d24cdstip7q8pz.cloudfront.net...(iii) a parallelogram whose all sides are equal d. rhombus (iv) a...

2

X – Std. Quadratic Equations – Classnotes I

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Instructions:This booklet given to you is your Class Notes. Keep filling this sheet as the class

proceeds. At the end of this session, you will have your notes ready.

1. Introduction

Question1. Perform this activity : Take four points in a plane such that no three of them are collinear

and try to join them. Can you name the figure obtained?

Let A, B, C and D be four points in a plane such that:

(1) No three of them are collinear

(2) The line segment AB, BC, CD and DA do not intersect except at their end points

Then, the figure made up of the four line segment is called the ‘quadrilateral’ with vertices A,

B, C and D.

Figure 1:Quadrilateral ABCD

Did you know? The word ‘quad’ means four and the word ‘lateral’means sides

Important terms related to quadrilateral:

Adjacent sides: Two sides of quadrilateral are consecutive or adjacent sides, if they have a

common point.

Opposite sides: Two sides of quadrilateral are said to be opposite if they don’t have any

common end- point(vertex).

Consecutive angles: The consecutive angles of a quadrilateral are two angles which include

a side in their intersection.

Opposite angles: Two angles of a quadrilateral are said to be opposite angles if they don’t

have any common side.

Question2. Recognize the following in the given quadrilateral ABCD

(i) Adjacent sides

(ii) Opposite sides

(iii) Consecutive angles

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(iv) Opposite angles


a. Angle sum property of a quadrilateral

Question3. Check the following table and try to find the sum of angles of different polygons where

‘𝑛’ is the total number of sides in that polygon.

Figure

Side(𝒏)

Number of triangles a

given polygon can be

divided into

Angle sum

Triangle

3

1

1 × 180° = 180°

Quadrilateral

___

2

2 × 180° = (𝑛 − 2) × 180°

Pentagon

___

3

3 × 180° = (𝑛 − 2) × 180°

Hexagon

___

____

_____________________

Question4. Prove that the sum of the four angles of a quadrilateral is 360°.

Question5. The angles of a quadrilateral are respectively 120°, 82° and 42°. Find the fourth angle.

Question5. In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 2: 3: 4: 6. Find the

measure of each angle of the quadrilateral.

Question6. In a quadrilateral ABCD, AO and BO are the bisector of ∠A and ∠B respectively. Prove

that ∠AOB = 1

2 (∠C + ∠D).

(Refer to the subtopic Introduction _ Introduction)

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2. Types of Quadrilateral – Visualised

Question7. Match the following quadrilaterals with their respective properties:

a. Parallelogram

(i) One pair of opposite sides of

quadrilateral is parallel

b. Rectangle

(ii) Both pair of opposite sides of

quadrilateral are parallel

c. Trapezium

(iii) A parallelogram whose all sides are

equal

d. Rhombus

(iv) A parallelogram whose all sides are

equal and one angle is 90°

e. Square

(v) A quadrilateral whose two pairs of

adjacent sides are equal and it is not

a parallelogram

f. Kite

(vi) One angle is 90° and opposite sides

are parallel and equal

Note: A given quadrilateral in 1st column can be matched with more than one property in the

2nd

column.

(Refer to the subtopic Types of Quadrilateral – Visualised _ Types of Quadrilaterals)

3. Properties of a Parallelogram

Question8. A parallelogram ABCD is given and it is divided into two triangles by its diagonal. Prove

that the triangles obtained are congruent triangles (Theorem 1)

(Refer to the subtopic Properties of Parallelogram _ All about quadrilaterals)

Question9. Prove that: In a parallelogram,opposite sides are equal.(Theorem 2)

Question10. Prove that: the opposite angles of a parallelogram are equal (Theorem 3)

Question11. Prove that: The diagonals of a parallelogram bisect each other.(Thoerem 4)

(Refer to the subtopic Properties of Parallelogram _ Theorem 1)

Question12. If a diagonal of a parallelogram bisects one of the angles of the parallelogram, it also

bisects the second angle. Also, prove that it is a rhombus.

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Question13. Prove that the angle bisectors of a parallelogram form a rectangle.

Question14. In a parallelogram ABCD, ∠ C = 110°, determine the measure of ∠A and ∠B.

Question15. Find the four angles ∠A, ∠B , ∠C and ∠D in the parallelogram ABCD.


Question16. AN and CP are perpendicular to the diagonals BD of a parallelogram ABCD. Prove that:

(i) ∆AND ≅ ∆CBP

(ii) AN = CP

Condition for a quadrilateral to be a parallelogram

(i) A quadrilateral is a parallelogram if its opposite sides are equal.

(ii) A quadrilateral is a pralleogram if its opposite angles are equal.

(iii) If the diagonals of a quadrilateral bisect each other , then the quadrilateral

is a parallelogram.

(iv) A quadrilateral is a parallelogram, if its one pair of opposite sides are

equal and parallel.

a. Properties of special parallelograms

We know that, rectangle, rhombus and square are parallelograms. Since they are

parallelograms they satisfy all the properties of a parallelogram. They also satisfy some

special properties of their own.

Let us discuss these special properties of special parallelograms.

Question17. Prove that : If the two diagonals of a parallelogram are equal, it is a rectangle.

Note: The diagonals of a rectangle are equal in length.

Question18. Prove that: If the diagonals of a parallelogram are perpendicular, then it is a rhombus.

Note : The diagonals of a rhombus are perpendicular to each other.

Question19. Show that if the diagonals of a parallelogram are equal and bisect each other at right

angle, then it is a square.

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Hint: Sum of consecutive interior angles on the same side of transversal is 180°

(Refer to the subtopic Properties of Parallelogram _ Theorem 3)

4. The Mid- Point Theorem

We proved many theorems on a parallelogram. Taking the help of the various theorems on

parallelogram let us discuss some interesting and useful facts about a triangle. These facts are

stated and proved as theorems of a triangle.

Note: A theorem is a statement that can be demonstrated to be true by accepted mathematical

operations and arguments.

Activity on Mid-Point theorem.

1. Draw a large scalene triangle on a sheet of paper.

2. Name the vertices A, B and C. Find the mid-points (D and E) of two sides and connect them.

3. Cut out △ABC and cut along line DE.

4. Place △ADE on quadrilateral BDEC with vertex E on vertex C. Write down your

observations.

5. Shift △ADE to place vertex D on vertex B. Write down your observations.

6. What do you notice about the lengths DE and BC?

Theorem : Prove that the line segment joining the mid-point of any two sides of a triangle is

parallel to the third side and equal to the half of it.

Given: ∆ ABC, D and E are the mid-points of sides AB and AC respectively. DE is joined

To Prove : DE ǁ BC and DE = 1

2 BC

Construction : Produce the line segment DE to F, such that DE = EF. Join FC.

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Proof : In ∆s AED and CEF , we have

AE = ________ [Since E is the mid-point]

∠ ________ = ∠CEF [Vertically opposite angles]

DE = EF [By construction]

∴ ∆ AED ≅ ∆ CEF [By ____________ congruency]

AD = CF [ _____________________ ]------------- (i)

∠ADE = ∠ _________ ------------- (ii)

D is the mid-point of AB

⇒ AD = DB

⇒ DB = CF [From (i)] ------------- (iii)

Now, DF intersects AD and FC at D and F respectively such that

∠ADE = ∠CFE [From (ii)]

i.e, alternate interior angles are equal

∴ AD ǁ FC

⇒ DB ǁ CF ------------- (iv)

From (iii) and (iv), we find that DBCF is a quadrilateral such that one pair of sides are equal

and parallel.

∴ DBCF is a parallelogram.

⇒ DF ǁ _____ and DF = _______ [Opposite sides of a ǁgm

are equal and parallel]

But DE = EF

∴ DE ǁ BC and DE = 1

2 BC

(Refer to the subtopic Mid- point Theorem _ Mid-point Theorem)

Question20. Prove that the line drawn through the mid – point of one side of a triangle, parallel to

another side, intersects the third side at its mid – point. (Converse of Mid- point Theorem)

(Refer to the subtopic Mid- point Theorem _ Converse of Mid-point Theorem)

Question21. D, E and F are respectively the mid – points of sides BC, CA and AB of equilateral

triangle ABC.Prove that DEF is also an equilateral triangle.

Question22. Show that the line segments joining the mid – points of opposite sides of a quadrilateral

bisect each other.

Use of Quadrilateral in our daily life :

Nearly all papers and magazines are quadrilaterals, as are the footprints of most boxes, the shapes

of many rooms, and so on. Quadrilaterals tend to pack nicely, so are used to tile large areas and

small. They tend to be a preferred shape for many things for that reason. You can put them side-

by-side without wasting space.

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X – Std. Quadratic Equations – Homework

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Instructions: This booklet given to you is your Homework Sheet. Solve these problems at home. You

shall be exam-ready if you can finish all the problems.

Questions from 1 to 5 are MCQ’s

1. The angles of a quadrilateral are respectively 120°, 98° and 92°. Find the fourth angle.

A. 70°

B. 50o

C. 77°

D. 180°

2. Diagonals necessarily bisect opposite angles in a

A. Rectangle

B. Parallelogram

C. Isosceles trapezium

D. Square

3. The bisectors of any two adjacent angles of a parallelogram intersect at

A. 30°

B. 45°

C. 60°

D. 90°

4. In a rhombus ABCD, if ∠ACB = 40°, then ∠ADB =

A. 70°

B. 45°

C. 50°

D. 60°

5. Diagonals of a quadrilateral ABCD bisect each other. If ∠A = 45°, then ∠B =

A. 115°

B. 120°

C. 125°

D. 135°

Questions from 6 to 20 are subjective questions

6. The diagonals of a parallelogram ABCD intersect at O. A line through O intersects AB at X

and DC at Y. Prove that OX = OY.

7. If diagonal of a parallelogram bisects one of the angles of the parallelogram, then prove that it

also bisects the second angle. Also, prove that it is a rhombus.

8. ABC is a triangle. D is a point on AB such that AD = 1

4 AB and E is a point on AC such that

AE = 1

4 AC. Prove that DE =

1

4 BC.

9. BM and CN are perpendiculars to a line passing through the vertex A of a triangle ABC. If L

is the mid- point of BC, prove that LM = LN.

10. Prove that a diagonal of a parallelogram divides it into two congruent triangles.

11. Show that the line segments joining the mid- points of the opposite sides of a quadrilateral

bisect each other.

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X – Std. Quadratic Equations – Homework

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12. In the given figure, M, N and P are the mid- point of AB, AC and BC respectively. If MN

= 3cm, NP = 4 cm and MP = 2.5 cm, Calculate BC, AB and AC.

13. Prove that the four triangles formed by joining in pairs, the mid- points of three sides of a

triangle, are congruent to each other.

14. In a parallelogram ABCD, if ∠A = (2x - 20°), ∠B = (y + 15°), ∠C = (x + 40°), then find the

value of x and y.

15. In the given figure, ABCD is an isosceles trapezium. Find x and y.

16. In a parallelogram ABCD, the bisector of ∠A also bisects BC at X. Find AB: AD

17. Find the value of x, y and z in the given figure if given PQRS is a rhombus and PR is

produced to T.

18. If the angles of quadrilateral are in the ratio 3: 5: 9: 13, then find the measure of the smallest

angle.

19. If the bisector of two adjacent angles A and B of a quadrilateral ABCD intersect at a point O

such that ∠C + ∠D = k∠AOB, then find the value of k.

20. Prove that the line segment joining the mid- point of two sides of a triangle is parallel to the

third side.

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IX – Std. Constructions – Class Notes I

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1. Introduction

Question1.The two most common instruments which are often used for construction are:

Question2.What is the meaning of ‘congruence of the figures’? How construction and congruence are

related?

Question3. Construct an angle of 60° and write down the steps of construction.

2. Construction of Angle Bisectors

Construction of angle bisectors is useful in making angles such as 150,30

0,45

0, etc…. The

concept behind this construction lies in congruence of triangles. Following are the steps

required for the construction of angle bisector:

i. Start with ∠ PQR that is to be bisected ( Construct ∠PQR)

ii. Place the compass point on the angle's vertex Q.

iii. Adjust the compasses to a medium wide setting. The exact width is not

important.(consider a width which is convenient to you)

iv. Without changing the compasses' width, draw an arc across each leg of the angle.

v. The compasses' width can be changed here if desired. But it would be great if you can leave

it the same.

vi. Place the compasses at the point where an arc intersects a leg and draw an arc in

the interior of the angle.

vii. Without changing the compasses setting repeat for the other leg so that the two arcs

intersect.

viii.Using a ruler, draw a line from the vertex to the point where the arcs cross (intersect).

Done. This is the bisector of the ∠ PQR.

Question4. Can you construct the corresponding figures for the above steps?

(Refer to the subtopic Angle Bisector_ Angle Bisector)

Question5. Prove the above method of construction of an angle bisector using congruence of

triangles.

(Refer to the subtopic Angle Bisector_ Angle Bisector)

Question6. Construct angular bisectors for the following angles. (i) 300

(ii) 900.

Note : An angle bisector, is a line passing through the vertex of the angle that cuts it into two

equal smaller angles.

http://www.mathopenref.com/vertex.html

http://www.mathopenref.com/arc.html

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3. Construction of Perpendicular Bisector

Perpendicular bisector of a line segment AB is the line that bisects the line segment AB and is

also perpendicular to it.

(Refer to the subtopic Perpendicular Bisector_ Perpendicular Bisector)

Question7. Based on the figure given below, write down the steps of construction for perpendicular

bisector of line segment PQ.

Helping Hand: While drawing a perpendicular bisector of a given line segment, when you place the

compass at one end of the line segment, make sure that you adjust the compass to slightly longer than

half the length of the line segment. If you take the compass length less than half the length of a line

segment, you may end up getting the following figure:

and the bisectors (arcs) will not intersect. Also, there is one more property associated with it. Can you

state the property?

(Refer to the subtopic Perpendicular Bisector_ Perpendicular Bisector)

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Did You Know?

Constructing a perpendicular bisector is actually

constructing angle bisector of 180° at the middle

of the line.

4. Construction of various angles

a. Construction of an angle of measure 600

An angle of measure 60o is easiest angle to construct. First, let us know the steps of

construction and then we will try to see the logic behind it. For a given a ray PQ, we need

to construct an ∠RPQ of measure 60o.

Draw a ray PQ.

Set the compasses on P, and set its width to any convenient setting. Draw an arc

across PQ and up over above the point P.

Let the arc cut the ray PQ at point A.

Without changing the compass’s width, move the compass to the point A, and make

an arc that intersects the first arc at a point. Let it be B.

Join PB and produce it till R to get a ray PR

The ∠RPQ so obtained is the angle measure of 60°

(Refer to the subtopic Constructing a 60o_ Constructing angles)

Question8. Using the above steps construct an angle of measure 60°

Let us understand why it works?

Join AB, and try to reason the following statements.

Statement

Reason

1 Line segments AB, PB, PA

are congruent

2 Triangle APB is anequilateral

triangle

3 ∠ APB has a measure of 60°

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b. Construction of an angle of measure 90o

We already know how to construct a perpendicular bisector of a line segment.

Constructing a 90o angle is nothing but the construction of a perpendicular to a line at a

suitable point,say A.

(Refer to the subtopic Constructing a 60o_Constructing an Angle of 90

o)

Question9. Construct an angle 90° at the initial point of a given ray and write down the steps for the

construction. Also justify your construction.

Construction:

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Steps of construction:

(Refer to the subtopic Constructing a 60oangle_Constructing an Angle of 90

o)

c. Construction of an angle of measure 75o

Construction of an angle of measure 75o is simple if you have understood all the concepts

till now. 75o can be constructed in the following two ways:

(i) 75o is the sum of angles 60

o and 15

o. You know how to construct 60

o and 15

o is

the bisector of 30o.

(ii) 75o is the bisector of 150

o. 150

o, on the other hand can be constructed as sum of

angles 90o and 60

o.

(Refer to the subtopic Constructing a 60oangle _ Constructing an Angle of 75

o)

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Question 10. Construct an angle of measure 75o using any one of the methods stated above and write

the steps of construction. Also justify your construction.

Question 11. Construct an angle of measure 135o and write down the steps of construction.

Question12. Construct an angle of measure 7.5o

Helping Hand: 7.5° is basically half of 15°, which is half of 30°. You already know how to construct

30°.

5. Construction of Triangles

Misconception: The congruency conditions SSS, SAS, ASA and RHS are randomly created

conditions for proving congruency.

Clarification: Congruency conditions SSS, SAS, ASA and RHS have a reason behind their

creation. Using any of these conditions, an identical triangle can be created for a given

triangle, which will be exactly same in terms of length of sides and measures of angles. But, if

other conditions are given, then exactly same triangles can not be created. For e.g. AAA,

SSA, etc.

a. Construction of Triangle when the sum of two sides is given.

Let us move one step further. We know how to construct a triangle, when 3 sides are

given, two sides and an involving angle are given, two angles and an involving side are

given. Let us recollect all those construction methods and try constructing triangle when

different conditions are given. Suppose, for a triangle ABC, following measures are

given:

- Side BC

- ∠ B

- AB+AC

Now, we are required to construct the triangle ABC.

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Steps of construction are as follows:

Draw a line BX and cut off a line segment BC from it.

At B, construct given ∠XBY.

From BY, Cut a line segment BD = AB+AC cm.

Join CD.

Draw the perpendicular bisector of CD, intersecting BD at point A.

Join AC.

∆ ABC is the required triangle.

Did You Know?

When a line or line segment is folded along its perpendicular

bisector, we get the mirror image of the line or line segment. The

perpendicular bisector acts as a mirror.

Question14. Construct a triangle ABC with the following measures:

- Side BC = 10 cm

- ∠ B = 45o

- AB+AC = 16 cm

(Refer to the subtopic Triangle Construction 1_Triangle Construction 1)

b. Construction of Triangle when the difference of the two sides is given.

Given a triangle ABC, in which length of side BC and measeure of ∠ B is given. Now,

following two cases are possible:

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(i) AB > AC

Question15. Based on the above figure, can you write down the steps of construction?

(Refer to the subtopic Triangle Construction 2 _ Triangle Construction 2)

(ii) AB < AC

Following are the steps of construction:

a. Draw the base BC.

b. Make ∠XBC and extend ray BX in opposite direction.

c. Mark the point D on the extended ray BX, such that BD = AC - AB

d. Join DC.

e. Draw the perpendicular bisector of DC such that, it intersects the ray BX at point A.

f. Join AC.

g. ∆ ABC is the required triangle.

Question16. Based on the above steps, can you construct a ∆ ABC with given measures base

AB=5cm, ∠A=30° and AC - BC = 2.5 cm

(Refer to the subtopic Triangle Construction 2 _ Triangle Construction 2)

c. Construction of Triangle when the perimeter and two base angles of the triangle are

given.

Given:

AB+BC+CA = a cm

∠ B

∠ C

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The steps of construction are as follows:

Draw the line segment XY = ‘a’cm. (Given perimeter of the triangle)

Draw the ray XL at X maing an angle same as ∠B with XY.

Draw the ray YM at Y maing an angle same as ∠C with XY.

Draw angle bisector of ∠LXY.

Draw angle bisector of ∠MYX such that it intersects the angle bisector of ∠LXY at a

point A.

Draw the perpendicular bisector of AX such that it intersects XY at a point B.

Draw the perpendicular bisector of AY such that it intersects XY at a point C.

Join AB and AC.

∆ ABC is the required triangle.

Question17. Construct a triangle ABC with perimeter 20 cm, ∠ B = 30o and ∠ C = 45

o and write down

the steps of construction.

(Refer to the subtopic Triangle Construction 3_Triangle Construction 3)

Special cases of construction of triangles:

Question18. Construct an isosceles triangle ABC with BC = 6.2 cm and altitude = 4.4 cm.

Question19. Construct an isosceles triangle ABC with BC = 6.8 cm and ∠A = 90°

Question20. Construct an equilateral triangle whose altitude is 4.6 cm.

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IX – Std. Constructions – Homework

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1. The following are the steps of construction of a triangle in which the base, the sum of other two

sides and one base angle is given.

1. At B, ∠XBY = γ

2. Draw a ray BX and cut off a line segment BC of length a cm.

3. Join C and D.

4. Join A and C.

5. With B as centre and radius x cm draw an arc to meet BY at D.

6. Draw the perpendicular bisector of CD intersecting BD at A.

The proper order of steps of construction is ____.

a. 6, 5, 4, 3, 1 and 2

b. 3, 2, 1, 4. 6 and 5

c. 1, 2, 3, 4, 5 and 6

d. 2, 1, 5, 3, 6 and 4

2. The following are the steps of construction of a triangle in which base BC = 4.5 cm, ∠ B = 60° and

the sum of other two sides is 8 cm.

1. Along BX, cut off BP = 8 cm

2. Join AC

3. Draw the perpendicular bisector PC to intersect BP at A

4. Draw BC = 4.5 cm

5. Join PC

6. At B, ∠ CBX = 60°


a. 1, 2, 3, 4, 5 and 6

b. 4, 6, 1, 5, 3 and 2

c. 2, 1, 5, 3, 6 and 4

d. 6, 5, 4, 3, 1 and 2

3. The following are steps of construction of a triangle given its base, difference of the other two sides

and one base angle.

1. From BY, cut off BD of d cm

2. Join C and D

3. Draw a ray BX and cut off a line segment BC of a cm

4. Draw perpendicular bisector of CD intersecting BY at A

5. Join A and C

6. At B, construct ∠XBY = α


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a. 3, 6, 1, 2, 4 and 5

b. 6, 5, 4, 3, 1 and 2

c. 1, 2, 3, 4, 5 and 6

d. 3, 1, 6, 2, 4 and 5

4. The following are steps of construction of a right triangle when one side is 6 cm and sum of other

side and hypotenuse is 10 cm.

1. From BY cut off a line segment BD of 10 cm

2. Draw the perpendicular bisector of CD intersecting BD at A

3. Draw a ray BX and cut off a line segment BC of length 6 cm

4. Join A and C

5. Construct ∠ XBY = 90°

6. Join C and D

The proper order of steps for construction is ____.

a. 1, 2, 3, 4, 5 and 6

b. 2, 4, 6, 5, 1 and 3

c. 3, 5, 1, 6, 2 and 4

d. 5, 2, 1, 3, 6 and 4

5. The following are the steps of construction of a triangle in which BC = 3.8 cm, ∠ B = 45° and AB +

AC = 6.8 cm.

1. Draw ∠ CBX = 45°.

2. Join C and D.

3. Join C and A.

4. From ray BX, cut-off line segment BD equal to AB + AC i.e., 6.8 cm.

5. Draw BC = 3.8 cm.

6. Draw the perpendicular bisector of CD meeting BD at A.

The proper order of steps of construction is _____.

a. 5, 2, 4, 1, 6 and 3

b. 4, 1, 5, 2, 3 and 6

c. 4, 1, 5, 3, 6and 2

d. 5, 1, 4, 2, 6 and 3


6. Construct a triangle ABC in which BC = 4.6 cm, ∠B=45° and AB + CA=8.2 cm

7. Construct a right triangle when one side is 3.5 cm and the sum of other side and hypotenuse is

5.5 cm

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8. Construct a triangle ABC in which BC = 6.5 cm, ∠B=60° and AB + CA = 10 cm

9. Construct a triangle ABC in which BC = 4.5 cm, ∠B=45° and AB – AC = 2.5 cm

10. Construct a triangle ABC in which BC = 5 cm, ∠B=30° and AC – AB = 2 cm

11. Draw the perpendicular bisector of a line segment of length 10 cm.

12. Draw an angle bisector of an angle of measure 135°.

13. Construct an angle of measure 15°. Also, write down the steps of construction.

14. Draw an angle of measure 3.75° and write down the steps of construction.

15. Write down the steps for the construction of perpendicular bisector of any line. Also, write the

significance of a perpendicular bisector.

16. Write down the steps of construction of an angle of measure 270°.

17. Draw the perpendicular bisector of a line segment of measure 15 cm. Make an angle of 45° on

the bisector and write the steps of construction.

18. Construct a triangle ABC with perimeter 15 cm, ∠B =30oand ∠C= 45

oand write down the

steps of construction.

19. Construct a triangle ABC with perimeter 25 cm, ∠B =45o

and ∠C= 30o

and write down the

steps of construction.

20. Construct an angle of measure 48.75°. Also, write down the steps of construction.

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IX – Std. Areas of Parallelograms and triangles –Class Notes I

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1. Basics of area

The area of a plane figure refers to the amount of space within the boundaries of a two-

dimensional shape and will be measured in square units.

e.g.𝑚2, 𝑐𝑚2, 𝑖𝑛𝑐ℎ𝑒𝑠2, 𝑘𝑚2, 𝑒𝑡𝑐.

We cannot calculate the area of a line.Area is calculated for a closed two-dimensional object,

a line is not a two-dimensional figure. Hence, length of the line can be calculated but not the

area.

Congruency ⇒ Equal Area, But, Equal Area ⇏ Congruency

2. Figures on the Same Base and Between the Same Parallels

Question1.Mention the twocondition based on which we can decide that the two figures are on the

same base and between the same parallels.

(Refer to the subtopic Introduction_Introduction)

Question2. Identify the figure, which lie on same base and between the same parallelsfrom the

following figures:

Figure 1.

Figure 2.

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Figure 3.

Figure 4.

3. Parallelograms on the same base and between the same parallels

Theorem :Parallelograms on the same base and between the same parallels are equal in area.

Figure 5.parallelograms

Proof: Two parallelograms 𝐴𝐵𝐶𝐷and 𝐸𝐹𝐶𝐷, on the same base 𝐷𝐶 and between the same

parallels 𝐴𝐹 and 𝐷𝐶.

To prove:𝑎𝑟(𝐴𝐵𝐶𝐷) = 𝑎𝑟(𝐸𝐹𝐶𝐷)

In ∆𝐴𝐷𝐸 and ∆𝐵𝐶𝐹

∠𝐷𝐴𝐸 = ∠𝐶𝐵𝐹[Corresponding angles from 𝐴𝐷||𝐵𝐶 and transversal 𝐴𝐹]− − −(1)

∠______ = ∠______[Corresponding angles from 𝐸𝐷||𝐹𝐶 and transversal 𝐴𝐹]− − −(2)

Also, 𝐴𝐷 = ____ [ Opposite sides of the parallelogram ABCD] − − −(3)

So, ∆𝐴𝐷𝐸 ≅ ∆𝐵𝐶𝐹 [By ______ rule, using equations(1),(2) and (3)]

Therefore, 𝑎𝑟(𝐴𝐷𝐸) = 𝑎𝑟(𝐵𝐶𝐹) [Since, _______________________________]− − −(4)

𝑎𝑟(𝐴𝐵𝐶𝐷) = 𝑎𝑟(𝐴𝐷𝐸) + 𝑎𝑟(𝐸𝐷𝐶𝐵)

𝑎𝑟(𝐴𝐵𝐶𝐷) = _______ + 𝑎𝑟(𝐸𝐷𝐶𝐵) [ Using equation(4)]

𝑎𝑟(𝐴𝐵𝐶𝐷) = 𝑎𝑟(𝐸𝐹𝐶𝐷)

Hence, the area of two parallelograms 𝐴𝐵𝐶𝐷 and 𝐸𝐹𝐶𝐷 are equal.

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(Refer to the subtopic Theorems_Theorem-I)

Question3. Show that area of a parallelogram is the product of its base and the corresponding

altitude.

Hint: construct of rectangle with the base of parallelogram

Question4. Parallelogram 𝐴𝐵𝐶𝐷 and rectangle 𝐴𝐵𝐹𝐸 have samebase 𝐴𝐵, and the length and

breadth of the rectangle are as 9 cm and 5 cm. find the area of the parallelogram.

Hint: Apply formula of area of parallelogram

Question5. If a triangle anda parallelogram are on the same base and between the same parallels,

show that the area of the triangle is equal to half of the area of the parallelogram.

Hint: Construct a line such a way that it forms another parallelogram, then use the concept of

parallelograms on the same base and between the same parallels are equal in area.

Question6. The side 𝐴𝐵 of a parallelogram 𝐴𝐵𝐶𝐷 is produced to any point 𝑃. A line through 𝐴 and

parallel to 𝐶𝑃 meets 𝐶𝐵 produced at 𝑄 and then parallelogram 𝑃𝐵𝑄𝑅 is completed as shown in the

figure. Show that 𝑎𝑟(||gm𝐴𝐵𝐶𝐷) = 𝑎𝑟(||gm𝑃𝐵𝑄𝑅).

Figure 6.

Hint: Join 𝐴𝐶 and 𝑃𝑄

Question7. In the given figure, 𝑃𝑄𝑅𝑆 is a rectangle. If 𝑃𝑆 = 8 𝑐𝑚and 𝑆𝑅 = 4𝑐𝑚, then find the area

of ∆𝐴𝐵𝐶.

Figure 7.

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Question8.Triangle 𝐴𝐵𝐷 and parallelogram 𝐴𝐵𝐶𝐷 are on the same base 𝐴𝐵. If base and altitude of the

parallelogram are 15 cm and 10 cm, find the area of the triangle.

Did You Know?

The diagonals of parallelograms divide it into four

triangles of equal area.

4. Triangles on the Same Base and Between the Same Parallels

Theorem :Triangles on the same base and between the same parallels are equal in area

Figure 8.Triangles

Given: Two triangles 𝐴𝐵𝐶 and 𝑃𝐵𝐶 on the same base 𝐵𝐶 and between the same parallel lines

𝐵𝐶 and 𝐴𝑃.

To prove:𝑎𝑟(∆𝐴𝐵𝐶) = 𝑎𝑟(∆𝑃𝐵𝐶)

Construction:Through 𝐵, draw 𝐵𝐷||𝐶𝐴 intersecting 𝑃𝐴 produced in 𝐷 and through 𝐶, draw

𝐶𝑄||𝐵𝑃, intersecting line 𝐴𝑃 in 𝑄.

Proof:𝐵𝐷||𝐶𝐴 [By construction]

𝐵𝐶||𝐷𝐴[Given]

Therefore, 𝐵𝐶𝐴𝐷 is a parallelogram

Similarly, 𝐵𝐶𝑄𝑃 is a parallelogram

𝑎𝑟(||gm𝐵𝐶𝑄𝑃) = 𝑎𝑟(||gm______)− − −(1) [ Since, Parallelograms are on the same base

and between the same parallels]

The diagonals of a parallelogram divides it into two triangles of equal area

∴ 𝑎𝑟(∆𝑃𝐵𝐶) =1

2 𝑎𝑟(_______) − − − (2)

𝑎𝑟(________) = 1

2𝑎𝑟(||gm𝐵𝐶𝐴𝐷) − − − (3)

𝑎𝑟(||gm𝐵𝐶𝐴𝐷) = 𝑎𝑟(||gm𝐵𝐶𝑄𝑃) [from equation (1)]

⇒1

2 𝑎𝑟(||gm𝐵𝐶𝐴𝐷) =

1

2𝑎𝑟(||gm𝐵𝐶𝑄𝑃)[ multiplying

1

2 on both sides]

⇒ar(∆ABC) = ar(_______)[Using equations (2) and (3)]

Hence proved.

(Refer to the subtopic Theorems_Theorem-II)

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Question9.Prove that the area of a triangle is equal to half of product of its base and corresponding

altitude.

Hint: construct a parallelogram with the base of triangle

Did You Know?

Median of a triangle divides it into two triangles

of equal area.

Question10. In the given figure,𝐴𝐵𝐶𝐷 is a quadrilateral, it is being given that 𝑀 is the mid-point

of𝐴𝐶.Prove that 𝑎𝑟(quadrilateral 𝐴𝐵𝑀𝐷) = 𝑎𝑟(quadrilateral 𝐷𝑀𝐵𝐶)

Figure 9.

Theorem :Two triangles with the same base(or equal bases) and equal areas will have equal

corresponding altitudes.

Figure 10.

Given :𝑎𝑟(∆𝐴𝐵𝐶) = 𝑎𝑟(∆𝑃𝑄𝑅) and 𝐴𝐵 = 𝑃𝑄

To prove:𝐶𝑁 = 𝑅𝑇

In ∆𝐴𝐵𝐶, 𝐶𝑁 is the altitude corresponding to side 𝐴𝐵.

∴ 𝑎𝑟(∆𝐴𝐵𝐶) = 1

2 (____ × 𝐶𝑁) − − − (1)

Similarly,

𝑎𝑟(∆𝑃𝑄𝑅) = 1

2 (𝑃𝑄 × ____) − − − (2)

Since, 𝑎𝑟(∆𝐴𝐵𝐶) = 𝑎𝑟(∆𝑃𝑄𝑅)

⇒1

2(𝐴𝐵 × 𝐶𝑁) =

1

2(𝑃𝑄 × 𝑅𝑇) [Using equations (1) and (2)]

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⇒ ____________ = 𝑃𝑄 × 𝑅𝑇

⇒ ____ × 𝐶𝑁 = ____ × 𝑅𝑇 [Given]

⇒ 𝐶𝑁 = 𝑅𝑇

Hence, proved

(Refer to the subtopic Theorems_Theorem-III)

Helping Hand: If two triangles with the same base(or equal bases) and equal areas will have equal

corresponding altitudes.For having equal corresponding altitudes, the triangles must lie between the

same parallels.

Question11. 𝐴𝐷is the median of ∆𝐴𝐵𝐶and ∆𝐴𝐷𝐶. 𝐸is any point on 𝐴𝐷. Show that area of

∆𝐴𝐵𝐸 =area of ∆𝐴𝐶𝐸.

Hint: Median of a triangle divides it into two triangles of equal area.

Question12. 𝑂is any point on the diagonal 𝐵𝐷 of the parallelogram 𝐴𝐵𝐶𝐷. Prove that 𝑎𝑟(∆𝑂𝐴𝐵) =

𝑎𝑟(∆𝑂𝐵𝐶)

Hint: Join 𝐴𝐶 and use the concept that the median of a triangle divides it into two triangles of equal

area

Application:

Bridges are constructed by combining triangles so, designers need to have a very good knowledge of

the properties of triangles, rectangles and parallelograms to make sure the structure is strong enough.

Activity:

Students are requested to go for ‘Shape Scavenger Hunt’. Being in the class, try to find different

figures and classify them as squares, rectangles, triangles, parallelograms and other shapes. Once this

is done, use scale to find their dimensions and calculate their area. The student who finally has the

largest value of area wins the game.

Figure 11.

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IX – Std. Areas of Parallelograms and triangles – Homework

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1. 𝑃 and 𝑄 are the points on sides 𝐴𝐵 and 𝐴𝐶 in a ∆𝐴𝐵𝐶, such that:

Area of ∆𝐴𝐵𝐶 =1

2×Area of ∆𝐴𝑃𝑄Then,

A) ∆𝑃𝐵𝐶 is congruent to ∆𝑄𝐵𝐶

B) 𝑃𝑄 =1

2𝐵𝐶

C) 𝑃𝑄||𝐵𝐶 D) All options are correct

2. In the given figure, the angles 𝐵𝐴𝐷 and 𝐴𝐷𝐶 are right angles and 𝐴𝑋 || 𝐵𝐶.If 𝐴𝐵 = 𝐵𝐶 = 5cm and𝐷𝐶 = 8 𝑐𝑚, then the area of 𝐴𝐵𝐶𝑋 is

Figure 12.

A) 20 𝑐𝑚2 B) 64 𝑐𝑚2 C) 42 𝑐𝑚2 D) 32 𝑐𝑚2

3. If a triangle and parallelogram are on the same base and between same parallels, then the

ratio of the area of the triangle to the area of the parallelogram is A) 1: 1 B) 2: 1 C) 1: 2 D) 1: 4

4. The base 𝐵𝐶 of triangle 𝐴𝐵𝐶 is divided at 𝐷 such that 𝐵𝐷 =1

2 𝐷𝐶, then

𝑎𝑟(∆𝐴𝐵𝐷): 𝑎𝑟(∆𝐴𝐵𝐶) is A) 2:3 B) 1:2 C) 1:4 D) 1:3

5. The median of a triangle divides it into two

A) Triangles of equal area B) Scalene triangles C) Right angle triangles D) Isosceles triangles


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6. In figure, parallelograms 𝐴𝐵𝐶𝐷 and 𝑃𝐵𝐶𝑄 are given. If 𝑅 is a point on 𝑃𝐵, then show that

𝑎𝑟(∆𝑄𝑅𝐶) =1

2(||gm𝐴𝐵𝐶𝐷).

Figure 13.

7. 𝐷 and 𝐸 are points on sides 𝐴𝐵 and 𝐴𝐶, respectively of ∆𝐴𝐵𝐶 such that 𝑎𝑟(∆𝐷𝐵𝐶) =

𝑎𝑟(∆𝐸𝐵𝐶). Prove that 𝐷𝐸||𝐵𝐶.

8. In the given figure, 𝐴𝐵𝐶𝐷 is a parallelogram and BC is produced to a point 𝑄 such that 𝐴𝐷 =

𝐶𝑄. If 𝐴𝑄 intersects𝐷𝐶 at 𝑃, then show that 𝑎𝑟(∆𝐵𝑃𝐶) = 𝑎𝑟(∆𝐷𝑃𝑄).

(Hint: Join 𝐴𝐶)

Figure 14.

9. Show that if a triangle and a parallelogram are on the same base and between the same

parallels, then the area of the triangle is equal a half the area of the parallelogram.

10. Show that a median of a triangle divides it into two triangles of equal areas.

11. Diagonals 𝐴𝐶 and 𝐵𝐷 of a quadrilateral 𝐴𝐵𝐶𝐷 intersect at 𝑂 in such a way

that 𝑎𝑟(∆𝐴𝑂𝐷) = 𝑎𝑟(∆𝐵𝑂𝐶). Prove that 𝐴𝐵𝐶𝐷 is a trapezium.

12. In the given figure, 𝐵𝐶||𝑋𝑌, 𝐵𝑋||𝐶𝐴 and 𝐴𝐵||𝑌𝐶. Prove that: 𝑎𝑟(∆𝐴𝐵𝑋) = 𝑎𝑟(∆𝐴𝐶𝑌)

(Hint:Join𝑋𝐶)

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Figure 15.

13. 𝑃 and 𝑄 are respectively the mid- points of sides 𝐴𝐵 and 𝐵𝐶 of a triangle 𝐴𝐵𝐶 and 𝑅 is

the mid- point of 𝐴𝑃, show that𝑎𝑟 (∆𝑃𝑅𝑄) = 𝟏

𝟐 𝑎𝑟 (∆𝐴𝑅𝐶)

14. If 𝐸, 𝐹, 𝐺 and 𝐻 are respectively the mid-points of the sides of a parallelogram 𝐴𝐵𝐶𝐷. Show

that 𝑎𝑟(𝐸𝐹𝐺𝐻) =1

2𝑎𝑟(𝐴𝐵𝐶𝐷).

15. In the given figure,𝐴𝐵𝐶𝐷 , 𝐷𝐶𝐹𝐸 and 𝐴𝐵𝐹𝐸 are parallelograms. Show that 𝑎𝑟(∆𝐴𝐷𝐸) =

𝑎𝑟(∆𝐵𝐶𝐹).

Figure 16.

16. In the given figure, diagonals 𝐴𝐶 and 𝐵𝐷 of a quadrilateral 𝐴𝐵𝐶𝐷 intersect each other at P.

Show that 𝑎𝑟(∆𝐴𝑃𝐵) × 𝑎𝑟(∆𝐶𝑃𝐷) = 𝑎𝑟(∆𝐴𝑃𝐷) × 𝑎𝑟(∆𝐵𝑃𝐶).

(Hint: from 𝐴 and 𝐶, draw perpendiculars to 𝐵𝐷)

Figure 17.

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17. In the given figure, 𝐴𝐵𝐶𝐷 is a parallelogram. Points 𝑃 and 𝑄 on 𝐵𝐶 trisect 𝐵𝐶 in three equal

parts. Prove that 𝑎𝑟(∆𝐴𝑃𝑄) = 𝑎𝑟(∆𝐷𝑃𝑄) =1

6𝑎𝑟(||gm𝐴𝐵𝐶𝐷)

(Hint: Join 𝐵𝐷)

Figure 18.

18. In the given figure, 𝐴𝐵𝐶𝐷𝐸 is any pentagon.𝐵𝑃 drawn parallel to 𝐴𝐶 meets 𝐷𝐶 produced at

𝑃 and 𝐸𝑄 drawn parallel to 𝐴𝐷 meets 𝐶𝐷 produced at 𝑄. Prove that 𝑎𝑟(𝐴𝐵𝐶𝐷𝐸) =

𝑎𝑟(𝐴𝑃𝑄).

Figure 19.

19. In the given figure, 𝐴𝐵𝐶𝐷 is a parallelogram. 𝑃 is any point on 𝐶𝐷. If 𝑎𝑟(∆𝐷𝑃𝐴) = 15 𝑐𝑚2

and𝑎𝑟(∆𝐴𝑃𝐶) = 20 𝑐𝑚2, find the 𝑎𝑟(∆𝐴𝑃𝐵).

Figure 20.

20. In the given∆𝐴𝐵𝐶, if 𝐿 and 𝑀 are the points on 𝐴𝐵 and 𝐴𝐶, respectively such that 𝐿𝑀||𝐵𝐶.

Prove that 𝑎𝑟(∆𝐿𝑂𝐵) = 𝑎𝑟(∆𝑀𝑂𝐶).

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Figure 21.

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IX – Std. Statistics – Class Notes I

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1. Introduction

Did You Know?

Thomas Elva Edison patented almost 1,300 inventions

in his lifetime.

Did You Know?

About 39,000 gallons of water are used to produce

the

average car.

Any information when represented using numerical data is Statistics. It deals with the

collection, presentation, analysis and interpretation of numerical data. We generally use

statistics when we have huge collection of data.

(Refer to the subtopic Introduction_Clip 1)

Did You Know?

A new baby usually deprives each of it’s parents around

350-400 hours of sleep in the first year.

Did You Know?

The most children born to one women was 69, she was a peasant who lived a

40 year life, in which she has 16 twins, 7 triplets and 4 quadruplets.

Question1.What do you mean by data?

Question2.Can you give 2 practical examples of statistical data?

Arranging data in a order to study their salient features is called presentation of data.

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Question3.The maximum recorded temperature of Bengaluru on 26th April, 27

th April, 28

th April, 29

th

April, 30th April and 1

st May was 37°,39°,39°,39°,38° and 37° respectively. Can you think of any way

to show this date more clearly.

Data arranged in ascending or descending order is called arrayed data or an array.

When an investigator or surveyer, with a definite plan or design in mind collects data first

handedly, it is called primary data.

Data when collected by a surveyer, comes to you, is known as the secondary data.

Range of the data is the difference between the maximum and the minimum values of the

observations.

The small groups obtained on dividing all the observations are called classes or class

intervals and the size is called the class size or class width.

Question4.What is the range of data observation given on Question no.3?

2. Frequency Table

a. Frequency:

Frequency is a number which tells how many times a particular data is present in a given

set of data.

(Refer to the subtopic Frequency table_ Grouping)

Question5. Write frequency of all the numbers in the following data:

8, 6, 8, 5, 6, 4, 7, 9, 7, 4, 3, 5, 3, 5, 4, 8

b. Frequency distribution: It is a tabular arrangement of data showing their corresponding

frequencies. This table is called frequency distribution table.

Ungrouped frequency distribution

Question6. Consider the following data which gives the number of goals scored by 16 players in a

football tournament.

6, 8, 8, 6, 5, 4, 4, 8, 10, 8, 10, 6, 9, 8, 4, 8

Complete the following frequency table:

Number of Goals Tally Marks Frequency

4

5

6

8

9

10

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Grouped frequency distribution

Question7. Given below are the marks obtained by 30 students in an examination:

08 18 35 42 46 24 20 36

07 17 45 10 30 19 29 10

36 47 40 25 23 04 16 21

16 34 46 42 33 01

Taking class interval as 1 – 10, 11 – 20, ….., 41 – 50; make frequency table for the above

distribution.

Marks Tally Marks Frequency

1 – 10

11 – 20

21 – 30

31 – 40

41 – 50

(Refer to the subtopic Frequency table_ Grouped and Ungrouped frequency)

Note: In the above question; If the interval is taken as 1 – 10, 10 – 20, 20 – 30, 30 – 40, 40 –

50, then marks 10 are included in the interval 10 – 20 not in 1 – 10. Similarly, marks 30 are

included in 30 – 40 not in 20 – 30.

(i) Inclusive frequency distribution (non-overlapping): Upper limit of lower class does not coincide with the lower limit of the next

class. (Question7 is an example)

(ii) Exclusive frequency distribution (overlapping): Upper limit of one class coincides with the lower limit of the next class.

(Refer to the subtopic Frequency table_Overlapping groups)

c. Class Interval and Class Limits:

Question8. In case of an exclusive frequency distribution; If 20 – 30 is a class interval which is

bounded between 20 and 30, then

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1. 20 is called _______________

2. 30 is called ________________

Similarly, in case of an inclusive class distribution, 1 – 10 is a class interval whose

3. Lower limit is _____________

4. Upper limit is ______________


Note: To get continuity and to get correct class limits, exclusive distribution is adopted.

Following adjustment should be done to convert inclusive class intervals to exclusive

intervals.

(1) Find the difference between the upper limit of one class and lower limit of the next

class.

(2) Divide the difference by 2. The value obtained is called adjustment factor.

(3) Subtract the adjustment factor from all the lower limit.

(4) Add the adjustment factor to all the upper limit.

The class limits obtained after adjustment is done is called actual (true) class limits.


Question9. Complete the following table using adjustment factor.

Marks before adjustment Marks after adjustment

1 – 10 0.5 – 10.5

11 – 20

21 – 30

31 – 40

41 – 50

(a) The difference between the actual lower limit and the actual upper limit is called _______

(b) Class size of the interval 10.5 – 20.5 is ______

(c) What is Class-Mark of a class interval?

(d) Class Mark of 10.5 – 20.5 is ______

d. Cumulative Frequency:

Cumulative frequency of a class interval is the sum of frequencies of all classes up to that

class including the frequency of that particular class.

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Question10. Complete the following cumulative frequency distribution table.

Class Interval Frequency Cumulative Frequency

0 – 8 8

8 – 16 12

16 – 24 10

24 – 32 8

32 – 40 5

3. Bar graph

There are three types of graphical ways to represent data.

Bar Graphs

Histograms

Frequency Polygons

The type of representation will depend on the type of data you have or basically the type of

variable you have.When the variable you measure is not a number, we use bar graph and

when the variable we use is a number we either use Histograms or Frequency Polygons.A bar

graph is the diagram showing a system of connections or interrelations between two or more

things by using bars.

(Refer to the subtopic Bar Graph_Types of Bar Graph)

Consider the following example:

A survey was done in a class in which students were asked about their favourite colours.

Following was the data collected:

Number of Students FavouriteColour

40 Red

35 Pink

20 Yellow

55 Purple

10 Orange

15 Green

45 Black

The data can be represented in the following way as bar graph:

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Question11. (a) As per the bar graph. What is the total strength of the class?

(b) Which is the least and most liked colour?

4. Histogram

A histogram is a form of bar graph which is used for continuous class intervals.

In a histogram:

• the bars do not have gaps between them.

• the width of the bars is proportional to the class intervals of data.

• the height of the bars represents the different values of the variable.

• the area of each rectangle is proportional to its corresponding frequency.

(Refer to the subtopic Histogram_Histogram and Frequency Polygon)

Procedure to construct a histogram:

Take a graph and draw two perpendicular lines. Mark them as OX and OY.

Take horizontal line OX as x-axis and vertical line OY as y-axis.

Choose suitable scale for class intervals on x-axis and represent class limits.

Choose a suitable scale along y-axis to represent frequencies.

Construct rectangle with class intervals as base and heights proportional to the

frequencies.

For a histogram, following two cases are possible:

a. When the size of class intervals are equal.

b. When the size of class intervals are not equal.

Question12. What is the procedure for creating a histogram in case the class intervals are not equal?

(Refer to the subtopic Histogram_Example with Histogram)

Question13. Consider the following example: (Equal Class Intervals)

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In a hospital, there are total 200 patients of different age group. The following data is given:

Age(in years)

0-10

10-20

20-30

30-40

40-50

50-60

No. of patients

45

30

25

40

50

( )

Prepare a histogram for the same.

Consider another example of unequal class interval.

In an examination, 200 students appeared in total. The maximum marks was 100.The result

published by the teacher was as follows:

Marks No. of students Grade

0-10 10 FAIL

10-30 30 D

30-40 40 C

40-70 20 B

70-80 50 A

80-90 30 A+

90-100 ( ) A++

In the above table, we can see that the class widths are not of equal length.So, firstly we need

to make them equal. Say, for instance, in the class interval 10-30, the class size is 20, no. of

students is 30, so by unitary method we can say that when the class size is 10, no. of students

will be ___________________.So, the new table will be as follows:

Marks Frequency Width of class Length of Rectangle

0-10 10 10 10

10× 10 = _____

10-30 30 20 30

20× 10 = _____

30-40 40 10 40

10× 10 = _____

40-70 20 30 20

30× 10 = _____

70-80 50 10 50

10× 10 = _____

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80-90 30 10 30

10× 10 = _____

90-100 20 10 20

10× 10 = _____

Since we have calculated these lengths for an interval of 10 marks in each case, we may call

these lengths as “proportion of students per 10 marks interval”.

So, the correct histogram with varying width can be given as follows:

Proportion of

students per

10marks interval

M

arks

(Refer to the subtopic Bar Graph_Example with Histogram)

a. Frequency Polygon

(Refer to the subtopic Bar Graph_Example-Frequency Polygon)

A frequency polygon is formed by joining the midpoints (class mark) of the adjacent

rectangles in a histogram with line segments.

Class mark for a class interval = Upper class limit + Lower class limit

2

Did You Know?

A frequency polygon can also be formed by joining the class marks of the given data

with line segments.

0

10

20

30

40

50

60

0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100

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Frequency polygons are used to represent the data when the data is continuous and very large.

The midpoints at each end are joined to the immediately lower or higher assumed class

interval of zero frequency.

Did You Know?

The area of a histogram is equal to the area enclosed by its corresponding

frequency polygon.

Question14. Draw the histogram of the following data given. Also, plot the frequency polygon of the

given data in the same graph.

Salary(in lakhs)

0-10

10-20

20-30

30-40

40-50

50-60

No. of

employees

50

90

105

30

20

10

(Refer to the subtopic Bar Graph_Example-Frequency Polygon)

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5. Mean, Median, Mode

a. Mean:

It is also called arithmetic mean, average.

To find mean of a set of observations, divide the sum of observations by the number

of observations.

(Refer the video clip Mean, mode, median in Mean, Median, Mode of Statistics)

Question15. 𝑥1 + 𝑥2 + 𝑥3 + ⋯ + 𝑥𝑛−1 + 𝑥𝑛 is denoted as _____

Question16. What is the mean of 𝑥1, 𝑥2, 𝑥3, … 𝑥𝑛−1, 𝑥𝑛?

Question17. Find the mean of the even numbers between 10 and 25.

Properties of mean:

If each observation is increased by a quantity 𝑎, then their mean is also increased by 𝑎.

If each observation is decreased by a quantity 𝑎, then their mean is also decreased by 𝑎.

If each observation is multiplied by a quantity 𝑎, then their mean is also multiplied by 𝑎.

If each observation is divided by a quantity 𝑎, then their mean is also divided by 𝑎.

Did you know?

Sum of deviations from each observation will be zero.

If 𝑥1, 𝑥2, 𝑥3, … 𝑥𝑛−1, 𝑥𝑛 are ‘n’ observations and �̅� is the mean, then; ∑(𝒙 − �̅�) = 𝟎

Question18. Find the mean of 8, 4, 7, 6, 12 and 5.

Find the resulting mean if each observation, given above is :

(1) increased by 5 (2) Decreased by 3 (3) Multiplied by 2 (4) Divided by 6

Question19. The mean of 6, 7, P, 5, 3 is 5. What is the value of P?

If 𝑥1, 𝑥2, . . . , 𝑥𝑛 are observations with respective frequencies 𝑓1, 𝑓2, . . . , 𝑓𝑛, then this means observation 𝑥1

occurs 𝑓1 times, 𝑥2 occurs 𝑓2 times, and so on. Now, the sum of the values of all the observations = 𝑓1𝑥1 + 𝑓2𝑥2 + . . . + 𝑓𝑛𝑥𝑛, and sum of the number of

observations = 𝑓1 + 𝑓2 + . . . + 𝑓𝑛. So, the mean x of the data is given by

�̅� =𝑓1𝑥1 + 𝑓2𝑥2 + . . . + 𝑓𝑛𝑥𝑛

𝑓1 + 𝑓2 + . . . + 𝑓𝑛

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�̅� =∑ 𝑓𝑖𝑥𝑖

𝑛𝑖=1

∑ 𝑓𝑖 𝑛𝑖=1

Question20. The marks obtained by 30 students of Class X of a certain school in a Mathematics paper,

consisting of 100 marks are presented in table below. Find the mean of the marks obtained by the students.

Marks Obtained 10 20 36 40 50 56 60 70 72 80 88 92 95

𝑥𝑖

Number of Students 1 1 3 4 3 2 4 4 1 1 2 3 1

𝑓𝑖

Solution: To find the mean marks, we require the product of each 𝑥𝑖 with the corresponding frequency𝑓𝑖.

Marks obtained (𝑥𝑖)

No. of students (𝑓𝑖)

𝑓𝑖𝑥𝑖

10 1 1 × 10 = 10

20 1

36 3

40 4

50 3 3 × 50 = 150

56 2

60 4

70 4

72 1

80 1

88 2

92 3 95 1

Total ∑ 𝑓𝑖 = ∑ 𝑓𝑖𝑥𝑖 =

�̅� =∑ 𝑓𝑖𝑥𝑖

𝑛𝑖=1

∑ 𝑓𝑖 𝑛𝑖=1

=

= __________

b. Median:

It is the middle term of the data arranged in ascending or descending order.

If number of observations (n) is odd, then median =

(𝑛+1

2)

𝑡ℎ term

If number of observations (n) is even, then median =

1

2[(

𝑛

2)

𝑡ℎ𝑡𝑒𝑟𝑚 + (

𝑛

2+ 1)

𝑡ℎ𝑡𝑒𝑟𝑚]

(Refer to the subtopic Mean, Median, Mode _Mean, mode, median)

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Question21. Find the median of 25, 16, 26, 16, 32, 31, 19, 28, 35

c. Mode:

An observation with maximum frequency is called mode.

(Refer to the subtopic Mean, Median, Mode _Mean, mode, median)

Question22. Find the mode of the following data.

2, 3, 4, 5, 0, 1, 3, 3, 4, 3

Space for Extra Notes and Rough Work

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IX – Std. Statistics – Homework

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Questions from 1 to 7 are MCQ’s.

1. Which of the following is a measure of central tendency?

A) Frequency

B) Cumulative frequency

C) Mean

D) Class limit

2. Find the range of the following data: 26, 21 33, 17, 14, 15, 27.

A) 13

B) 15

C) 17

D) 19

3. The mean of five observations is 12 . If the mean of first 3 is 10. What is the mean of other

two?

A) 10

B) 12

C) 12.5

D) 15

4. If we need to graphically represent data in which the values of the variables are not

quantifiable

(and hence not comparable), then which of the following is used?

A) Bar graph

B) Frequency polygon

C) Histogram

D) Any of the above

5. If two bars in a histogram have unequal widths but same height, then which class interval will

have a higher density of observations?

A) The one with higher class width

B) The one with lower class width

C) Both have same density

D) Can’t say

6. Which set of data has a mean of 15, a range of 22, a median of 14, and a mode of 14?

A) 25, 15, 14, 3, 7

B) 3, 14, 19, 25, 14

C) 14, 22, 15, 15, 9

D) 14, 22, 14, 15, 4

7. Which number is not the mean, median, or mode of the data set 4, 3, 15, 11, 3, 8, 7, 5?

A) 5

B) 6

C) 7

D) 3

Questions from 8 to 26 are subjective questions.

8. The numbers of roses in a garden on 30 days of a month are recorded as below.

40, 45, 43, 50, 62, 56, 50, 45, 40, 50, 52, 65, 62, 56, 45, 50, 60, 62, 68, 62, 55, 42, 45, 50, 60

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(a) What type of collection of data is given?

(b) What is the range of the collection?

(c) What is the frequency of 62?

(d) Write the data in arrayed form.

9. Write the tally marks of a frequency 14.

10. The weights (in kg) of the students in a class are recorded as shown below.

60, 65, 63, 70, 65, 62, 65, 63, 64, 60, 68, 58, 62, 65, 63, 65, 64, 60, 62, 63

If the collection of data be grouped into the class intervals 56 – 59, 59 – 62, 62 – 65, 65 – 68,

68 – 71, then answer the following:

(a) What is the type of these class intervals?

(b) Find the frequency of the class interval 59 – 62.

(c) Write the tally marks for the frequency of variable 65 and the class interval 62 – 65.

(d) What are the class limits and class boundaries of the class interval 68 – 71?

(e) What is the class size and the class mark of the class interval 68 – 71?

11. For the collection of numbers 12, 15, 8, 13, 12, 15, 9, 16, 24, 20, 20, 16 and 10, answer the

following:

(a) What is the cumulative frequency of 16?

(b) If 15 – 20 be an overlapping class interval when the numbers are grouped, find the

cumulative frequency of the class interval.

(c) If 15 – 20 be a non overlapping class interval when the numbers are grouped, find the

cumulative frequency of the class interval.

12. If the class marks of two consecutive overlapping intervals of equal size in a distribution are

94 and 104, then find the corresponding intervals.

13. State true or false.

(a) In the context of gathering data for a statistical study, if the data is obtained from a

source which had collected it for some other purpose, then data is called secondary

data.

(b) The marks obtained by 10 students in a mathematics test are:

52, 46, 92, 78, 62, 44, 34, 58, 52.

When the data is presented in this way, it is called ungrouped frequency table.

14. A cumulative frequency distribution is given below. Rewrite it as a frequency distribution.

Class interval Cumulative frequency

0 – 15 5

15 – 30 12

30 – 45 20

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45 – 60 29

60 – 75 50

15. State whether the following are true or not.

(a) If the classes in a grouped frequency distribution are not continuous, we can make

them continuous by finding half of the difference between the upper limit of a class

and the lower limit of the next class, and then adding the same to all the lower limits

and subtracting it from all the upper limits in the distribution.

(b) The difference between the highest and lowest observations in some data is called the

range.

(c) The classes in a histogram need not be of the same widths.

16. Answer the following questions.

(a) What do you call the middle value when you arrange all the observations in

ascending or descending order?

(b) What is the ratio of the sum of all observations to the number of observations called?

(c) What do you call the most frequent observation in the data?

(d) Which measure of central tendency is not a good representative of the data when you

have a few extreme values?

17. A frequency distribution is given below:

Class

interval

10 – 20 20 – 30 30 – 40 40 – 50 Total

Frequency 50 40 30 40 160

Draw a histogram for the data.

18. Draw a frequency polygon for the above question.

19. Draw a histogram and the frequency polygon in the same figure for the following distribution.

Class

interval

0 – 30 30 – 60 60 – 90 90 – 120 120 – 150 Total

Frequency 7 4 5 10 6 32

20. Find the mean of all the prime numbers between 30 and 60.

21. Find the value of 𝑝 if the mean of 6, 16, 𝑝, 12 is 14.

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22. The mean weight of 4 members of a family is 60 kg. Three of them have the weights 56 kg,

68 kg and 72 kg. Find the weight of the fourth member.

23. Find 𝑝 if 15, 12, 11, 𝑝 + 6, 7, 5, 4 are in order and their median is 10.

24. Find the median of the following data:

Number 10 11 12 13 14

Frequency 1 2 3 4 5

25. Find the mode of the following data:

110, 120, 130, 120, 110, 140, 130, 120, 140, 120.

26. Find the mean:

Observation 3 6 7 9 10

Frequency 8 10 15 9 8

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IX – Std. Surface Areas and Volumes – Class Notes I

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1. Introduction

Whenever you wrap gifts for your friends or parents, how do you choose the size of the wrapping

paper?

First, you will measure the size of the gift, i.e. area of the outer surface which you can touch, then

you choose a wrapping cover whose size is slightly more than that of the gift (to take care about

folding etc.)

Question1. Write down the difference between plane figures and solid figures. Also, give some

examples.

Plane figure Solid figure

(Refer to the subtopic Introduction_Part I)

Question2. Classify the following figures into plane figures and solid figures.

Figures (To be created) Plane figure/Solid figure

Triangle

Cylinder

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Cuboid

Circle

Rectangle

Cone

Question3. Match the following table which tells how these solid figures are formed.

Figures (to be created) How they are formed?

Right circular cylinder a. 6 squares joined together

Right circular cone b. Two circles and a bended reactangle/A

rectangle rotated about one of its sides.

Cube c. A circular disc rotated about its diameter.

Sphere d. A right triangle rotated about its base.

(Refer to the subtopic Introduction_Part I)

2. Surface Area of a Cuboid and Cube

Question4. What are the different units of area?

Question5. Fill in the blanks.

(a) 1 𝑚2 = _________𝑐𝑚2

(b) 1 𝑐𝑚2 = __________𝑚2

(c) 15 𝑚2 = ________________𝑚2

a. Cuboid

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Question6. Dimensions of a cuboid is represented as ____________ × ___________ × ___________

Question7. Number of rectangular faces in a cuboid is _______

Question8.Fill in the blanks:

a. Area of a rectangle = ________×__________

b. Number of rectangular faces in a cuboid is _______

c. So, Sum of areas of all rectangular faces = ________________

d. Sum of areas of all rectangular faces excluding top and bottom = ________________

e. An open cuboid with dimension 𝑙 × 𝑏 × ℎ is given below. All the 6 faces are numbered from

1 to 6. Fill the blank spaces with the symbols.

Area of the face (1) = ________





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Total surface area of the cuboid = Area of [(1) + (2) + (3) + (4) + (5) + (6)]

=__________________

=__________________

(Refer to the subtopic Surface area of cuboid and cube_Surface area of cuboid)

Total Surface Area : The surface area or Total Surface Area of a three-dimensional figure is the sum

of the areas of all its faces.

Lateral Surface Area : The lateral surface of an object is the area of all the sides of object excluding

area of its base and top. For a cuboid, the lateral surface area would be the area of four sides.

Curved Surface Area : Curved Surface Area is the surface area of the curved part of the object. For

objects having curved surfaces, their lateral or curved surface areas are the same.

Also, Total surface area of a cuboid = Lateral surface area + 2 × Area of the base

Question9. Total surface area of a cuboid with dimension 3 𝑐𝑚 × 𝑥 𝑐𝑚 × 4 𝑐𝑚 is 94 𝑐𝑚2. Find the

value of 𝑥.

Hint: Form an equation in 𝑥 and solve.

Did you know?

Length of the diagonal of a cuboid is equal to √𝑙2 + 𝑏2 + ℎ2 [By Pythagoras theorem]

b. Cube

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Length, breadth and height of a cube are same. Therefore, each face will have equal area.

Area of one face of a cube of side length 𝑎 𝑢𝑛𝑖𝑡𝑠 = ____×_____=_____

Total surface area of the cube = _______

Did you know? The length of the longest rod that can be

placed in a rectangular box or room is equal to its diagonal.

Question10. Surface area of a cube is 150 𝑚2. Find the side length of the cube.

Question11. A birthday gift is 55 cm long, 40 cm wide and 5 cm high. You have one sheet of

wrapping paper that is 75 cm by 100 cm. Is the paper large enough to wrap the gift? Explain.

Hint: Find total area of the gift and wrapping paper, then compare the two.

Question12. The floor of a rectangular hall has a perimeter 200 m. If the cost of painting the four

walls at the rate of Rs 10 per m2 is Rs 10000, find the height of the hall.

Hint: Area of the four walls is equal to the lateral surface area.

3. Surface Area of a Right Circular Cylinder

Lateral or curved surface area of the figure = Perimeter of the base circle × Height of the

cylinder

= ___________×_________

= __________

Total surface area of the cylinder = __________+_____________ = __________

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(Refer to the subtopic Surface area of a right circular cylinder_Surface area of a cylinder)

Question13. Mary is wrapping a can of paint as a gag gift for a friend. If the can is 11 cm high and

has a diameter of 7 cm, how many sq.cm of wrapping paper will she use in completely covering the

can?

Question14. A cylindrical tube needs to be painted on the inside as well as on the outside. It is open

on both ends. If the diameter of the tube is 18 cm, and it is 110 cm in length, what is the total surface

area of this tube which will be painted?

Hint: Note that the tube is opened on both ends.

Question15. A metal pipe is 77 cm long. The inner diameter of a cross section is 7 cm, the outer

diameter being 14 cm. Find its total surface area.

Hint: Total surface area = Curved surface area of inner surface + Curved surface area of outer

surface + Area of both circular ends

4. Surface Area of a Right Circular Cone

A cone with height ℎ, slant height 𝑙 and base radius 𝑟 is given below.

Following figure shows how an opened sheet of paper which is used to make a cone looks.

If the sheet is divided into many triangles, whose height is the slant height of the cone ‘𝑙’ and

the bases 𝑏1, 𝑏2, 𝑏3 … ..

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We know that, area of each triangle = 1

2× Base of each triangle × _________

Total area of the sheet, which is the lateral surface area of the cone,

=1

2× 𝑙(𝑏1 + 𝑏2 + ⋯ )

Since base of the cone is a circle with radius 𝑟,

𝑏1 + 𝑏2 + 𝑏3 … . = circumference of the base circle = ___________

Therefore, Lateral surface area = 1

2× ____ × ______ = _____________

Now, if the base of the cone is closed. Then,

Total surface area of the cone = Lateral surface area + Area of the base

= ________+______ = ___________

(Refer to the subtopic Surface area of a right circular cone_Surface area of a cone)

By Pythagoras theorem,

Relation between ℎ, 𝑙 and 𝑟 is 𝑙2 = ____ + _____

Did you know? Traffic cones can be found along highways and sidewalks

throughout the world. The traffic cone's circular base provides stability

to keep the cone upright.

Question16. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is

24m.

Question17. How many meters of cloth of 5m width will be required to make a conical tent, the

radius of whose base is 7 m and height is 24 m? What is the cost of the cloth required to make the tent

if 1 m2 cloth cost Rs. 50.(Take 𝜋 =

22

7)

5. Surface Area of Sphere

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Surface area of a sphere whose radius is 𝑟 = _______

Surface area of the sphere in terms of diameter ‘𝑑’ = _________

(Refer to the subtopic Surface area of a sphere_Surface area of a sphere1,2)

Did you know? The earth is approximately a sphere (actually it is sphere

slightly flattened at the poles).

Question18. Find the radius of a sphere whose surface area is 616 cm2.

Hint: Take 𝜋 =22

7

Did you know? The surface area of the Earth is 510 million square

kilometers or 5.1 × 108 𝑘𝑚2.

Question19. The diameter of earth is 4 times that of moon. What is the ratio of their surface areas?

(Refer to the subtopic Surface area of a sphere_Diameter of the moon)

Hemisphere

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Curved surface area of the hemisphere = 1

2× ____________________

= _____________

Total surface area = ________+_________=________

(Refer to the subtopic Surface area of a sphere_Q-Hemispherical bowl)

Question20. A hemispherical bowl is made of steel, 1 cm thick. The inner radius of the bowl is 4 cm.

Find the outer curved surface area of the bowl.

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IX – Std. Surface Areas and Volumes – Class Notes II

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Instructions: This booklet given to you is your Class Notes. Keep filling this sheet as the class


6. Volume of a Cuboid

Volume is the measure of the amount of space inside of a solid figure, like a cube, ball,

cylinder or pyramid. It's units are always "cubic".

Volume of a cuboid = Area of the base × Height

=________×______×_________

If 𝑙, 𝑏 and ℎ represents length, breadth and height of a cuboid, then its volume,

𝑉 = _______

In case of a cube,

Volume of a cube with side 𝑎, 𝑉 = ____× ______ × ______ = _________

(Refer to the subtopic Volume of a cuboid_Surface area of a cuboid)

Question21. Find the surface area of a cube whose volume is given as

(a) 125 cm2

(b) 216 cm2

Question22. Two cubes each of volume 64 cm3 is joined end-to-end. Find the total surface area and

volume of the resulting cuboid.

Question23. A 2 cm thick iron open box is made whose external dimensions are 32cm × 22cm ×

12cm. Find the weight of the box if its density is 80gm

cm3.

Hint: Volume of the iron used will be the difference between external volume and internal volume.

Relation between density, weight and volume is, 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =𝑤𝑒𝑖𝑔ℎ𝑡

𝑣𝑜𝑙𝑢𝑚𝑒

Question24. How many litres of water flows out of a pipe of cross section area 5 cm2 in 1 minute, if

the speed of the water in the pipe is 30 cm/s?

(Hint: The volume of water that flows in unit time = Area of cross section×speed of flow of water)

7. Volume of a Cylinder

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For a cylinder with radius 𝑟 and height ℎ,

Base area = ________

Volume of the cylinder = Base area × Height = ____________

(Refer to the subtopic Volume of a cylinder_Volume of a cylinder)

Did you know? Engines as all motorcars have, cylinders to compress the fluids for combustion. These

cylinders have a very practical volume and in fact is exactly why we have the "size" of a car

mentioned as a 2L or a 1.4L vehicle. That volume of the cylinders combined is directly related to

the power and how expensive the car would be to drive.

Question25. A cylindrical swimming pool is 4.5 m high and has a diameter of 24 m. If the pool is

filled right up to the top edge how many cubic metre of water can the pool hold?

Hint: Volume of the water filled in will be equal to the volume of the cylindrical pool.

Question26. A 20 m deep well with diameter 14 m is dug up and the earth from digging is evenly

spread to form a platform 22m × 14m. Find the height of the platform.

Hint: Volume of the well will be equal to volume of the platform.

Question27. A small can of soup has a radius of 3.5 cm and a height of 12 cm. A family size can has

a radius of 5 cm and is 15 cm high. Which contains more soup, one family sized can, or two small

cans?

a. Hollow Cylinder

R, 𝑟 and h are the external radius, internal radius and height of the cylinder respectively.

Thickness of wall in terms of R and 𝑟 is ______

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Area of cross section is _____________

Total surface area = External curved surface area + Internal curved surface area + 2 (Cross

section area)

Total surface area = ______+_____+_______

Volume of the material = External volume – Internal volume

Volume of the material = ________ - _________= _______

Question28. A hollow copper pipe of inner radius 3 cm and outer radius 4 cm is melted and changed

into a solid right circular cylinder of the same length as that of the pipe. Find the area of the cross

section of the solid cylinder.

(Hint: The volume of hollow pipe = volume of solid cylinder)

8. Volume of a Cone

Misconception: Volume of a shape is not always (𝑎𝑟𝑒𝑎 𝑜𝑓𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡).

For example: In case of volume of a cone; Volume of cone is not equal to (𝜋𝑟2 × ℎ).

Clarification: In case of cylinder, the circumference is constant along the height. But in case of cone,

the radius of the circle varies from r to 0.)

If a cone and a cylinder have same base radius and height, then;

Volume of cone = ___ × (Volume of cylinder)

Volume of a cone with radius 𝑟 and height ℎ is ___________

(Refer to the subtopic Volume of a cone_Volume of a sphere)

Question29. What is the volume of a cone of radius 7 cm and height is 10 cm?

Question30. A tent is cylindrical in shape with a conical top above it. The radius of the base of the

tent is 7 m. The height of the cylindrical part is 20 m and the height of the conical part is 4 m. Find the

volume of the air in the tent.

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Question31. A right triangle ABC with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm.

Find the volume of the solid so obtained.

Hint: Find the radius and height of the solid obtained.

9. Volume of a Sphere

Volume of a sphere with radius 𝑟 = ________

Since a hemisphere is half of a sphere,

Volume of a hemisphere with radius 𝑟 = __________

(Refer to the subtopic Volume of a sphere_Volume of a sphere)

Did you know? Volume of earth is 1.08321 × 1021 𝑘𝑚3

Question32. Find the volume of a sphere whose radius is 14 cm.

Question33. The surface area of a sphere and the curved area of a cylinder are in the ratio 2:1. Find

the ratio of their volumes, if their radii are equal.

Question34. Find the volume of the greatest sphere that can be cut of a cube of side 14 cm. Find also

the volume of the remaining piece.

Hint: The diameter of the greatest sphere will be equal to the side length of the cube.

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1. The curved surface area of a right circular cylinder of height 14 cm is 88cm2. Then the

diameter of the base of the cylinder is

A) 1 cm

B) 2 cm

C) 3 cm

D) 4 cm

2. The diameter of the moon is approximately one fourth of the diameter of the earth. Ratio of

their surface areas is: (considering both of them to be perfectly spherical)

A) 1

2

B) 1

4

C) 1

8

D) 1

16

3. Height and radius of a cylinder is doubled, then its lateral surface area increases by ____

times.

A) 2

B) 4

C) 8

D) 6

4. Total surface area of cube is 150 cm2. Volume of the cube is

A) 64 cm3

B) 216 cm3

C) 100 cm3

D) 125 cm3

5. The radii of two cylinders are in the ratio of 2:3 and their heights are in the ratio of 5:3. The

ratio of their volume is:

A) 10:17

B) 20:27

C) 17:27

D) 20:37

6. The number of planks of dimensions (4m × 50cm × 20cm) that can be stored in a pit which

is 16m long, 12m wide and 4m deep is

A) 1900

B) 1920

C) 1800

D) 1840

7. The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped

into it. The ratios of the surface areas of the balloon in the two cases is

A) 1:4

B) 1:3

C) 2:3

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D) 2:1

Questions from 8 to 22 are subjective questions.

8. The surface area of the six faces of a rectangular solid are 4, 4, 8, 8, 18 and 18 sq. cm. Find

the volume of the solid.

9. External dimensions of a wooden cuboid are 30 cm × 25 cm × 20 cm. If the thickness of the

wood is 2 cm all around, find the volume of the wood contained in the cuboid formed.

10. A large box is a cube with sides of length 80 cm. There are smaller cubical boxes of sides 20

cm. What is the volume of the larger box and the smaller box? How many small boxes will fit

in the large box?

11. The 'staircase' below is built from wooden cubes with sides of length 6 cm. What is the

volume of the staircase? A similar staircase is 4 blocks high instead of 3.What is the volume

of this staircase?

12. There is a 40 m long tunnel. The cross-section of the tunnel is a semicircle with a radius 6.

Find the cost of painting the tunnel at the rate of Rs.2/m2.

13. A metal pipe is 77 cm long. The inner diameter of a cross section is 4cm, the outer diameter

being 4.4 cm. Find its

(i) inner curved surface area.

(ii) Outer curved surface area.

(iii) Total surface area.

14. Find the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2m

in diameter and 4.5 m high.

15. The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28cm, find

(i) height of the cone.

(ii) Slant height of the cone.

(iii) Curved surface area of the cone.

16. A cone of radius 4 cm is divided into two parts by drawing a plane through the mid-point of

its axis and parallel to its base. Compare the volumes of the two parts.

17. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter

of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface

area of the vessel.

18. A right circular cylinder circumscribes a sphere completely. On the two flat surfaces of

cylinder, right circular cones are placed. Volume of a right circular cone is equal to the

volume of the right circular cylinder. Find volume of sphere in terms of the radius and/or the

height of the cone, if height of the cone is h.

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19. The diameter of a metallic ball is 4.2cm. What is the mass of the ball, if the density of the

metal is 8.9 g per cm3?

20. A heap of the wheat is in the form of a cone whose diameter is 10.5m and height is 3m. Find

its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the

canvas required.

21. A river 3m deep and 40m wide is flowing at the rate of 2km per hour. How much water will

fall into the sea in a minute?

22. A plastic box 1.5 m long wide and 65 cm deep is to be made. It is opened at the top. Ignoring

the thickness of the plastics sheet, determine:

(i) The area of the sheet required for making the box.

(ii) The cost of the plastic box, if the cost of the sheet is Rs.100/m2.

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IX – Std. Introduction to Trigonometry – Classnotes I

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1. Introduction

a. Have you seen a civil engineer measuring the elevation of a building or height of a tower

using some device?

b. How can you find the depth of a sea without exploring it?

c. Can you calculate the height of the building, being at the roof?

These are some of the applications of trigonometry in our daily life.Just imagine how hard it

would be to measure the depth of a ocean if trigonometry would not be there.

Trigonometry is composed of two words: ‘Trigon’ which means ‘triangle’ and ‘meteron’

which means ‘to measure’. Combined it means measurements of sides or angles of a triangle

and that is what trigonometry is all about.

Did you know? Early study of triangle can be

traced to the 2nd

millennium BC

(Refer to the subtopic Introduction _ Introduction(clip1))

2. Trigonometry Ratios

Question 1: Consider a right angled ∆𝐴𝐵𝐶 right angled at B. Write all the possible ratios of the sides

of the given right angled triangle.

Figure 22: A right angled triangle

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Now that you have all the possible ratios, these ratios are given specific names with respect to

a reference angle, which is ∠C here. w.r.t. ∠C, let’s call side 𝑝 as opposite (𝑜) and side 𝑏 as

adjacent (𝑎).

𝒐

𝒉= 𝒔𝒊𝒏𝒆 𝑪

𝒂

𝒉= ____________

𝒐

𝒂= ____________

𝒉

𝒐= ____________

𝒉

𝒂= ____________

= 𝒄𝒐𝒕𝒂𝒏𝒈𝒆𝒏𝒕 𝑪

For the convenience these are referred as sin, cos, tan, cosec, sec and cot.

These are the ratios of a right angled triangle which relate the angles of triangle to the lengths

of its sides.

(Refer the subtopic Trigonometric Ratios_Possible Trigonometric Ratios,Naming the Ratios)

Helping Hand :The three basic ratios can be easily remembered using a simple trick. We

know that in right angled triangle 3 sides are known as opposite(O), Adjacent(A) and

Hypotenuse(H)

Figure 23: Short-cut

So, ratios can be remembered OH-AH-OA which are sin, cos and tan respectively.

Another method to remember facts and relationship in trigonometry is:

Sine = Opposite /Hypotenuse

Cosine = Adjacent /Hypotenuse

Tangent = Opposite /Adjacent

You can remember is as: “Some Old Houses Can Always Hide Their Old Age”

Misconception: Sin A is the product of ‘Sin’ and ‘A’.

Clarification: The symbol Sin A is used as an abbreviation for the ‘Sin of angle A’.

Question2. If 𝜃 =12

5 , Find the values of other trigonometric ratios.

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Fun Activity: Let’s check your understanding of Trigonometric ratios with this activity.

Figure 24: Activity

a. Reciprocal Relations:

Write the ratios of sin 𝐶 and 𝑐𝑜𝑠𝑒𝑐 𝐶 from above ratios you have written and try to relate the

ratios.

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sin 𝐶 =𝑜

ℎ=

1

ℎ𝑜

= _____________

Can we conclude that sin 𝐶 is ___________ of 𝑐𝑜𝑠𝑒𝑐 𝐶.

Question 3: Write the remaining reciprocal relations.

b. Quotient Relations:

Divide the ratios sin 𝐶and cos 𝐶,

𝑠𝑖𝑛 𝐶

𝑐𝑜𝑠 𝐶 =

𝑜

ℎ𝑎

ℎ

= _____ = ______

Question 4: Do you find any other similar relation?

Note: The values of the trigonometric ratios of an angle do not vary with the lengths of the

sides of the triangle, if the angle remains the same.

Let us understand this through the given figure:

Figure 25: ∆PQR

From the figure, In ∆𝐴𝐵𝑅

sin 𝜃 =

and In ∆𝑃𝑄𝑅,

sin 𝜃 = =3

5

Question 5: Find all trigonometric ratios of ∠A and ∠C, based on the given figure.

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Figure 26: ∆ABC

Did you know?

The modern sine convention is first mentioned

in the Surya Siddhanta, and its properties were

further documented by Indian astronomer

and mathematician Aryabhata.

Question 6: If 𝑥 is an acute angle of a right angled triangle and 𝑠𝑖𝑛 𝑥 = 3

7 , find the exact value of

trigonometric functions of 𝑐𝑜𝑠 𝑥 and 𝑐𝑜𝑡 𝑥.

Question7. If 𝑠𝑒𝑐 Ө = 5

4 find the value of

𝑠𝑖𝑛Ө−2 tan Ө

2𝑐𝑜𝑠𝑒𝑐 Ө.

Question8: If tan Ө = 12

13, find the value of

𝑠𝑖𝑛2Ө−2 cotӨ

2𝑠𝑖𝑛ӨcosӨ.

3. Trigonometric ratios of some specific angles.

We can find the ratios of some standard angles like 30°, 45°, 60°, 90° and 0°.

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You can take help from the given two triangles, for finding the ratios of specific angles, 45°,

30°and 60°.

Figure 27: Trigonometric ratios of specific angles

(Refer to the subtopic Trigonometric Ratios of some specific angles _ Special angles-

Tabulated)

For the angles 0° and 90°, you can take the help of unit circle.

Figure 28: Unit circle

𝜽

𝟎°

𝟑𝟎°

𝟒𝟓°

𝟔𝟎°

𝟗𝟎°

sin 𝜽

cos𝜽

tan 𝜽

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cosec 𝜽

sec 𝜽

cot 𝜽

Helping Hand: Following table can be used as a cheat sheet to remember the above values:

θ 0° 30° 45° 60° 90°

sin θ

√0

4= 0 √

1

4=

1

2 √

2

4=

1

√2 √

3

4=

√3

2 √

4

4= 1

The values of cos 𝜃 can be obtained by writing the 𝑠𝑖𝑛 𝜃 values in reverse order.

The values of tan 𝜃 can be obtained by dividing 𝑠𝑖𝑛 𝜃 by cos 𝜃.

Remaining rows can be obtained by using the reciprocal relations.

(Refer to the subtopic Trigonometric Ratios of some specific angles_ Special angles-

Tabulated)

Did you know?

The trigonometry emerged in the Hellenistic world

during the 3rd century BC from applications of

geometry to astronomical studies.

Question 9: It is given that, 𝑡𝑎𝑛2𝜃 = 2𝑡𝑎𝑛𝜃

1−𝑡𝑎𝑛2𝜃. Can you find the value of tan 15°?

Question10. Find the value of: 4

3tan2 30° + 𝑠𝑖𝑛260° − 3 cos2 60° +

3

4tan2 60° − 2 tan2 45°.

Question11. Evaluate 5𝑠𝑖𝑛230°+𝑐𝑜𝑠245°−4𝑡𝑎𝑛230°

2𝑠𝑖𝑛30°𝑐𝑜𝑠30°+𝑡𝑎𝑛45°

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Question12. Find the value of the : (i) 𝑡𝑎𝑛𝑥 = sin 45°𝑐𝑜𝑠45° + 𝑠𝑖𝑛30°

(ii) cos 2𝑥 = 𝑐𝑜𝑠60°𝑐𝑜𝑠30° + 𝑠𝑖𝑛60°𝑠𝑖𝑛30°

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4. Trigonometric ratios of complementary angles

Question13. What are complementary angles?

In a right angled triangle, one angle is 90° and the remaining angles are complementary to

each other. Let us find out the relation of trigonometric ratios with complementary angles.

Figure 29: Trigonometric ratios of complementary angles

Consider ∠C = θ, then ∠A = ________.

𝑠𝑖𝑛 𝐴 = _____and 𝑐𝑜𝑠 𝐶 = ______

Question14: What can you conclude from the above values? Write the remaining similar relations.

(Refer the subtopic Trigonometric Ratios of complementary angles _complementary angles)

Question15: Now, can you figure out why we wrote trigonometric ratios of cosine in the reverse order

of trigonometric ratios of sine in the 𝑐ℎ𝑒𝑎𝑡 𝑠ℎ𝑒𝑒𝑡?

Did you know?

The 3rd

century astronomers first noted that, if at least

the length of one side and the value of one angle of a right triangle

is known, then all other angles and lengths can be determined.

Question16. Find the value of the given expression: 𝑠𝑖𝑛𝜃 × 𝑐𝑜𝑠(90° − 𝜃) + 𝑐𝑜𝑠𝜃 × 𝑠𝑖𝑛(90° − 𝜃).

Question17. In figure 6, Can you find the value of the given expression below?

𝑐𝑜𝑠(90° − 𝐴)

𝑠𝑖𝑛𝐴+

𝑠𝑖𝑛 𝐴

𝑐𝑜𝑠(90° − 𝐴) (𝐴 ≠ 0)

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Question18. Prove that, 𝑡𝑎𝑛1°𝑡𝑎𝑛2°𝑡𝑎𝑛3° … … … 𝑡𝑎𝑛89° = 1

Question19. If 𝑠𝑖𝑛5𝜃 = 𝑐𝑜𝑠4𝜃, where 5𝜃 and 4𝜃 are acute angles, find the value of 𝜃.

5. Trigonometric Identities :

You know that an equation is called an identity when it is true for all values of variables

involved in it. Similarly, in trigonometry we have few identities that are true for all values of

angles. I hope you remember Pythagoras theorem,

Figure 30: Trigonometric identities

In ΔABC,

𝑜2 + 𝑎2 = _____….(1)

Question20. Write the result we get on dviding equation (1) by ℎ2(in terms of sine, cosine).

Question21. Write the results we get on dividing equation (1) by 𝑎2and 𝑜2, respectively.

Note: For the above identities, remember that the value of 𝜃 lies between 0o and 90

o (including both).

Question22. Solve for 𝑥, if 𝑥 is a positive acute angle. 𝑠𝑒𝑐 𝑥 × 𝑐𝑜𝑡 𝑥 =2

√3

Question23. Prove the following trigonometric identities: 𝑠𝑒𝑐 𝜃 + 𝑡𝑎𝑛 𝜃 = 1

𝑠𝑒𝑐 𝜃−𝑡𝑎𝑛 𝜃

Question24. Prove the following trigonometric identities: (1 + tan2 𝜃)(1 + 𝑠𝑖𝑛𝜃)(1 – 𝑠𝑖𝑛𝜃) = 1

Applications of Trigonometry in our day-to-day life:

a. Much of architecture and engineering relies on triangular supports. When you drive

across a suspension bridge, you are benefiting from trigonometry, which an engineer used

to calculate the correct length of support cables, the height of support towers, and the

angle between the two.

b. Trigonometry plays a major role in musical theory and production. Sound waves travel in

a repeating wave pattern, which can be represented graphically by sine and cosine

functions.

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c. Electrical engineers use trigonometry to model this flow and change of direction, with the

sine function used to model voltage.

d. Engineers rely on trigonometric relationships to determine the sizes and angles of

mechanical parts used in machinery, tools and equipment.

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1. The value of cot 75° is equal to

A. cot 15°

B. tan 5°

C. tan85°

D. None of these

2. The value of cos 37°

sin 53° is

A. 2

B. 1

C. -1

D. 0

3. The value of cos1°cos 2°cos 3°……..cos 180° is

A. 1

B. 0

C. -1

D. None of these

4. The value of cot12°cot38°cot52°cot 60°cot 78°

A. 1

√3

B. 1

2

C. √3

D. 3

2

5. Which expression is not equivalent to sin 150°

A. sin 30°

B. –sin 210°

C. cos 60°

D. –cos 60


6. If 𝑐𝑜𝑠9𝜃 = 𝑠𝑖𝑛 𝜃 and 9𝜃 < 90°, find the value of 𝑐𝑜𝑡5𝜃.

7. If 4 𝑡𝑎𝑛Ө = 3, then find the value of (4 sin θ– cosӨ)

(4 sin θ+cosӨ)

8. If tan 𝐶 =12

5, find the value of

1 + 𝑠𝑖𝑛 𝐶

1− 𝑠𝑖𝑛 𝐶

9. If 𝑥𝑐𝑜𝑠𝜃 – 𝑦𝑠𝑖𝑛𝜃 = 𝑎, 𝑥𝑠𝑖𝑛𝜃 + 𝑦𝑐𝑜𝑠 𝜃 = 𝑏, prove that 𝑥2 + y2 = a2 + b2

10. If 𝑡𝑎𝑛𝐴 + 𝑠𝑖𝑛𝐴 = 𝑚 and 𝑡𝑎𝑛𝐴 – 𝑠𝑖𝑛𝐴 = 𝑛, show that 𝑚2 – 𝑛2

= 4√𝑚𝑛

11. If cosA

cosB = 𝑚 and

cosA

sinB = 𝑛 , show that (𝑚2

+ 𝑛2)𝑐𝑜𝑠2𝐵 = 𝑛2

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12. If 𝑐𝑜𝑠 𝑥 = 𝑐𝑜𝑠60° 𝑐𝑜𝑠30° + 𝑠𝑖𝑛60° 𝑠𝑖𝑛30°, then find the value of 𝑥.

13. If 𝑠𝑖𝑛 𝜃 + 𝑐𝑜𝑠 𝜃 = 𝑝 and 𝑠𝑒𝑐 𝜃 + 𝑐𝑜𝑠𝑒𝑐 𝜃 = 𝑞, show that 𝑞(𝑝2 − 1 ) = 2𝑝

14. If 𝑥 = 𝑟 𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐶 , 𝑦 = 𝑟𝑠𝑖𝑛𝐴𝑠𝑖𝑛𝐶, 𝑧 = 𝑟𝑐𝑜𝑠𝐴, Prove that 𝑟2

= 𝑥2

+ 𝑦2

+ 𝑧2

.

15. Prove that (𝑠𝑖𝑛𝜃 – 𝑐𝑜𝑠𝑒𝑐𝜃) (𝑐𝑜𝑠𝜃 – 𝑠𝑒𝑐𝜃) = 1

𝑡𝑎𝑛 Ө+𝑐𝑜𝑡 Ө

16. Prove that tan2 𝐴 – tan2 𝐵 = 𝑐𝑜𝑠2𝐵− 𝑐𝑜𝑠2𝐴

𝑐𝑜𝑠2𝐵𝑐𝑜𝑠2𝐴

17. Evaluate the following

: 𝑠𝑖𝑛225° + 𝑠𝑖𝑛265° + √3 (𝑡𝑎𝑛5° 𝑡𝑎𝑛15° 𝑡𝑎𝑛30° 𝑡𝑎𝑛75°𝑡𝑎𝑛85°)

18. Prove that 𝑐𝑜𝑠𝑒𝑐2𝜃 + sec2 𝜃 = 𝑐𝑜𝑠𝑒𝑐2𝜃 sec2 𝜃

19. If t𝑎𝑛 Ө + 𝑠𝑖𝑛 Ө = 𝑚 and 𝑡𝑎𝑛 Ө – 𝑠𝑖𝑛 Ө = 𝑛, show that 𝑚2 – 𝑛2 = 4√𝑚𝑛

20. If 𝑠𝑒𝑐 Ө + 𝑡𝑎𝑛 Ө = 𝑝, show that 𝑝2− 1

𝑝2+ 1 = sin Ө

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1. Cartesian System

a. Introduction

Question1. Suppose there is a dot on a piece of paper whose dimensions are given below in the

following figure. Write the best statement which describes the exact position of the dot.

Figure 1: Piece of Paper

Question2. Let’s perform a simple activity: Can you describe your position in the classroom with

respect to rows and columns (of benches)

(Hint: Count the total number of rows and columns of benches)

Question3. A flocking bird is trapped in the net which consist of square boxes of dimension 1 cm.

Can you represent the position of the bird from the given figure?

Figure 2: Bird trapped in the net

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Now, from the above examples you can observe that the position of any object whether it is a

bird or ball can be represented with the help of two perpendicular lines. The position of any

object can be represented by its position, with respect to horizontal and vertical line.

Co-ordinate geometry is a merger of geometry and algebra. It was the French mathematician

Rene Descartes who lay the idea of a co-ordinate system. In this system every point(a

geometric concept) is assigned a pair of numbers (algebra concept) as its unique “address”

We all have studied number line. On the number line, distance from a fixed point are marked

equally in positive direction and as well as in negative direction. The point from which the

distance is marked is called origin.

Question4. Represent -4, 3, -4.5 on the number line.

Descartes’ idea is based on two real lines intersecting at right angles. In Cartesian system, two

perpendicular lines are placed on the plane such that the location of any point is referred to

these lines.

Did you know?

Eventhough Rene Descartes is given the sole credit for the invention

of co-ordinate geometry, another mathematician Pierre de Fermat also developed

the similar system independently.The key difference between them is , Fermat

always

started with an algebraic equation and then described the geometric curve

which

satisfied it. Descartes started with geometric curves and produced

their equations.

Question5. State whether the following statement is true or not: The perpendicular lines can be in any

direction and we can choose these two lines to locate a point.

Figure 3: Perpendicular lines

Question6. In the given figure two lines AC and DB are intersecting at right angle.Using the figure,

try to answer the following questions:

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Figure 4: Two perpendicular lines

(i) DB is known as _________ axis.

(ii) AC is known as _________ axis.

(iii) O is called____________.

(iv) OB is called ___________ ____axis.

(v) AO is called ___________ ____ axis.

(vi) OD is called ___________ ____ axis.

(vii) OC is called ___________ ____ axis.

Note: The system consisting of the x-axis, the y-axis and the origin is also called Cartesian co-

ordinate syatem. The x-axis and the y-axis together are called co-ordinate axes.

(Refer to the sub topic Cartesian System_Coordinate Geometry )

b. Co-ordinates of points

In a co-ordinate plane the position of each point is represented by an ordered pair with

respect to the co-ordinate axes.

(i) From the origin ‘O’, measure of the distance of the point along 𝑥-axis is called

𝑥-coordinate or abscissa of the point.

(ii) From the origin ‘O’, measure of the distance of the point along y-axis is called

𝑦-coordinate or ordinate of the point.

c. Quadrants and sign convention

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Question7. From the above figure4, you can see that the axes divide the plane into four parts. These

parts are called quadrants . Can you name the quadrant with their specific sign?

Figure 5: Two perpendicular lines

Question8. Define the following terms:

(i) 𝑋 – 𝑎𝑥𝑖𝑠

(ii) 𝑌 – 𝑎𝑥𝑖𝑠

(iii) Origin

Note: The coordinates of origin are (0, 0 )

(Refer to the sub topic Cartesian System_Coordinate Geometry )

Did you know ?

Using the Cartesian coordinate system, geometric shapes

(such as curves) can be described by Cartesian

equations

Question9. From the given points, name the abscissa and ordinates.

(2,5), (-7,-6), (4,-5), (-7,9), (0,0) and (3,-6)

Question10. Using the given figure below, answer the following questions:

(i) Co-ordinate of A

(ii) Co – ordinate of B

(iii) Co-ordinate of C

(iv) The abscissa of point C.

(v) The ordinate of point B

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Figure 6: Graph

2. Plotting a Point on the Graph

If the coordinates of a point (𝑥, 𝑦) are known, we can place the point in a co-ordinate

plane(graph). This process is called as plotting the point.

Note: The coordinate of a point (x, y) tell us that the distance of this point from y – axis along the

positive x – axis is x units and the distance of this point from x – axis along the positive y – axis is y

units.

Question11. Locate the points (2, 5), (-2, -4) and (-4, 5) in the Cartesian plane.

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Question12. In which quadrant do each of these points (-2, 8), (-3, 4) and (-6, -3) lie? Verify your

answer by locating them on Cartesian plane.

(Refer to the subtopic Plotting on a Graph_Coordinate Geometry)

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Question13. Plot the point P(7, -3). From the point P, draw PQ perpendicular to 𝑥-axis and PR

perpendicular to 𝑦 –axis. Write the co-ordinates of point Q and point R.

a. Graph of a linear Equation

Question14. The graph of a linear equation is always ________________.

Question15. The equation of the 𝑥-axis is ___________________.

Question16. The equation of the 𝑦-axis is ___________________.

Note: If the linear equation is in one variable, then it is either parallel to x – axis or y – axis.

If the linear equation is in two variables, it is a straight line which is neither parallel to 𝑥 – axis

nor

𝑦 – axis.

𝑥 = ±𝑎 is the equation of a line parallel to the 𝑦 – axis and at distance of ‘𝑎’ units from it.

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𝑦 = ±𝑎 is the equation of a line parallel to 𝑥 – axis and at a distance of ‘𝑎’ units from it.

Question17. Find the equations of the lines given in the following figure.

Figure7 Figure 8

Solution: In the figure 7,

AB is parallel to 𝑦-axis and it is at a distance of ‘-3’ units from the 𝑦-axis

∴ Equation of AB is 𝑥 = −3 i.e. 𝑥 + 3 = 0

To draw the graph of a linear equation:

(1) Plot a few points, which satisfy the given equation.

(2) Draw a straight line passing through these points.

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(i) To draw the graph when the linear equation of the form y = mx

Question18. Draw the graph for 𝑦 = −5𝑥.

Note :The graph of the linear equation of the form y = mx always passes through the origin.

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Question19. Draw the graph for 4𝑥 − 𝑦 = 0

(ii) To draw the graph when the linear equation of the form y = mx+c; where c is

rational but not zero

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b. Inclination and Slope

(i) Inclination:

The angle of inclination of a straight line is the angle the line makes with the x-axis, with

the angle being measured anti-clockwise.

In the following figures, 𝛼 and 𝛽 are the angles of inclination.

Note: For x –axis and every line parallel to x – axis, the inclination α = 0°.

For y – axis and every line parallel to y – axis, the inclination α = 90°

(ii) Slope

If α is the inclination of a line, the slope of line is tan α and is usually denoted by

letter m.

Slope = 𝑚 = 𝑡𝑎𝑛 𝛼

Example:

If the inclination of a line is 60°, then α = 60°

The slope of the line = m = tan 60° = √3

Note : The slope is also called as gradient.

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Question21. We know that, For 𝑥 –axis and every line parallel to 𝑥 – axis, the inclination α = 0°.

Then what is the slope of 𝑥 – axis?

Question22. We know that, For y – axis and every line parallel to y – axis, the inclination α = 90°.

Then what is the slope of 𝑦-axis?

c. Y – Intercept

If a straight line meets 𝑦 – axis at a point, the distance of this point from the origin is

called

𝑦 – intercept and is usually denoted by c.

In the figures given above, 𝑦-intercepts are OP and OR

For 𝑥 – axis, 𝑦 intercept = 0

For every line parallel to 𝑦 – axis, 𝑦 – intercept = 0

𝑦 – Intercept is:

(1) Positive, if measured above the origin.

(2) Negative, if measured below the origin.

d. Finding Slope and the Y – intercept of a given line

1. Consider the line of the form 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0

2. Change the subject of the equation in terms of 𝑦. i.e. 𝑦 =−𝑎

𝑏𝑥 −

𝑐

𝑏

3. Make y, the subject of the equation.

4. The coefficient of x is the slope and the common term is the y – intercept:

∴Slope (m) = −𝑎

𝑏 and y- intercept (c) =

−𝑐

𝑏

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Question23. Find the slope and 𝑦 – intercept of the line 3𝑥 + 4𝑦 = 7

Question24. Find the slope and y – intercept of (−2𝑥) = 4𝑦 – 2

Question25. Find the equation of a line whose :

(i) Slope = 5 and 𝑦 – intercept = -8

(ii) m = 8 and c = -6

Question27. Find the inclination of the line whose slope is:

(i) 0

(ii) √3

(iii) 1

√3

Question26. Draw the line 2𝑥 – 3𝑦 – 18 = 0 on a graph paper.From the graph paper, read the y –

intercept of the line.

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e. Solution of Simultaneous Linear Equations in Two Variables Graphically

An equation of the form 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0 is called a linaer equation in two variables with

𝑥 and 𝑦 as the variables.

Follow the following steps to solve simultaneous linear equations graphically

1. Draw the graph of each given equation on the same graph paper.

2. Find the co-ordinates of the point of intersection of the two lines.

3. The co-ordinates give the solution of the given equations.

Question28. Solve graphically 𝑥 − 2𝑦 − 4 = 0 ; 2𝑥 + 𝑦 = 3

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Question29. The sides of a triangle are given by the equations 2𝑦 − 𝑥 = 8 ; 5𝑦 − 𝑥 = 14 and

𝑦 − 2𝑥 = 1

respectively. Find graphically, the area of triangle and the co-ordinates of the vertices of the triangle.

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1. The measure of an angle between the coordinate axes is

A. 0°

B. 180°

C. 90°

D. 45°

2. The abscissa of any point on y –axis is

A. 0

B. 1

C. -1

D. None of these

3. The perpendicular distance of a point P(5, 3) from x – axis is

A. 5

B. 3

C. 4

D. 2

4. The distance of the point P(2, 3) from the origin is

A. 2

B. 3

C. √13

D. None of these

5. The quadrant in which point P(4, - 3) will lie is

A. 1st quadrant

B. 2nd

quadrant

C. 3rd

quadrant

D. 4th quadrant


6. Find the quadrant in which the following points lie?

(i) (2, -3)

(ii) (-2, -7)

7. Plot the following points on a graph sheet.

(i) (5, -6)

(ii) (0, -7)

8. Find the distance of point P (4, -5) from the origin.

9. Find the perpendicular distance of the point P (-4, -5) from y – axis.

10. Find the perpendicular distance of the point P (2, -7) from x-axis.

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11. Find the area of triangle formed by the points A (2, 0), B (6, 0) and C (4, 6).

12. Plot the following points on the graph sheet and name the geometrical figure obtained.

P (0, 1), Q (0, 5) and R (3, 4)

13. Find the quadrant in which the point P lie whose abscissa and ordinate are 3 and -5.

14. Write the sign convention for all four quadrants.

15. Find the area of triangle formed by the points A (0, 0), B (4, 0) and C (2, 2).

16. Find the equation of a line whose slope is -3 and y – intercept is 4.

17. Draw the graph of 2x – 4 = y

18. Draw the graph of equation: 𝑥

4 +

𝑦

5 = 1

19. Draw the graph of equation 3x – 4y = 12 and use the graph drawn to find

(i) Y1, the value of y, when x = 4

20. Find the slope and y – intercept of the line:

(i) 2x – 3y + 7 = 0

(ii) 2y + 5x – 9 = 0

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1. Introduction

You have seen weather forecasting news in television news channels. They predict weather

for next days using the past details as shown in the figure.

They make comments like

“It will probably rain today”

“There is a chance of heavy rain for next three days” etc.

(Refer to the subtopic Introduction_Introduction 1)

Question1. Write down some statements related to day to day life where we use the words probably,

likely, chance etc.

In our day to day life, we generally come across various statements like this,

Indirectly, we are always associated with probability.


Did you know? Probability theory was invented by gamers. People wanted to understand the odds in games of chance. Pioneers included Gerolamo Cardano in the sixteenth century, and Pierre de Fermat and Blaise Pascal in the seventeenth century. This whole branch of

mathematics grew up around gaming.

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2. Experimental Approach of Probability

a. Definitions of Various Terms

(i) Experiment

Question2. Define Experiment. Give examples.

There are two types of experiments.

For example;

In tossing of a coin one is not sure if a head (H) or a tail (T) will be obtained.

Therefore, it is a Random experiment.

If we mark Tail(T) on both sides of a coin and it is tossed, then we always get the

same outcome assuming that it does not stand vertically. Therefore, it is a Deterministic

experiment.

(Refer to the subtopic Introduction_Introduction-II)

Question3. From the examples given above, write the difference between ‘Random experiment’ and

‘Deterministic experiment’. Give more examples.

Random experiment Deterministic experiment

(ii) Trial

When we perform an experiment, it is called a trial of the experiment.

Question4. Write down trials of the following experiments.

(a) A die is thrown 6 times.

(b) A coin is tossed 1000 times.

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(c) Tossing of three coins simultaneously 50 times.

Note: If an experiment is repeated ‘n’ times, then we say that ‘n’ trials of the experiment have

taken place.


(iii) Elementary event

An outcome of a trial is an elementary event.

In tossing of a coin, the possible outcomes are _______ and ________.

If 𝐸1 and 𝐸2 are the two elementary events associated with the above experiment, then

𝐸1 =________________________________

𝐸2 =______________________________

Question5.A die is rolled. Define the six elementary events associated with this experiment.

𝐸1 =

𝐸2 =

𝐸3 =

𝐸4 =

𝐸5 =

𝐸6 =

Hint: Think about the outcomes of the experiment.


Did you know? The probability of living 110 years or more is about 1 in 7 million

Question6. A box contains 2 white balls, 1 black ball, 2 red balls and a green ball. 2 balls are taken

randomly from the box. What are the events associated with above experiment?

(iv) Compound event

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A collection of two or more possible outcomes of a trial of a random experiment is called

a compound event.

Consider the experiment of tossing 2 coins simultaneously. If we define the event

“Getting exactly one tail”, then it is a collection of elementary events (outcomes) ‘TH’

and ‘HT’. Therefore it is a compound event.

Similarly, 3 coins are tosses simultaneously. If we define the event “Getting atleast two

heads”, then the elementary events (outcomes) in the compound event are found as

follows.

Elementary events associated to the experiment are,

(𝐻𝐻𝐻, 𝐻𝐻𝑇, _____, 𝑇𝐻𝐻, 𝑇𝑇𝐻, ______, 𝐻𝑇𝑇, ________)

Let 𝐸 be the event “Getting atleast two heads”. Then,

𝐸 = __________________________________________

Question7. In a throw of a cubical die, prove that the following events are compound events.

(a) “Getting an odd number”

(b) “Getting an even number”

b. Emerical Probability

If we try to define probability mathematically, we can say that probability is used to

quantify the chances of occurrence of events.

In other words,

P(E) =Number of trials in which the event happpened

Total number of trials

(Refer to the subtopic Experimental approach to probability_ Experimental approach to

probability)

Did you know?

Athletes and coaches use probability to determine the best sports strategies for

games and competitions. A baseball coach evaluates a player's batting average

when placing him in the lineup.

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Question8. Find the chance of occurrence of prime numbers when you throw a die.

Hint: Probability of occurrence of each event is equally likely.


Did you know? Probability plays an important role in analyzing insurance policies to determine which plans are best for you or your family and what deductible amounts you need.

Let us take another example of die. We know that, when we throw a die, there are equal

chances that any of the six numbers can turn up. So, if you are asked, what are the chances of

occurrence of number ‘2’? By simple logic you can say that occurrence of ‘2’ is among one of

the six events. So, the probability of occurrence of ‘2’ will be ____

Similarly, the probability of occurrence of each of the faces will be ___ only.

Since, there are altogether 6 faces,

The probability of getting each face = 1

6.

Sum of all probabilities of getting each face = 1

6+

1

6+

1

6+

1

6+

1

6+

1

6= ____ = _____

Generalising, we can say that for any experiment, the sum of the probabilities of all the

elementary events will be 1.

0 ≤ 𝑃(𝐸) ≤ 1

For example, probability of occurrence of number ‘7’ on throw of a dice is 0. Also, the

probability of occurrence of a natural number on throw of a dice is ‘1’.


probability)

Question9. A die is thrown 200 times with the frequency for the outcomes 1, 2, 3, 4, 5, 6 as given in

the following table:

Outcome 1 2 3 4 5 6

Frequency 20 45 66 77 89 100

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Find the probability of getting each outcome.


probability)

Note: The probability of an event lies between 0 and 1 ( 0 and 1 inclusive )

Question10. Two dice are rolled. Then, what is the probability that the total score is a prime number?

Did you know? According to probability, The probability of you being born was about 1 in 400

trillion.

Question11. A jar contains 10 red balls, 4 blue balls and 4 green balls. What is the probability of

getting green balls?

Question12. If a coin is tossed two times, what is the probability of

(i) Getting head at least once?

(ii) Getting exactly one head?

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1. The probability of an impossible event is

A) 1

B) 0

C) Less than 0

D) Greater than 0

2. Which of the following cannot be the probability of an event?

A) 1

2

B) 4

5

C) 1

D) 5

4

3. Two coins are tossed simultaneously. The probability of getting atmost one head is

A) 1

4

B) 3

4

C) 1

2

D) 1

4

4. In a football match, Messi makes 6 goals from 14 penalty kicks. The probability of converting

a penalty kick into a goal by Messi, is

A) 1

7

B) 2

7

C) 3

7

D) 5

7

5. The total number of outcomes of tossing 4 coins simultaneously is

A) 3

B) 6

C) 8

D) 16

6. Probability of an event not happening is x. The probability of the event happening is

A) 1 + x

B) 1 − x

C) x

D) 1

x

Questions from 7 to are subjective questions.

7. A coin is tossed 500 times with the following frequencies:

Head = 375 and Tail = 125

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Compute the probability for each event.

8. Three coins are tossed simultaneously 100 times with the following frequency of different

outcomes:

Outcomes 3 heads 2 heads 1 heads 0 heads

Frequency 15 15 30 40

Find the probability of getting

(1) Three heads

(2) Zero heads

9. In a cricket match, a batsman hits a boundary 8 times out of 40 balls he plays. Find the

probability that he didn’t hit a boundary.

10. A box contains 4 green marbles, 6 white marbles and 8 yellow marbles. Find the probability

of getting yellow marbles if it is randomly picked.

11. The percentage of marks obtained by a student in a class test are given below :

Unit Test 1st 2nd 3rd 4th 5th

Percentage of marks

obtained

56 67 78 91 94

Find the probability that the student gets marks between 85 to 95 percentages in the above

unit tests.

12. A bag contains 6 white balls and some red balls. If the probability of drawing a white ball

from the bag is 2

5, find the number of red balls in the bag.

13. A bag contains 50 coins and each coin is marked from 51 to 100. One coin is picked at

random. Find the probability that the number on the coin is not a prime number.

14. Eleven boys of wheat flour, each marked 5 kg, actually contained the following weights of

flour (in kg):

4.97, 5.05, 5.08, 5.05, 5.14, 4.89, 4.45, 5.99, 5.02

Find the probability that any of these bags chosen at random contains more than 5 kg of flour.

15. A geometry box consists of 16 markers having 4 different colors i.e. red, blue, yellow and

green. What can be the range of probability of getting blue marker when picked randomly?

16. Given below is the frequency distribution of daily wages of some employee.

Wages 150 - 160 160 - 170 170 - 180 180 - 190 190 - 200 200 -210

No. of

employee

3 6 8 12 4 2

An employee is selected at random. Find the probability that his wages are at least Rs. 180

17. Following table shows the birth month of students of class 9th.

Jan Feb March April June July August September Oct. Nov. Dec

3 6 8 9 10 2 1 8 12 0 3

Find the probability that a student was born in October.

d24cdstip7q8pz.cloudfront.net...(iii) a parallelogram whose all sides are equal d. rhombus (iv) a...

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