallel to it. Name the line x = 3

iii) Since it is given that y = -5, draw a line below the X-axis at a distance

of 5 units from it and parallel to it. Name the line y = -5

Que 3 : Solve the following sub questions.

1) How many lines are there which are parallel to Y-axis and having distance

6 units from it? Write the equations of those lines.

2) Draw the graph of the following equation.

i) x = 5 ii) y = 4 iii) x= -3 iv) y = -2

Link:



49121682




97

DAY : 42nd

Topic : Co-ordinate geometry

Subtopic : The graph of Linear equation in general form.

Competency Statement: To draw the graph of Linear equation in general form.

Let’s Recall

i) Graphs of linear equation in one variable are parallel to axes.

ii) Write the equation of line parallel to the X-axis at a distance of 7 units from it and

below the X-axis.

Important Points:

1) Graph of linear equation in the form of ax + by + c = 0 (a ≠ 0) (b ≠ 0) intersect the

both the axes.

2) The coordinates of any point on the line satisfy the equation of the line.

3) If the coordinate of a point satisfy the equation then the point lies on the line.

To draw the graph of a linear equation in general form, three or four solutions of

that equation are to be obtained. Each solution gives co-ordinate of a point on the

line. When the point are plotted then the line can be draw.

Exercise

Qu 1 : Choose the correct alternative answer for the following questions.

i) To draw the graph of x - y = 7, find the value of y if x = 3

A) 10 B) -10 c) 4 d) -4

ii) To draw the graph of 2x + y = 12, if the value of y = 4 then find x

A) 8 B) 4 c) 6 d) 2

Qu 2 : Complete the following activity to draw the graph of x - y = 1.

98

Solution :

Plotting the above ordered pairs on graph paper and draw a line passing through

the three points.

Qu 3 : Solve the following sub questions.

1) Plot the points (1, 2) (3, 6) (-2, -4) on the graph paper. Decide whether the points

are collinear or not, if so, draw a line passing through those point and write the

equation of that line.

2) Draw the graph of following equations.

i) 2x + y = 5 ii) y = x + 3 iii) x = y

Link :

1) Parallel line to axes

https://diksha.gov.in/play/collection/do_312528194785001472250597?referrer=utm_source%3Dmobile%26


2) Graph of linear equation



x 2 –2

y 1 2

(x,y) (2, 1)

(i) x – y = 1

Put y = 1

∴ x – 1 = 1

∴ x = 1 + 1

∴ x = 2

ii) x – y = 1

Put y = 2

∴ x – = 1

∴ x = 1 +

∴ x = 3

(iii) x – y = 1

Put y = –3

∴ x – (-3) = 1

∴ x + 3 = 1

∴ x = 1 −

∴ x =





99

DAY : 43rd

Topic: Trigonometry

Subtopic: Introduction to Trigonometry and Trigonometric Ratio


i) Tell the different Trigonometric ratios using similar triangles and Pythagoras

theorem.

ii) Use of trigonometric ratios for solving problems.

Let’s Recall

i) Right angled triangle has a hypotenuse, Opposite to right angle and there are two

acute angles for a right angled triangle.

Important Points:

Terms related to triangle.

In right angled triangle ∆ABC,

if m∠B = 900 then ∠A and ∠C are acute angles.

100

Trigonometric ratios :-

sin ϴ =AB

AC=

Opposite side of ∠θ

hypotenuse

cos ϴ =BC

AC=

Adjacent side of ∠θ

hypotenuse

tan ϴ =AB

BC=

Oppositeside of ∠θ

Adjacent side of ∠θ

tan ϴ =AB

BC=

AB

ACBC

AC

=sinθ

cosθ

sin(90 − θ) =BC

AC= cosθ

cos(90 − θ) =AB

AC= sinθ

tan(90 − θ) =BC

AB= cotθ

Important equation in trigonometry

sin2𝛳 + cos2𝛳 = 1

Solved the Example :

Ex. ∆ABC is a right angled triangle

∠C= 900, AC = 6, BC = 8, then

i) Find AB

ii) write the three trigonometric

ratios of ∠A and ∠B

Solution :- In ∆ABC is a right angled triangle

By using Pythagoras theorem

AB2 = AC2 + BC2

= 82 + 62

= 64 + 36

= 100

AB = 10

101

sin A = BC

AB =

8

10 sin B =

AC

AB =

6

10

cos A = AC

AB =

6

10 cos B =

BC

AB =

8

10

tan A = BC

AC =

8

6 tan B =

AC

BC =

6

8

Exercise X

1) In right angled ∆XYZ

XY = 5, ∠Y = 900, YZ= 12

then find sin x , cos X, tan X

and also find sin z, cos Z, tan Z Y Z

2) Fill in the boxes with correct answers

(i) sin 20° = cos °

(ii) tan 30° tan ° = 1

(iii) cos 40° = sin °

Link:













102

DAY : 44th

Topic : Trigonometry

Subtopic : Trigonometric Ratios of specific measured angles


Determining the trigonometric ratios of an angle of a particular measure.

Let’s Recall

Observe the following figures and see what you remember

Important Points:

Trigonometric Ratios of specific measured angles

Measured

angle 00 300 450 600 900

Ratio

sin 0 1

2

1

√2

√3

2 1

cos 1 √3

2

1

√2

1

2 0

tan 0 1

√3

1 √3 Not define

103

Solved the Example:

Ex. 1 : Find the value of 2 tan 450 + cos 30 - sin 60

Solution : 2 tan 450 + cos 300 − sin 600

= 2 × 1 +√3

2−

√3

2

= 2 + 0

= 2

Ex. 2 : Find the value of cos 470

sin 430

Solution : 470 + 430 = 900 therefore 470 and 430 are pair of complementry angles

sin θ = cos (90 - θ)

sin 43° = cos (90 - 43°) = cos 47°

cos 47°

sin 43° = cos 47°

cos 47° = 1

Exercise

Qu 1 : Choose the correct alternative answer for the following multiple choice

questions.

i) What is the value of sin 00

A) 1 B) 0 C) 1

2 D)

1

√2

ii) 4 tan 450 +2 cos 450 - 2 sin 450 = ?

A) 0 B) 1 C) 4 D) -4

Qu 2 : Find the value of

(i) 2 cos 60° + 5 tan 45° (ii) 4

5 tan2 30° + 3 sin2 30°

(iii) cos2 45° + sin2 30° (iv) 2 sin 30° + cos 0° + 3 sin 90°

(v) cos 60° cos30°+ sin 60° sin 30°

Link:

https://diksha.gov.in/play/collection/do_312528194785001472250597?referrer=utm_source%3Dmo

bile%26utm_campaign%3Dshare_content&contentId=do_3130185632768655361172

https://diksha.gov.in/play/collection/do_312528194785001472250597?referrer=utm_source%3Dmo

bile%26utm_campaign%3Dshare_content&contentId=do_3130185632642908161171





104

Final Test DAY – 45th

Std : 10 th Sub : Mathematics – I & II

Time :2 hrs Marks :30

Qu 1: Choose the correct alternative answer for the following questions.(6 marks)

i) Which of the following is the set of natural numbers?

A) {...-2, -1, 0, 1, 2...} B) {1,2,3....} C) {0,1,2,3....} D) {-1,-2,-3,....}

ii) If P = 3 then find the value of polynomial P2 - 9

A) 18 B) -18 C) 0 D) 1

iii) x + y = 4 is the linear equation in two variables and when x = 2 then the value

of y is .............

A) -2 B) 4 C) -4 D) 2

iv) If all pairs of adjacent sides of a quadrilateral are congruent then it is called .........

A) Rectangle B) Parallelogram

C) Trapezium D) Rhombus

v) How many lines are determined by three distinct points?

A) two B) three C) one or three D) six

vi) If two sides of a triangles are 5cm and 1.5cm, the length of its third side can

not be ---------

A) 3.7 cm B) 4.1cm C) 3.8 cm D) 3.4cm

Qu 2: Solve the following sub questions. (4 marks)

i) Find the Geometric mean of 2 and 8.

ii) Factorise x2 - 25

iii) The measure of one of the angles of a parallelogram is 400 then find the measure

of its adjacent angle?

iv) In ∆PQR if ∠R >∠Q then which of the following statements is true

A) QR > PR B) PQ > PR C) PQ < PR D) QR < PR

105

Qu 3: Solve the following sub questions. (14 marks)

i) If 𝑎

𝑏=

5


𝑎+𝑏

𝑎−𝑏

ii) Write two solutions of the equation x + y = 7

iii) Find the factors of polynomialx2 -3x - 28

iv) Find the value of 5 sin300 + 3 tan 450

v) State in which quadrant or on which axes the following points lie.

i) A(12, 0) ii) B(-5, -2) iii) C(2, 10) iv) D(15, -18)

vi) Radius of a circle is 10cm. The length of a chord of a circle is 16cm. Find the

distance of the chord from centre.

vii) In the adjoining figure

line l || line m and line n || line P.

Find ∠a, ∠b and ∠c from

the given measure of an angle.

Qu 4 : Solve the following sub questions. (6 marks)

i) Solve the following simultaneous linear equation.

2x +3y = 18 ;x + y = 7

ii) In right angle ∆YXZ, ∠X = 900

XZ = 8cm, YZ = 17cm then

find the value sin Y,cos Y, tan Y

sin Z, cos Z, Tan Z.

106

Answers

DAY 1st (Set – Concept, methods of writing set and its elements)

Qu.1 (i) A (ii) C (iii) B Qu.2 1) A = {2, 3, 5, 7, 11, 13, 17, 19} 2) B = {c, r, I, k, e, t} 3) C = {a, e, i, o, u} 4) D= {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48} 5) E= {1, 4, 9, 16, 25, 36, 49}

DAY 2nd (Subset) Qu.1 (i) B (ii) C Qu.2 1) {2}, {3}, {5}, {7}, {11}, {2,3}, {2,5}, {2,7}, {2,11}, {3,5}, {3,7}, {3,11}, {5,7}, {7,11}, {2,3,5}, { } In this way many subsets can be written. 2) D C 3) Y X, Z X, T X, Y T

DAY 3rd (Basic Concepts in geometry) 1) Infinite 2) One and only one 3) One and only one 4) Yes 5) Yes 6) One Point 7) One line

DAY 4th ( Euclidean geometry) 1) Option 1 - when A – B – C then 2, Option 2 - when A – C – B then 18, 2) (-8, 2) 3) 9

DAY 5th (Line and angle)

1) Yes 2) No 3) No 4) If the sum of the measures of two angles is 90°. 5) No.

DAY 6th (Euclidean geometry)

1) If quadrilateral is a cyclic then Opposite angles of its are supplementry. 2) If the diagonals of a quadrilateral are congruent, then that quadrilateral is a rectangle. 3) Given – In ∆ ABC, sideAB sideBC sideAC, To Prove - A B C

107

DAY 7th (Properties of rational numbers) Qu.1 (i) B (ii) C Qu.2 i) 30, 80, 40, 50

7 2) - 28 , - 4

DAY 8th (Operations on similar surds: Addition and Subtraction)

Qu.1 (i) D (ii) C Qu.2 1) 23√7 2) 7√5 3) −25√3

DAY 9th (Operations on similar surds : Multiplication, division) Qu.1 (i) C (ii) A Qu.2 1) 36√5 2) 8 3) 11 −4√10 4) (i) √11

11 ii) 3√7

35

DAY 10th ( Properties of Parallel lines)

Qu.1 : 1. C , 2. B , 3. D

Qu.2 : 1. ∠CSQ , ∠CSQ , 120° , 180°

2. Pairs of alternate interior angles, 70° , 180° , 110°

3. ∠b , ∠c , ∠b , ∠b

4. Side SR , ∠s , 65° , 115°

Qu. 3 : (i) ∠a = 50° , ∠q = 130° , ∠r = 50° , ∠s = 130° (ii) x = 36°

DAY 11th ( Tests for parallel lines)

प्र. 1 : 1. A , 2. B

प्र. 2 : 1. line l ∥ line m , ∠a, ∠b, line m

2. 130° , Properties of opposite angles, 130° , Interior angles test

प्र. 3 : (i) Line p is not parallel to line q.

DAY 12th ( Degree of the polynomial, Standard form, Coefficient form of a

polynomial)

Qu. 1 : 1) C 2) A 3) D

108

`Qu. 2 : Table 1 : 1) 5 2) 8 3) 9 Table 2 : 1) 5d4+4d3-3d2+1 2) -3y5-y4+7y3+6y2+10 3) 8p3+9p2+5p-9 Table 3 : 1) (7,-8,3) 2) (9,7,-5,9) 3) (10,-4) Table 4 : 1) (x2 – 2x + 3) 2) (x3 +0x2 + 0x – 3 ) /(x3-3) 3) (-4x4+ 0x3+1x2+2x+ 9) / (-4x4+1x2+2x+ 9)

DAY 13th ( Operation on Polynomials )

Qu.1 : 1) Yes 2) No 3) Yes 4) No

Qu.2 : 1) a) 7b2 – 4b b) 13w2 + 15w3 + 2w – 18

2) a) –a4-20a2+1 b) 3a2+5a-11

3) a) 9p2 + 27 b) -45a3 + 15a2 c) 15q2 – 1q – 6 d) -6b3-7b2+24b

DAY 14th ( Polynomials.)

Qu. 1: 1) 7 2) 2 3) -13

Qu. 2: 1) 3 ,4, 18, 84 2) -2, 8, 4, 12

Qu. 3: 1) 3 2) 23 3) 1

DAY 15th ( Factors of the Polynomial.)

Qu. 1: 1) a 2) b 3) c

Qu. 2: 1) 15x, 3x, (x-3), (5x+3) 2) 3n, 2n, (n-3), (n-2) 3) p2, (p+5), 2, (p+2)

Qu. 3: 1) 6x (x – 9) 2) 7 (p-3) (p+3) 3) (a+14) (a-3) 4)2(h+3)2 5) 3(k-5) (k-1)

DAY 17th ( Types of Triangle )

Qu. 1) (1) D (2) C

Qu. 2) ∠ACB, 50°, ∠DBC, 60°

Qu. 3) Obtuse Triangle

Qu. 4) 5.5 cm

DAY 18th ( Congruence of triangles)

Qu. 1) (1) B (2) B (3) A Qu. 2) (a) LM (b)UV (c) TV, LN (d) T (e) M (f) V, N

Qu. 3) 1) S-A-S Test 2) hypotenuses side test. 3) S-A-A Test Qu. 4) ∠CAD, Side AD, S—A-S Test, ∠C

109

DAY 19th (Property of triangle 30°-60°-90° and 45°-45°-90°)

Qu. 1) (1) B (2) D

Qu. 2) 45°, 1

√2 ,

1

√2 , 2

Qu. 3) 8 cm, 4√3 cm

DAY 20th ( Similar triangles)

Qu. 1) (1) D (2) B

Qu. 2) (a) ∠X, ∠Y, ∠Z (b) ST, YZ, RS

Qu. 3) (a) ∠U, 70° (b) ∠V, 50° (c) 180°, 60°

DAY 21st (Some properties of triangle)

Qu. 1) (1) A (2) D

Qu. 2) (a) ∠ACD (b) ∠A, ∠B (c) Theorem of remote interior angle (d) 125° Qu. 3) 5cm, 7.5cm

DAY 22nd and 23rd ( Basic Construction)

Students should construction themselves using geometric compass box.

DAY 24th (Equation in two variables- concept)

Qu.1) : 1) like this (15, 6), (17, 4), (25, -4), (-6, 27), 2) like this (2, 14), (8, 8), (20, -4), (-2, 18) 3) like this (27, 2), (30, 5), (26, 1), (35, 10) 4) like this (15, 1), (18, 4), (16, 2), (20, 6) Qu.2) : x , y ,12, 10, 6, 4, 12, 10, 6, 4 Qu.3) : i) like this (-2, 14), (8, 4), (10, 2), (3, 9) ii) like this (6, 2), (4, 0), (12, 8), (20, 16) Qu.4) : like this x + y = 14 , x – y = 6 , x + y = 5 , x + y = 7 , x – y = 13

DAY 25th (Concept of simultaneous equations

and to find their solutions.) Qu.1) : 1) x + y = 58 , x – y = 4 2) x + y = 74 , x – y = 10 3) x + y = 56 , x – y = 18 प्र.2) : i) (3, 1) ii) (2, -3) iii) (6, 3)

110

DAY 26th(To solve simultaneous equations by substitution method.) Qu.1) : 1) ( 2, 1) 2) (8, 3) 3) (3, -1) 4) (5, -2) DAY 27th(solutions of ax + by = p and bx + ay = q types of simultaneous equations.)

Qu.1) : i) x + y = 4 ; x – y = -2 ii) x + y = 6 ; x – y = 2 Qu.2) : 1) m = 2 ; n = -1 2) x = 3 ; y = 1

DAY 28th (Word problems based on simultaneous equations part - I) Qu.1) : i) C ii) A

Qu.2) : 1) Age of Yusuf = 33 years and age of Ajay = 18 years 2) That Fraction = 4

11 3) That Number = 75

DAY 29th (Word problems based on simultaneous equations part - II)

Qu.1) : i) C Qu.2) : 1) 2 books = 12 Rs. and 3 Pens = 21 Rs. 2) 840 and 960

3) length 16 and breadth 10

DAY 31st (Quadrilateral and types of quadrilateral. ) Qu.1) : i) D ii) C Qu.2) : 1) Square , rectangle 2) Square, rhombus 3) Square 4) rectangle Qu.3) : a) RS b) PS c) R d) 1800

Qu.4) : length 8 cm and perimeter 28 cm

DAY 32nd (Properties of quadrilateralsand triangles.) Qu.1) : i) B Qu.2) : i) False ii) False iii) True iv) True Qu.3) : Theorem of midpoints of two sides of a triangle, EF, 5.6 , 11.2

Qu.4) : 1600

Qu.5) : length 10cm , breadth 6cm

111

DAY 33th (Various parts of circle ) Qu.1) : i) B ii) D iii) C Qu.2) : 1) 16 2) 3850 Sq.m. 3) 88 cm

DAY 34th (Properties of arc of a circle) Qu.1) : i) C ii) B Qu.2) : 1) a) central angle AOQ, minor arc : arc AYQ, major arc : arc AXQ b) 700, 2900

2) 1100 , 1600

DAY 35th (Properties of chord of a circle)

Qu.1) : i) D ii) B Qu.2) : 1) 7.5 cm 2) 8 cm 3) 6 cm

DAY 36th (Operations on equal ratios)

1) a) 2

3 b)

5

2

2) a) 1

2 b)

5

1

3) 19

11

4) 165

129

DAY 37th (Proportion and Continued Proportion)

1) 10 2) Yes 3) 6

DAY 38th (Word problems based on Ratio & Proportion part I)

1) 216, 120 2) 48, 64 3) 15 years and 20 years

112

DAY 39th (Word problems based on Ratio & Proportion part II)

1) 30O, 60O, 90O 2) 20 cm, 8 cm 3) 10 cm, 10 cm, 15 cm

DAY 40th (Point, Co-ordinates and Plotting Points)

Qu. 1 : (i) B , (ii) C Qu. 2 : (i)

(ii)

Qu. 3 : Rectangle

DAY 41st (Linear equation)

Qu. 1 : (i) D , (ii) C Qu. 2 : (i) Qu. 3 : (1) 2 , x = 6 व x = - 6

DAY 42nd (The graph of Linear equation in general form)

Qu. 1 : (i) D , (ii) B Qu. 2 : 2 , 2 , 3 , - 2 , 3 , ( 3 , 2 ) , - 3 , ( - 2 , - 3 )

Qu. 3 : the points are collinear, equation : y = 2x or 2x – y = 0

1.quadrant–III 2. Y – axis 3.quadrant– I 4. X – axis 5. quadrant–II 6. Y – axis 7. X – axis 8.quadrant– IV

P ( 6, - 3 ) J ( – 4, - 3 ) E ( – 4, 4 ) A ( 3, 3 ) H ( – 8, 0 ) C ( 0, 7 ) L ( 0, - 6 ) Q ( 4, 0 )

(1) X–axis (2) X – axis (3) Y– axis (4) Y– axis (5) X– axis (6) Y – axis (7) Y– axis (8) X– axis

113

DAYS 43th (Introduction to Trigonometry and Trigonometric Ratio)

1. sin x = 12

13 , cos x =

5

13 , tan x =

12

5 , sin z =

5

13 , cos z =

12

13 , tan z =

5

12 .

2. (i) 70° (ii) 60° (iii) 50°

DAY 44th (Trigonometric Ratios of specific measured angles)

1. (i) B (ii) C

2. (i) 6 (ii) 61

60 (iii)

3

4 (iv) 5 (v)

√3

2

Documents

GOVERNMENT OF MAHARASHTRA...2 Bridge Course for Std – 10th Maths (Part 1& 2) • Promotor : Department of School Education, Government of Maharashtra. • Publisher : State Council