Oasis Math 9

(5)2 = OM2 + (4)2

or, 25 – 16 = OM2

∴ OM = 9 cm = 3cm

∴ Required distance of the chord from the centre is 3 cm.

Example: 2 A
In a circle of centre O, AB and CD are two parallel chords of C
lengths 16cm and 12 cm respectively, and radius of the circle is
10cm. Find the length of PQ. PQ
O
Solution:
D
B

Hence, AB||CD draw OP and OQ perpendicular to CD and AB respectively. POQ is a
straight line. So, OQ bisects AB and OP bisects CD.

Here,

Radius of circle = 10 cm

AB = 16 cm and CD = 12 cm

AQ = 12 AB = 1 × 16 cm = 8 cm.
2

Now, in right angled ∆OQA CP = 1 CD = 1 × 12 cm = 6 cm
2 2
or,
or, AO2 = AQ2 + OQ2
(10)2 = (8)2 + OQ2
And in right angled ∆OPC 100 – 64 = OQ2
OQ = 36 = 6 cm
or,
or, OC2 = OP2 + CP2
∴ (10)2 = OP2 + (6)2
From the figure, 100 – 36 = OP2

OP = 64 = 8 cm
PQ = PO + OQ
PQ = (8 + 6)cm

= 14 cm

Oasis School Mathematics-9 245

Example: 3

In the given figure, AB is a diameter, AB⊥CD. Prove that ∠COM = ∠DOM.

Solution: A

Given : O is centre of the circle, the diameter AB is perpendicular to C O D
the chord CD at M. M
B
To prove : ∠COM = ∠DOM

Contraction : J oin C and O, O and D.

Proof:

Statements Reasons
1. In ∆COM and ∆DOM 1.
(i) ∠OMC = ∠OMD (R) (i) Being both right angles.
(ii) CO = OD (H) (ii) Radii of the same circle.
(iii) CM = DM (S) (iii) A perpendicular drawn from centre of a circle

2. ∆ CMO ≅ ∆DOM bisects the chord.
3. ∴ ∠COM = ∠DOM 2. By R.H.S. axiom.
3. Corresponding angles of congruent triangles.

Example: 4 Hence, proved

In the given figure, O is the centre of the circle, chords MN and RS intersect at A. If OA
is the bisector of ∠MAR, prove that MN = RS.

Solution: M
Given: 'O' is the centre of the circle two chords MN and RS intersect PS
at A and OA is bisector of ∠MAR i.e. ∠MAO = ∠RAO.
To prove : MN = RS. OA
Construction : Draw OP⊥MN and OQ⊥SR. QN

Proof. R

Statements Reasons
1. In ∆OPA and ∆∆OQA 1.
(i) Being both 90º
(i) ∠OPA = ∠OQA (A)
(ii) ∠OAP = ∠OAQ (A) (ii) Given
(iii) OA = OA (S) (iii) Common side
2. ∆OPA ≅ ∆OQA 2. By A.A.S. axiom
3. OP = OQ 3. Corresponding sides of congruent triangles
4. MN = RS 4. Being two chords MN and RS equidistant from

the centre.
Hence, proved
246 Oasis School Mathematics-9

Example: 5

In the given figure, O is the centre of the circle. OX⊥BP and OY⊥DP. If OX=OY, prove

that BP = DP. P

Given : In a circle of centre O, OX⊥BP and OY⊥DP and OX = OY . A

To prove : BP = DP. x C
Construction : Join O and P. B

O

Proof : y

Statements Reasons D

1. OX = OY 1. Given

2. AB = CD 2. Equal chords are equidistant form the centre

3. BX = 1 AB, DY = 1 CD 3. A perpendicular drawn from centre of circle
2 2 bisects the chord.

4. ∴BX = DY 4. From statements (2) and (3)

5. In ∆PXO and ∆PYO

(i) ∠OXP = ∠OYP (R) (i) Being both are 90º

(ii) OP = OP (H) (ii) Common sides

(iii) OX = OY (S) (iii) Given

6. ∆PXO ≅ ∆PYO 6. By R.H.S. axiom

7. XP = YP 7. Corresponding sides of congruent triangles

8. BX + XP = DY + YP 8. Adding (4) and (7)

9. BP = DP 9. Whole part axiom

Hence, proved

Example: 6

In the given figure, O is the centre of the circle. AB = BC, prove that ∠ABD = ∠DBC.

Solution: A B
Given : 'O' is the centre of the circle, and AB = BC.
To prove : ∠ABD = ∠CBD M
Construction : From O, draw OM ⊥ AB and ON ⊥ BC.
DO
N

C

Proof.

Statements Reasons
1. In ∆OMB and ∆∆ONB
1.

(i) ∠OMB = ∠ONB (R) (i) Being both right angles

Oasis School Mathematics-9 247

(ii) OB = OB (H) (ii) Common sides
(iii) OM = ON (S) (iii) Equal chords are equidistant from the centre.
2. ∆OMB ≅ ∆ONB 2. By R.H.S. axiom
3. ∠OBM = ∠OBN 3. Corresponding angles of congruent triangles.
i.e. ∠ABD = ∠CBD
Hence, Proved

Exercise 13.1



1. (a) What is the relation of a chord with the line joining the centre of the circle to the
mid point of chord?

(b) What happens when the perpendicular is drawn from the centre of the circle to
the chord?

(c) If two chords are equidistant from the centre of the circle, what is their relation?

(d) If chord AB and CD are equal in length and distance of AB from the centre is 'a'
units, what is the distance of CD from the centre? Why?

(e) In the given figure 'O' is the centre of the circle, M is the O

mid point of AB, what is the relation between OM and AB?

A MB

C

(f) In the given figure, CD = 10cm, if OE⊥CD, what is the EO
length of CE and ED? D

PR

(g) In the given figure, PQ = RS and OM = 6cm, what is the MN
length of ON? Why? O

QS

(h) In the given figure, OP = OQ and AB = 9cm, what is the APB
length of CD? Why?
O

CQD

2. (a) In the given figure, AB = 16 cm and OM = 6cm. O
Find the radius of the circle. A MB

248 Oasis School Mathematics-9

CO
(b) In the given figure, OM = 5cm, OC = 13cm, find the A MB

length of the chord AB.

3. (a) In the given figure, O is the centre of a circle of radius O
10 cm, OP⊥AB, OQ⊥CD, AB||CD, AB = 12cm and CD
= 16 cm. Find the distance between two chords. CQ D
AP B

(b) In the given figure, O is the centre of the circle. PM Q
PQ//RS, OM⊥PQ, ON⊥RS. If PQ=8cm,
RS = 6cm, and r adius 5cm, find the length of MN. O

(c) In the given figure, O is the centre of the circle. AB and RN S
CD are two equal and parallel chords. If the distance A C
between two chords is 10 cm and the radius of the cir-
cle is 13 cm, find the length of the chords. POQ
BD
4. (a) In the given figure, the diameter CD meets the chord
AB at point P such that AP = PB = 4 cm, and CP = 3cm. D
Find the length of diameter of the circle.
O
(b) In a circle with centre O, an isosceles triangle ABC is
inscribed, AD⊥BC. If AB = AC = 15 cm and AD = 12 AP B
cm., find the length of AO. C
A

5. (a) In the given figure, O is the centre of the circle. OM⊥AB, O
BDC
ON⊥CD. If AB = 24 cm, OM = ON = 5 cm, show by cal-
culation that the lengths of the chords AB and CD are AC
equal.
MN
(b) In the given figure, 'O' is the centre of the circle. Radi- O
us of this c ircle is 13cm and PQ = RS = 24cm. Show by
calculation that equal chords are equidistant from the BD
centre.
PR

MN
O

QS

Oasis School Mathematics-9 249

6. In the given figure, the straight line ABCD cuts two O
concentric circles. Prove that AB = CD.
AB CD
7. Two equal chords AB and CD of the circle with
the centre O are produced to meet at P. Prove that; A B P
BP = PD. O D
S
8. In the given figure, O is the centre of the circle. Chords MN C
and RS intersect at P. If OP is bisector of ∠MPR, prove that M
MN = RS.
OP
9. In the given figure, O and P are the centre of two N
intersecting circles at A and B. Prove that
R
(i) AM = BM A
(ii) OM ⊥ AB.
OM P
10. Two chords AB and AC are equally inclined to the diameter
AD. Prove that: AB = AC. B

B

AO D

C

11. In the given figure, AC = DB, prove that OA = OB. O

AC D B

Answer

1. Consult your teacher 2. (a) 10 cm (b) 24 cm

3. (a) 2 cm (b) 7 cm (c) 24cm. 4. (a) 8.32 cm, b) 9.375 cm

Do You Know!

Pharaoh once asked Euclid to teach him Geometry the easy way to which the Mathematician
promptly replied, "There is no royal road to Geometry.

Project work

I. In a chart paper, draw a circle and show the following parts.
(i) diameter (ii) chord (iii) sector (iv) major arc (v) minor arc (vi) segment
II. By paper folding, prove that perpendicular drawn from the center of the circle bisects the chord.

250 Oasis School Mathematics-9

Unit

14 Construction

14.1 Warm-up Activities

Discuss the following in your class.
• What are the types of quadrilateral? • Recall the features of parallelogram.
• Recall the features of square, rectangle and rhombus.
• Recall the features of trapezium.
• If a line is given, how to draw another line parallel to this from the given point.
• If a line is given, how to draw another line perpendicular to this line from the

given point.

14.2 Construction of Quadrilaterals

A quadrilateral can be drawn if :
• all four sides and a diagonal are given. • three sides and two diagonals are

given.

• any one angle and four sides are given. • two sides and three angles are given.

Worked Out Examples

Type : I Rough Sketch

When any one diagonal and four sides are given: C

Example: 6 cm

Construct a quadrilateral ABCD in which AB = 5 cm, BC = 6.5 cm, D
CD = 6 cm, DA = 4.2 cm and diagonal AC = 8 cm.
A 5 cm B
C
4.2 cm
8 cm

6.5 cm

4.2 cm
8 cm

6.5 cm
6 cm Steps
D
• Draw a line segment AB = 5 cm.
A • From A, cut by the arc 8 cm and from B cut
5 cm
by the arc of 6.5 cm to get point C.
• Join AC and BC.
• Again from A, cut by arc length of 4.2 cm

and from C cut by an arc length of 6 cm to
get point D.
• Join AD and CD.
ABCD is the a required quadrilateral.

B

Oasis School Mathematics-9 251

Type : II Rough Sketch

When any three sides and both diagonals are given: C

D

7.5 cm
Example: 6.5 cm 5 cm
5 cm
Construct a quadrilateral ABCD in which AB = 3 cm, 6.5 cm
AD =5 cm, BC = 5 cm, diagonal AC = 6.5 cm and di- B
A 3 cm

agonal BD = 7.5 cm. C

Steps

D 5 cm • Draw a line segment AB = 3 cm.
7.5 cm
• From A, cut by an arc length of
5 cm
6.5 cm and from B, cut by an arc
length of 5 cm to get point C.

• Join AC and BC.

• From A, cut by an arc length of 5

cm and from B, cut by arc length
of 7.5 cm to get point D.

A 3 cm B • Join AD and BD.

Type: III ABCD is the a required quadrilateral.

When any one angle and four sides of the
quadrilateral are given:

Example: Rough Sketch
Construct a quadrilateral ABCD in which AB = 6.5 cm,
BC = 6.2 cm, CD = 4.2 cm, DA = 5.8 cm and ∠DAB = D 4.2 cm C
75º. P

5.8 cm

5.8 cm
D 4.2 cm C 6.2 cm

6.2 cm A 75º B

6.5 cm

75º 6.5 cm Steps
A • Draw a line segment AB = 6.5 cm.
• At A, draw ∠BAP = 75º.
• From A cut by an arc of length 5.8 cm to get

D.

• From D, cut by an arc length of 4.2 cm and

from B, cut by an arc length of 6.2 cm to get C.

B • Join DC and BC.

ABCD is the a required quadrilateral.

252 Oasis School Mathematics-9

Type: IV

When any three sides and two angles are given, Rough Sketch
Example:
Construct a quadrilateral in which AB = 6.5 cm, BC = 3.5 cm, D
AD = 4cm, ∠DAB = 60º and ∠CBA = 45º. C

P
4 cm
4 cm
4.5 cm60º 45º B
3.5 cm A 6.5 cm

DQ Steps

C • Draw a line segment AB = 6.5 cm.

• At the point A, draw ∠PAB = 60º and at
the point B, draw∠QBA = 45º.

60º 6.5 cm 45º • Take an arc of 4 cm and cut from A to D
A B and by the arc of 4.5 cm cut from B to C.

• Join DC.

ABCD is a required quadrilateral.

Type: V

Example: Rough Sketch

When any two sides and three angles are given, Q Q
Construct a quadrilateral ABCD in which AB = 5.7 cm, D
BC = 3.5 cm, ∠DAB = 60º, ABC = 105º and ∠BCD = 90º.
90º C

P 3.5 cm A 60º 105º B
D
3.5 cm5.7 cm
5.7 cm
60º Q Steps
A C
90º • Draw a line segment AB = 5.7 cm.
• At point A, draw ∠BAP = 60º and
105º
B at point B, draw∠ABQ = 105º.
• From B, cut by arc of 3.5cm to get

point C.
• At C, draw ∠BCD = 90º, which

meets AP at D.
ABCD is a required quadrilateral.

Oasis School Mathematics-9 253

Exercise 14.1

1. Construct a quadrilateral ABCD from the following data:
(a) AB = 5.8 cm, BC = 6.3 cm, CD = 5.2 cm, AD = 4.8 cm and diagonal BD = 6.9 cm.
(b) AB = 5.3 cm, BC = 6.2 cm, CD = 6.5 cm, DA = 6.1 cm and diagonal AC = 6.8 cm.

2. Construct a quadrilateral PQRS from the given data:
(a) QR = 5.6 cm, RS = 5.8 cm, PS = 6.3 cm, diagonal PR = 6 cm and diagonal QS = 7 cm.
(b) PQ = 5.4 cm, QR = 5.2 cm, PS = 6.5 cm, diagonal PR = 6.3 cm and diagonal QS = 6.8 cm.

3. Construct a quadrilateral ABCD from the given data:
(a) AB = 6.5 cm, BC = 5.4 cm, CD = 5.3 cm, AD = 6.6 cm and ∠DAB = 60º.
(b) AB = BC = 6.2 cm, CD = AD =4.9 cm and ∠ABC = 75º.

4. Construct a quadrilateral ABCD from the given data:
(a) AB = 6.7 cm, BC = 6.5 cm, CD = 6.1 cm, ∠ABC = 120º and ∠BCD = 45º.
(b) AB = 5 cm, CD = 6.1 cm, BC = 5.5 cm, ∠ABC = 45º and ∠DAB = 120º.

5. (a) Construct a quadrilateral ABCD in which AB = 6cm,
BC = 6.5cm, ∠A = 60º, ∠B = 120º and ∠C = 90º.

(b) Construct a quadrilateral PQRS in which PQ = 6.6cm,
QR = 7cm, ∠P = 90º, ∠Q = 60º and ∠R = 120º.

Answer
Consult your teacher.

14.3 Construction of Some Special Types of Quadrilaterals
I. Construction of a parallelogram:
Properties of a parallelogram

• Opposite sides are equal.
• Opposite angles are equal.
• Diagonals bisect each other.

Type: I

When the adjacent sides and the angle between them are given:

254 Oasis School Mathematics-9

Example: Rough Sketch D
Construct a parallelogram ABCD in which AB = 6 P
cm, BC = 5 cm and ∠ABC = 60º. A 5 cm

P6 cm 5 cm D
A 6 cm

6 cmB 60º5 cmC

6 cm
Think ! Opposite sides and angles of a
parallelogram are equal.

B 60º 5 cm C Steps
• Draw a line segment BC = 5 cm.
• At B draw ∠PBC = 60º.
• From B cut by an arc length of 6 cm to

get point A.
• From C cut by an arc of 6 cm and from

A cut by an arc of 5cm to get point D.
• Join, AD and CD.
ABCD is the a required parallelogram.

Type: II

When both the diagonals and angle between them are given:

Example: P Rough Sketch
Construct a parallelogram ABCD in which BD = 6 cm, A
AC = 8 cm and the angle between AC and BD is 45º. D 3 cm 4 cm C
4 cm 45º O 3 cm B

P D C Think ! Diagonals of a parallelogram
A 3 cm 4 cm bisect each other.

4 cm 45º O 3 cm If BD = 6cm then BO = OD = 3cm
B If AC = 8cm then AO = OC = 4cm.
Q
Steps

• Draw a line segment AC = 8cm.
• Find the mid-point of AC, by drawing

perpendicular bisector of AC.
• At O, draw ∠AOD = 45º (mid point of AC)

and produce the line on both sides of O.
• From O, cut by an arc of 3cm on both

sides to get B and D.
• Join AB, BC, CD and AD.
ABCD is the a required parallelogram.

Oasis School Mathematics-9 255

Type: III3 cm
When any one side and both the diagonals are given:
Example:
Draw parallelogram ABCD in which AB = 4 cm, di-
agonal AC = 10 cm and diagonal BD = 6 cm.

Q
P

DC

5 cm

O
5 cm

3 cm

A 4 cm B

Exercise 14.2

1. Construct a parallelogram ABCD from the given data:
(a) AB = 5 cm, BC = 6 cm and ∠ABC = 60º.
(b) BC = 6 cm, CD = 7.2 cm and ∠BAD = 75º.

2. Construct a parallelogram ABCD from the given data:
(a) AB = 5 cm, BD = 6.8 cm and ∠DBA = 45º.°
(b) AB = 6.2 cm, AC = 7 cm and ∠CAB = 30º.

3. Construct a parallelogram PQRS from the given data:
(a) PQ = 5.2 cm, QR = 4.5 cm and diagonal QS = 6.7 cm.
(b) QR = 5.4 cm, RS = 6.2 cm and PR = 7 cm

4. Construct the parallelogram ABCD from the given data:
(a) Diagonal AC = 5.8 cm, diagonal BD = 8 cm and angle between them is 60º.
(b) Diagonal AC = 6.4 cm, diagonal BD = 7cm and angle between hem is 45º.

256 Oasis School Mathematics-9

5. Construct the parallelogram ABCD from the given data:
(a) AB = 5.6 cm, diagonal AC = 7.8 cm and diagonal BD = 5.4 cm.
(b) CB = 5.6 cm, AC = 7 cm and BD = 6.4 cm.

Answer
Consult your teacher.

Construction of rectangle
Features.

• Each angle of a rectangles is a right angle.
• Its opposite sides are equal.
• Length of both diagonals are equal.

Type : I Rough Sketch P
When any one side and a diagonal are given:
Example: CD
Construct a rectangle ABCD in which side
AB = 5 cm and diagonal AC = 7.5 cm. A 90º B
5 cm
P
Think ! Each angle of a rectangle
is a right angle.
7.5 cm
D 7.5 cm
A C Steps
• Draw a line segment AB = 5 cm.
5 cm
• At B draw ∠ABP = 90º.

• From A cut by the arc length of
AC = 7.5cm which meets BP at C.

• Join AC.

• From A, cut by the arc equal to BC and
from C cut by the arc equal to AB to get D.

• Join AD and CD.

ABCD is a required rectangle.

B

Oasis School Mathematics-9 257

Type : II Rough Sketch

When two adjacent sides are given: P

Example: DC
Construct a rectangle ABCD in which AB = 7.2 cm and BC = 4.8 cm.

P
4.8 cmAB

4.8 cm 7.2 cm

DC Steps
• Draw a line segment AB = 7.2 cm.
• At B, draw ∠ABP = 90º.
• From B, cut by an arc of 4.8cm

to get C.
• From A cut by the arc of 4.8cm

and from C cut by the arc of
7.2cm to get point D.
• Join AD and CD.
ABCD is the a required rectangle.

A B
7.2 cm

Construction of a square:

Type : III Rough Sketch P
C
Example: 3 5.5 cm
When the length of a side is given:
D

Construct a square in which the perimeter is 22 cm. 5.5 cm
5.5 cm

Here, 4l = P A B

5.5 cm

or, l = P = 22 cm = 5.5 cm P
4 4

5.5 cmD 5.5 cm 5.5 cmC Steps

A • Draw a line segment AB = 5.5
5.5 cm cm.

• At point B, draw an ∠ABP=90º.
• From B, cut by the arc length of

5.5 cm to get C.
• From C and A cut by the arc of

5.5 cm to get point D.
• Join CD and DA.

ABCD is the a required square.

B

258 Oasis School Mathematics-9

Type : IV Rough Sketch3.6 cm 3.6 cm

When the length of a diagonal is given: P
Example: Construct a square in which the length D
of a diagonal is 7.2 cm.
7.2 cm
P
A OC
D
B
Q

Think ! Diagonals of square are
equal in length.

3.6 cm Steps

7.2 cm • Draw a line segment AC = 7.2 cm.

A C • Draw perpendicular bisector PQ of AC3.6 cm
which passes through O.

• From O, cut by an arc of 3.6 cm on both
sides of PQ to get points B and D .

• Join AB, AD, BC and DC.

B ABCD is the a required square.

Q Rough Sketch D

Construction of a Rhombus P
A 5.7 cm
Type : V
5.7 cm
When one side and any one angle is given: 5.7 cm
60º
Example: Construct a rhombus ABCD in which one B 5.7 cm C

side = 5.7 cm and ∠ABC = 60º. Think ! All sides of a
rhombus are equal.
P
Steps
A 5.7 cm D
• Draw a line segment BC =
5.7 cm 5.7 cm.
5.7 cm
• At B, draw an angle PBC
B 60º 5.7 cm C = 60º.

• From B, cut by an arc of
5.7 cm to get point A.

• From A and C cut by
equal arcs of 5.7 cm to
get point D.

• Join AD and CD.
ABCD is the a required

rhombus.

Oasis School Mathematics-9 259

Type : VI Rough Sketch2.6 cm
When both diagonals are given:
Example: P2.6 cm
Construct a rhombus in which the length of D
diagonals are 7.6 cm and 5.2 cm.
AC
P
7.6 cm

B
Q

Think ! Diagonals of rhombus bisect
each other at right angle.

D Steps

2.6 cm • Draw a line segment AC = 7.6 cm.
• Draw the perpendicular
A O
7.6 cm bisector PQ of AC which
passes through O of AC.
2.6 cm C • From O, cut by an arc of 2.6
cm on both sides of PQ to get
points B and D.
• Join AB, AD, CB and CD.
ABCD is the a required rhombus

.

B

Q

Exercise 14.3

1. (a) Construct a rectangle ABCD having AB = 5cm and diagonal AC = 6cm.
(b) Construct a rectangle PQRS where PQ = 6cm and diagonal PR = 8cm.
2. (a) Construct a rectangle ABCD where AB = 5.8cm and BC = 6.4cm.
(b) Construct a rectangle PQRS where PQ = 6.5cm, QR = 7.2cm.
3. (a) Construct a square ABCD in which AB = 4.4cm.
(b) Construct a square PQRS where PQ = 5.6cm.
4. (a) Construct a square ABCD where diagonal AC = 8cm.
(b) Construct a square PQRS where PR = 6.8cm.

260 Oasis School Mathematics-9

5. Construct a rhombus PQRS from the given information.
(a) QR = 6.4cm, and ∠PQR = 60º.
(b) PQ = 7cm. and ∠PQR = 75º.
6. Construct a rhombus ABCD from the given information.
(a) diagonal AC = 6cm and diagonal BD = 8cm.
(b) diagonal AC = 6.4cm and diagonal BD = 7.2cm.

Answer
Consult your teacher.

V. Construction of Trapezium

Type : I

When any three sides and one base angle are given: Rough Sketch

Example: Construct a trapezium ABCD in which P

AB = 5.4 cm, CD = 4.5 cm Q 60º D 4.5 cm C
DA = 3 cm, ∠ DAB = 60°º and DC || AB

P
3 cm A 60º B
5.4 cm
3 cm
D 4.5 cm C Steps
Q
• Draw a line segment AB = 5.4 cm.
60º • At A, draw an ∠PAB = 60º.

• From A, cut AP by an arc of 3cm to get
point D.

A 60º B • Draw DC//AB making ∠ADQ = 60º
5.4 cm = ∠DAB.

• Cut by arc of 4.5 cm to get point C.

• Join BC.

ABCD is the a required trapezium.

How to draw DC||AB ?
• Make alternate angles equal.
i.e. draw ∠PAB = ∠QDA = 60º.

Oasis School Mathematics-9 261

Type : II

When base angles, length of base and an arc of its Rough Sketch
non–parallel sides are given:
P CR
Example: QD

Construct a trapezium ABCD in which AB = 7cm, AD = 5
cm, ∠DAB = 60º, ∠CBA = 75º° and DC//AB.



PS
5 cm A 60º 7 cm 75º B
5 cm
Q C Steps
D R • Draw a line segment AB = 7cm.

• At A, draw an ∠PAB = 60º.

• From A, cut AP by an arc of 5 cm to
get point D.

• Draw DC//AB making ∠ADQ =
60º = ∠DAB.

A 60º 75º B • At B, draw ∠SBA = 75º.
ABCD is the a required trapezium.

7 cm

Exercise 14.4

1. (a) Construct a trapezium ABCD where BC||AD, BC = 5cm, AD = 4cm, AB =
4.5cm and ∠ABC = 60º.

(b) Construct a trapezium ABCD wher AB||DC, AB = 7cm, BC = 5cm, ∠ABC = 75º
and CD = 5.4cm.

2. (a) Construct a trapezium ABCD where AB||DC, AB = 7cm, BC = 5cm
AD = 6.5cm, ∠ABC = 60º.

(b) Construct a trapezium PQRS where QR||PS, QR = 6.8cm, ∠QRS = 75º,
RS = 5cm and SP = 4.8cm.

3. (a) Construct a trapezium EFGH where EF = 6.5cm, FG = 7.4cm, ∠EFG = 60º,
∠FGH = 75º and EH||FG.

(b) Construct a trapezium PQRS where SP = 7.6cm.∠ SPQ = 60º, ∠PQR = 90·,
PQ = 6.8cm and SR||PQ.

Answer ❏ ❏
❏Consult your teacher.

262 Oasis School Mathematics-9

Assessment Test Paper

Full marks: 40

Attempt all the questions.

Group-A [4 × 1 = 4]

1. Name the longest and the shortest side of A
∆ABC in the given figure.
800
A
B 700 C300

2. In the given figure, M and N are the mid points A
AB and AC respectively. What are the relations
M N

between MN and AC? BC

3. ABCD is a parallelogram. If DC = 7cm, BC = 6cm, find the A B
length of AB and AD. D 7cm
6cm

C

AC

4. AB and CD are two equal chords of OM and ON are their distance M N
from the centre of the circle and OM = 3.5cm, find the length of ON. O

Group-B [7 × 2 = 14] C BD

5. Find the value of x in the given figure. 300 x

E

A B D420 500 A B
6. In the given figure, AB = AC, find the value of x. 1100

A DCx D

7. In the given figure, ABCD is a rectangle, x F
find the value of x. 1200 600

BE C A

8. In the given figure, P and Q are the mid points of AB and P Q
C
AC respectively, if PQ = (2x+5)cm and BC = (3x+12)cm,
find the value of x and the length of BC. OC
B D

A

9. In the given figure, AB//CD, AO=8cm, OD=4cm,

CD=3cm, find the length of AB. B

Oasis School Mathematics-9 263

10. In the given figure, O is the centre of the circle. OC ⊥ AB, A
OC=5cm, AB = 24cm, find its radius. O

P 13cm S C

B

11. Find the value of 'x' in the given figure. x
12cm

Q 3cm R

Group-C [3 × 4 = 12] A B
12. In the given figure, AB=CD and AB//CD, prove that D C

AD=BC and AD//BC.

13. Verify experimentally that equal chords are equidistant
from the centre.

14. Construct a rhombus ABCD whose two diagonals AC and BD are 6cm and 8cm
respectively.

Group-D [2 × 5 = 10]

15. In the given figure, ABCD is a parallelogram, diagonal AC is P A B

produced in such away that PA = CQ, prove that PBQD is a C
Q
parallelogram. D

16. In the given figure, ABCD is a trapezium, P and Q are the mid AB
points of AD and BC respectively, prove that 2PQ = AB+CD.
P RQ

DC

264 Oasis School Mathematics-9

Trigonometry

88Estimated Teaching Hours

Contents

• Introduction of trigonometrical ratio on the basis of right
angled triangle

• Problems on measurement of the ratios of sine, cosine and
tangent

• Introduction and use of sine, cosine and tangent values
(00, 300, 450, 600 and 900)

• Problems on right angled triangle of sine, cosine and
tangent without using table

Expected Learning Outcomes

At the end of this unit, students will be able to develop the
following competencies:
• To find out p, b and h on a right angled triangle on the basis

of given angle of reference

• To solve the simple trigonometric identity

• To find the values of standard angle (0º, 30º, 45º, 60º, 90º) of
different trigonometric ratios

• To solve the problems related to standard angles of different
trigonometric ratios

• To solve the right angled triangle, using trigonometric ratios
and Pythagoras theorem

Teaching Materials

• Graph sheet, Models of right angled triangle, Chart paper, A4
size paper, Trigonometric table, etc.

Oasis School Mathematics-9 265

Unit

15 Trigonometry

15.1 Introduction

Trigonometry is the science that deals with the measurement of angles. Trigonometry
also measures the relation between the sides and the angles of a triangle. Trigonometry
has been extensively used in Engineering, Astronomy, Geography, etc.

Trigonometric Ratios A

Let us consider a right angled triangle ABC having Hypoatnenguleseof reference Opposite side
∠B = 90º. Let an acute angle C be θ. Then, 'θ' is called θ Perpendicular
the angle of reference. AC is a side opposite to right
angle. It is called hypotenuse. AB is a side opposite to C Adjacent side B
angle of reference. It is called a perpendicular (opposite
side). BC is a side adjacent to 'θ' and it is called a base (base)
(adjacent side).

The possible ratios of these three sides of ∆ABC are AB , ABCC, AB , AC , AC and BC .
AC BC AB BC AB

The first ratio AB i.e. the ratio of perpendicular to the hypotenuse is called sine of an
BC
angle 'θ'.
perpendicular
∴ sin θ = hypotenuse = AACB .................... (i)

Similarly, the ratio of adjacent side (base) to the hypotenuse of a right angled triangle is

called cosine of an angle θ. It is denoted by
base
∴ cos θ = hypotenuse = ABCC .................... (ii)

Similarly, the ratio of perpendicular to the base is called tangent of an angle θ.

∴ tan θ = perpendicular = AB .................... (iii)
base BC

The reciprocals of above ratios are respectively defined as,

Cosecant of an angle θ = cosec θ = hypotenuse = AACB
perpendicular

Secant of an angle θ = sec θ = hypotenuse = ABCC
base

Cotangent of an angle θ = cot θ = base = ABCB
perpendicular

The trigonometric ratios (ratios of sides of a right angled triangle) depend upon the
acute angle which is taken as angle of reference.

266 Oasis School Mathematics-9

Summary

In right angled triangle ABC, A

sin θ = p cos θ = b h
h cot θ h p
sec θ B θC
tan θ = p = b
b p b

cosec θ = h = h
P b



In right angled triangle ABC, if 'θ' is taken as the angle of reference,

AB = perpendicular, AC = hypotenuse, and BC = base.

If 'β' is taken as the angle of reference, A
BC = perpendicular, AC = hypotenuse and AB = base.

∴ sin θ = AACB and cos θ = ABCC β

sin β = ABCC and cos β = AB . B θC
AC

Worked Out Examples

Example: 1
Write all the trigonometric ratios with the given angle of reference.

(a) A (b) 4
P
α
3

B C Q 5 θR
Solution:

(a) In right angled triangle ABC, if α is taken as angle of reference,

perpendicular (p) = BC, h ypotenuse (h) = AC, base (b) = AB
p
therefore, sin α = h = ABCC

cos α = b = AACB
h
p BC
tan α = b = AB

cosec α = h = ABCC
p
h = AACB
sec α = b

cot α = b = AB
p BC

Oasis School Mathematics-9 267

(b) In ∆PQR, ∠P = 90º.

Taking 'θ' as the angle of reference,

perpendicular (p) = PQ = 3,

hypotenuse (h) = QR = 5, base (b) = PR = 4 .

We have, p 3
h 5
sin θ = =

cos θ = hb = 4
5

tanθ = pb = 3
4
5
cosec θ = ph = 3

sec θ = pb = 5
4

cot θ = b = 4
p 3

Example: 2 A
In the given right angled triangle ABC, ∠B = 90°, BD ⊥AC,

AB = 5, BC = 12. Find the value of sin θ, cos θ and tan θ. 5 D
θ 12
Solution:
B
In right angled triangle ABC, BD ⊥ AC and ∠ABD = θ C

Now, ∠BCD = 90º – ∠CBD

= 90º – (90º – θ)

=θ

In ∆ABC, taking ∠ACB = θ as the angle of reference,

then, p = AB = 5 Now,
= BC = 12
b = AC In ∆ABC,
= p² + b²
h
= (5)² + (12)²
Using Pythagoras theorem, h sin θ = AB = 5
AC 13

cos θ = ABCC = 12
13
= 25 + 144
tan θ = AB = 5
= 169 BC 12

= 13

268 Oasis School Mathematics-9

Example: 3
In the given figure, find the value of sinα and tanβ.

Solution: 13 D
In the right angled triangle ABC, taking 'α' as the angle of Aβ
reference, p = AB = 3, b = BC = 4, h = AC = ?
Using Pythagoras theorem, h2 = p2 + b2 3

B4 α C

AC2 = AB2 + BC2

AC = AB2 + BC2

= 32 + 42

= 9 + 16

= 25

AC = h = 5

sin α =ph = 3
5

Again, in right angled triangle ACD,

Taking 'β' as the angle of reference.

p = AC = 5, h = AD = 13 b = CD = ?

Using Pythagoras theorem, h2 = p2 + b2

AD2 = AC2 + CD2

CD = AD2 – AC2²

= (13)2 – (5)2

= 169 – 25

= 144

= 12

∴ tan β = p = AC = 5
b CD 12

Oasis School Mathematics-9 269

Exercise 15.1

1. Write all the trigonometric ratios in the given triangles with the given angle

of reference. P (c) M
(a) A
(b)

Rβ

L

B θC α N
Q

2 . Find the value of all trigonometric ratios in the given triangles with the given

angle of reference.

(a) P (b) (c)

A C D 13cm F
6cm β
6cm α
3cm 5cm12cm
8cm 10cm E
5cm
R θ Q B
3 3cm

3. In each of the given figures, find the values of all trigonometric ratios with the
given angle of reference.

(a) G (b) (c) P
A 13cm α C
2 2cm 8cm β
12cm R
B
θ I Q 10cm
H 2 2cm A

8cm E

4. (a) In the given right angled triangle, ∠B = 90°, EF ⊥ BC, α

∠CEF = α, AB = 8cm, and BC = 10cm. Find the B β C
value of sinα and cosβ. F 10cm

A

D

(b) In the given figure, ABC is a right angled triangle. Find 9cm

sinα, cosα, tanα, sinθ, cosθ and tanθ. B θ α
15cm
E C

D 8cm
C

5. (a) In the given figure, find the value of sinθ and cosα.

α 13cm θB
A

270 Oasis School Mathematics-9

P

(b) In the given figure, find the value of sinβ, and cosα. 5 cm aS
6. (a) In the given figure, find the value of tanα and tanβ. R
β 7cm
Q 12 cm

P

αβ 8 cm 15 cm

5 cm

Q 3cm S R
A C

(b) In the given figure, AD⊥BC, AB = 13cm, BD = 5cm, 13 cm β
AC = 15cm, find the value of sinα, tanα, cosβ and tanβ.

B α D
5cm

A

7. (a) In the given figure, AB = 6 cm, AC = 10 cm. 10cmD 6cm
BD ⊥ AC. Find the value of tanθ and secθ. 10 cm

C θ BA

(b) In the right angled triangle ABC, BD ⊥ AC. If AB = 10cm, D
BC = 15cm, find the value of sinα, cosα and tanα.

α 15cm C
B
Answer

1. Consult your teacher.

2. (a) sinθ = 1 , cosθ= 3 , tanθ = 1 , cosec θ = 2, sec θ = 2 , cot θ = 3
2 2 3 3

(b) sinα = 4 cosα = 3 , tanα = 43, cosecα = 5 , sec α = 35 , cot α = 3
5 5 4 4

(c) sinβ = 12 cosβ = 5 , tanβ = 152, cosec β = 13 , sec β = 153, cot β = 5
13 13 12 12

3. (a) sinθ = 1 , cosθ = 1 , tanθ = 1, cosecθ = 2 , sec θ = 2 , cot θ = 1
2 2
(b) sinα =153, cosα = 1132, tanα = 152, cosecα = 153, secα = 1123, 12
cot α = 5

(c) sinβ = 4 cosβ = 35, tanβ = 4 , cosecβ = 45, sec β = 53, cot β = 3
5 3 4
12 9
4. (a) sinα = 5 , cosβ = 5 , (b) sinα = 1125, cosα = 195, tanα = 9 , sinθ = 195, cosθ = 1125, tanθ = 12
41 41
(a) sin θ = 1132, cos α= 7
5. 5 (b) sinβ = 5 , cosα = 218 6. (a) tanα = 34, tanβ = 3
3 13

(b) sinα = 12 , tanα = 12 , cosβ = 12 , tanβ = 9 ,
13 5 15 12

7. (a) tanθ = 4 , secθ = 5 (b) sinα = 3 , cosα = 2 , tanα = 3
3 3 13 13 2

Oasis School Mathematics-9 271

15.2 Relation among the trigonometric ratios

In right angled triangle ABC, A

AB = perpendicular (p)

AC = hypotenuse (h) ph

BC = base (b) θ

We know that, B bC

sin θ = p and cosec θ = ph
h

∴ sin θ . cosec θ = ph. ph = 1

∴ sinθ. cosec θ = 1 ………………… (i)

From (i)

sin θ = 1 ………………… (ii)
cosec θ
cosec θ 1
Again, = sinθ ………………… (iii)

cos θ = hb and sec θ =hb
∴ cosθ. sec θ
cos θ. secθ = b . hb = 1
From (iv), h
cos θ
Again, sec θ = 1 ……………. (iv)

= 1 ……………. (v)
secθ

= 1 ……………. (vi)
cosθ

tan θ . cot θ = p . pb = 1
b

∴ tan θ. cot θ = 1 ……………. (vii)

tan θ = 1 ……………. (viii)
cot θ cotθ
Again, tan θ
= 1 ……………. (ix)
tanθ
p
= b

p

= h (Dividing both numerator and denominator by h)
b
h

= sinθ
cosθ

272 Oasis School Mathematics-9

∴ tan θ = csoinsθθ ……………. (x)

Similarly, cot θ = csoinsθθ ……………. (xi)



In a right angled triangle,

p² + b² = h² [Using Pythagoras theorem]

Dividing both sides by h2

p² + b² = 1
or, h² = 1
p² b²
or, h² + h²
or,
( ) ( )p2+ b 2 = 1
b h

sin²θ + cos²θ = 1 ……………. (xii)

From (xii), sin²θ = 1 – cos²θ ……………. (xiii)

sin θ = 1 – cos²θ ……………. (xiv)
Again, cos²θ = 1 – sin²θ ……………. (xv)

cos θ = 1 – sin² θ ……………. (xvi)
Again,

p² = h² – b² [ Using Pythagoras theorem]

Dividing both sides by p2

h² – b² = p²
p² p²

or, h² – b² = 1
p² p²

or, ( ) ( ) h2–b 2 = 1
p p

or, cosec2θ – cot2θ = 1 ……………….. (xvii)

From ..... (xvii)

cosec²θ = 1 + cot²θ ……………….. (xviii)

cosec θ = 1 + cot² θ ……………….. (xix)
cot²θ = cosec²θ – 1 ……………….. (xx)

cot θ = cosec² θ – 1 ……………….. (xxi)
Again, h² – p² = b²


Dividing both sides by b².
or,
h² – p² = b²
b² b²

Oasis School Mathematics-9 273

or, h² – p² = 1
b² b²
or,
or, ( ) ( )h 2– p 2 = 1
or, b b

sec²θ – tan²θ = 1 ……………. (xxii)

sec²θ = 1 + tan²θ ……………. (xxiii)

sec θ = 1+ tan² θ …………. (xxiv)
tan² θ = sec²θ – 1 .………. (xxv)

tan θ = sec²θ – 1 ……………. (xxvi)

Worked Out Examples

Example: 1

Prove the following identities.

(a) cos A. tan A. cosec A = 1

(b) (1 + tan²θ). (1 – sin²θ) = 1
Solution:

(a) L.H.S. = cos A . tan A. cosec A.
sinA 1
= cos A . cosA . sinA

= 1 = R.H.S.

(b) L.H.S. = (1 + tan²θ ) (1 – sin²θ)

= sec²θ . cos²θ
1
= cos2 θ . cos²θ.

= 1 = RHS.

Example: 2

Prove that : cosec4 θ – cosec2 θ = cot4 θ + cot2θ

Solution:

L.H.S. = cosec4θ – cosec2θ

= cosec2θ (cosec2θ – 1)

= cosec2 θ . cot2θ

= (1 + cot2θ). cot2θ

= cot2θ + cot4θ

= R.H.S.

274 Oasis School Mathematics-9

Example: 3

If cosA= 3 , find the value of sinA and tanA.
5
Solution:

Given, cosA = 3
5
We know that,

cosA = b
h

Let, b = 3, h = 5

Using Pythagoras theorem, h2 = p2 + b2 Alternative method:

or, (5)2 = p2 + 32 3
5
or, 25 = p2 + 9 Given, cosA =

or, 25–9 = p2 We have sinA = 1– cos2A

or, 16 = p2 =1–(3/5)2

or, p = 16 = 4 = 1– 2 95 = 12 56 = 4
3
We have, sinA = b = 4
h 5 then, tanA = sinA = 4/5 = 4
cosA 3/5 3
cosA = b = 3
h 5

tanA= b = 4
h 3

Exercise 15.2

1. Prove the following identities.

(a) tan A. cosec A = sec A (b) cot A . sec A = cosec A
cos A. tan A. cosec A = 1
(c) sin A. cot A. sec A = 1 (d)

(e) tan θ.cosec θ. sec θ – 1 = tan²θ.

2. Prove the following identities.

(a) (sin A + cos A)² = 1 + 2 sin A. cos A

(b) (sin A – cos A)² = 1 – 2 sin A . cos A

(c) 1 = seca – tana
(secα + tanα)

(d) 1 = cosecθ + cotθ
(cosecθ – cotθ)

(e) sec2θ – 1 = sin²θ
1+tan2θ

Oasis School Mathematics-9 275

(f) ( )1+ tanθ 2 = 1 + tan2θ
(g) 1 + cotθ 1 + cot2θ

1 + tan2θ = cosec²θ
tan2θ

(h) sec4 θ – sec²θ = tan4θ + tan²θ

(i) cot2θ – cos2θ = cot2θ. cos2θ

(j) cosec θ + 1 = 1 + sin θ
cosec θ– 1 1–sinθ

3. (a) If sin A = 153, find the value of cos A and tan A.

(b) If tan A = 3 , find the value of sec A.
4

(c) If sin A = 1132, and cos B = 3 find the value of cos A + sin B.
(d) 5
(e) a
If sin θ = b , find the value of tan θ.

If tan θ = p , find the value of p. sin θ + q cos θ.
q



Answer

3. (a) cosA = 1123, tanA = 5 (b) secA = 5 (c) 77 (d) a a² (e) p² + q²
12 4 65 b² –

15.3 Trigonometric ratios of some standard angles

The angles 0°, 30°, 45°, 60° and 90° have a huge importance in trigonometry. Such angles
are called the standard angles. Now, we are going to obtain the values of such angles
without using tables.

(i) Trigonometric ratios of 0°

Let ∠AOX be a very small angle having magnitude 0°. Take any point P on OA and
draw PM ⊥ OX. Now, ∠PMO = 90º. Then ∠OPM = 90º

Let, OP = OM = a, PA

Using Pythagoras theorem, O O0 X
M
PM = OP² – OM2

= a2 – a2
= 0.

In ∆POM,

276 Oasis School Mathematics-9

sin 0º = POMP = 0 = 0
cos 0º a
tan 0º
= OOMP = a = 1
a
Similarly,
= OPMM = 0 = 0
a

cosec 0º = ∞ sec 0º = 1 cot 0º = ∞

(ii) Trigonometric ratio of 30° and 60°

Let, ∆ABC be an equilateral triangle, where AC = AB = BC = 2a (suppose).
∠A = ∠B = ∠C = 60º.

AD is drawn perpendicular to BC. Since ∆ABC is an equilateral triangle,
∠BAD = ∠CAD = 30º and BD = DC = a.

In the right angled triangle ABD A

AD = (AB)2 – (BD)2 300 300
= (2a)2 – a2 2a 2a

= 4a2 – a2 600 600
= 3a2
= a 3 B a Da C

In the right angled triangle ABD,

sin 30º = BD = a = 1
AB 2a 2

cos 30º = AADB = a2a3 = 23
Similarly,
tan 30° = BD = a = 1
AB a3 3

cosec 30° = 2, sec 30° = 23, cot 30° = 3

Again, in the right angled triangle ABD,

sin 60º = AD = a2a3 = 3
AB 2

cos 60º = BD = 2aa = 1
AB 2

tan 60º = AD = aa3 = 3
BD

Similarly, 2 1
3 3
cosec 60º = , sec 60º = 2, cot 60º =

Oasis School Mathematics-9 277

(iii) Trigonometric ratios of 45° A

Let ∆ABC be a right angled isoscele triangle, right angled at C.

Then, ∠CAB = ∠ABC = 45º a2 a
450 C
Let, Ba

AC = BC = a

Using Pythagoras theorem,

AB = AC2 + BC2
= a2 + a2
= 2a2

= a 2

In the right angled triangle ABC,

sin 45º = AACB = a = 12
a2

cos 45º = BC = a = 1
AB a2 2
Similarly,
tan 45º = ABCC = a = 1
a

cosec 45º = 2 , sec 45º = 2 , cot 45º = 1

(iv) Trigonometric ratios of 90°

Let ∠AOX be an angle which is equal to 90º. Take any point P on OA and draw PM

perpendicular to OX. In the right angled triangle POM, ∠POM = 90º, ∠PMO = 90º, then

∠OPM = 0º. Y
Here, A

OP = PM = a. (suppose). P

Using Pythagoras theorem,

OM = OP2 – PM2

= a2 – a2

= 0° OM X

In the right angled triangle POM,

sin 90º = POMP = a = 1
a

cos 90º° = OM = 0 = 0
OP a

tan 90º = OPMM = a = ∞∞
0

Similarly, cosec 90º = 1, sec 90º = ∞, cot 90º = 0°

278 Oasis School Mathematics-9

Table: 0° 30° 45° 60° 90°

Angles →
Ratios ↓

sin 0 11 3 1
22 2

cos 1 3 1 1
2 2 20

tan 0 1 3∞
31

Worked Out Examples

Example: 1

Find the value of : 1
2
(a) sin0° + sin30° + cos60° – tan45° (b) tan²30° – sin²60° + cot²60° – sec²45°

Solution:

(a) Here,
sin0º + sin30º + cos60º – tan45º

= 0+ 1 + 1 – 1
2 2

= 1 – 1

= 0

(b) Here, tan²30° – sin²60° + 1 cot²60° – sec²45°
2
1 )2 23)2 )1 2 – ( 2) 2 ²
= ( 3 – ( + 1 (
2 3

= 1 – 3 + 1 . 1 – 2
3 4 2 3

= 31 – 3 + 1 –2
4 6

= 4 – 9 +2 – 24 = – 1272
12
-9
= 4

Oasis School Mathematics-9 279

Example: 2

Prove that :

1 + cot 30° 1 + cos 30°
1 – cot30° 1 – cos 30°
( ) 2=

Solution: ( )1 + cot 30º° 2 Again, R.H.S. = 1 + cos 30º°
L.H.S. 1 – cos 30º
= 1 – cot 30º°
3
( )1 + 3 2 1+ 2

= 1 – 3 = 1– 3
2
= 11 + 2 3 + 3
– 2 3 + 3 2+ 3
2
= 4 + 2 3 = 2– 3
4 – 2 3
2

= 22((22 + 3) = 2 + 3
– 3) 2 – 3

= 2 + 3 ∴ L.H.S. = R.H.S. Proved.
2 – 3

Example: 3

Prove that :

1 – cos30° = sec 60° – tan60°
1+cos30°

Solution:

L.H.S. = 1 – cos 30º°
1 + cos30º

= 1– 3
2
3
1+ 2

= 2– 3
2
=
= 2+ 3 (2 – 3 )2
2 = (2)2 – ( 3)2

2 – 3

2 + 3 = (2 – 3 )2
4–3
2 – 3 × 2 – 3
2 + 3 2 – 3 = 2 – 3

= sec 60 ° – tan 60° = R.H.S.
Hence, proved.

280 Oasis School Mathematics-9

Example: 4 A

In the given right angled triangle ABC, ∠B = 90°, B 300 C
∠C = 30°, BC = 2 3cm. Find the value of AB and AC. 2 3cm
Solution:

In ∆ABC, taking ∠C as the angle of reference.

tan C = AB
BC
or
tan30º = 2AB3
or,
or, 13 = 2AB3
Again,
AB = 2 cm
or,
sin C = AACB
or,
or, sin 30° = A2C
∴
1 = A2C
2

AC = 4 cm

AB = 2 cm

AC = 4 cm

Exercise 15.3

1. Find the value of : (b) sin30°.cos30°.tan30°
(a) sin0° + cos0° – tan0° (d) 2.sin60°. tan30°
(f) cot 60°. sin30°.cos30°
(c) 6 . sin45°. cosec 60° (h) 3tan²30°+ cot²45° + cot²30°

(e) 34.cosec60°.sec30°.tan45°
(g) sec45°– cosec60° + sin60° – tan30°

(i) 1 sin²60° + 4 cos²45° + 6cot²45°
2
tan60° – tan30°
(j) 3cot²45° – cos²30° – 1 tan²60° + sin²30° (k) 1 + tan60°.tan30°
2

(l) 2cot60°. cos60°. tan60°. (m) 1 – 2 sin²45°

(n) 2tan30° cosec 30° + sec²45° – tan²60°
1 + tan²30° (o) 3 sec30° + 2 cos²45° – cosec²45°

Oasis School Mathematics-9 281

2. Prove the following.

(a) cot²45° – sin²60°– 1 tan²60° + 1 sec²45° = –1
2 8

(b) cot²30° – 2 sin²45° + 3 cos²0° = 5

(c) sin²60°.(cos²30° + cos²60°) = 3
4

2tan30° ( )(e) 1 + cos30°
(d) 1+ tan²30° = sin60° 1 + tan30° 2 1 – cos30°
1 – tan30°
=

(f) sin45°.cos30°– cos45°.sin30° = 3–1
2 2

( )1 + sin60° 1 + cot60° 2 (h) 1 + sin30° = cosec60° + tan30°
1 – sin30°
(g) 1 – sin60° = 1 – cot60°

(i) cosec²30° – sec²30° = 1 cosec60° (j) ( 11 –+ ttaann 30° )2 1 + sin 60°
1 + 2 sin60°. tan60° 3 30° = 1 – cos 30°

3. In each of the following figures, find the values of 'x' and 'y' .

(a) A (b) P (c) G
y
x x x 600
C y y

300 5cm Q 450 R F 3cm E
4cm
B

4. Using the table of trigonometric ratios, find the values of 'x' and 'y' in each of the
following figures.

(a) (b) y (c) y
x 12cm
y 560 x 450
8cm
x

8cm 270

5. In the given triangle ABC, AD⊥BC, AB = 8 3 cm, A

∠ABC = 60° and ∠CAD = 45°. Find the measurement of 450
5cm
2 3cm
8 3cm
the sides AD, BD, CD and AC.

6. In each of the following right angled triangles, find the B 600 D C
value of 'x' and 'y'. F

(a) A (b) P (c) D 8cm
y x
x 10cm 6cm

yC Qx y 4 2cm

B R E

282 Oasis School Mathematics-9

Answer

1. (a) 1 b. 1 (c) 2 (d) 1 (e) 1 (f) 1 (g) 2 6 –3
4 4 23

(h) 5 (i) 67 (j) 1 (k) 1 (l) 1 (m) 0 (n) 3 (o) 1
8 3 2

3. (a) x = 4.33 cm, y = 2.5 cm (b) x = 4 2 cm, y = 4 cm (c) x = 2 3 cm, y = 3 cm

4. (a) x = 3.63 cm, y = 7.13 cm, (b) x = 8.09 cm, y = 14.47 cm

(c) x = 8 cm, y = 8 2 cm

5. AD = 12 cm, BD = 4 3 cm, CD = 12 cm, AC = 12 2cm

6. (a) x = 60° y = 30° (b) x = 60°, y = 30° (c) x= 45°, y = 450

Assessment Test Paper

Attempt all the questions Full Marks : 20

Group 'A' [4×2= 8] A

1. (a) In the given right angled triangle ABC, 3 units 5 units

find sinθ, cosθ and tanθ. θ
4 units
4 B C
5
(b) If sin A = , find the value of cos A.

2. (a) Prove that: sin A. cot A . sec A = 1

(b) Find the value of : 2tan30°
1 – tan²30°

Group 'B' [3 × 4 = 12] A D
B b
3. In the given figure, ∆ABC and ∆ACD are two right
angled triangles. Find the value of (i) sin α (ii) tan 12 cm
α (iii) sinβ (iv) cotβ. 4 cm

α C
3 cm
4. Prove that : 1 + sin 60° = 1 + tan 30° A
cos 60° 1 – tan 30°
y

5. In the given figure, ∆ABC is a right angled triangle. Find the 12 cm
length of AC, and the value of x and y.
B x C
Projec t work 4 3 cm

Draw different right angled triangle on the graph sheet. From the graph find base and
perpendicular. Calculate the ratio of perpendicular and base. Find the angles of the triangle
without actual measurement.

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Statistics

Estimated Teaching Hours 8

Contents

• Cumulative frequency distribution

• Line graph

• Pie-chart

• Histogram

• Cumulative frequency curve or ogive

• Measures of central tendency

• Measure of patrician values (Quartiles)

Expected Learning Outcomes

At the end of this unit, students will be able to develop the

following competencies:

• To make cumulative frequency distribution table from the
given data

• To draw the line graph from the given data

• To draw the pie chart from the given data and get the
information from the given pie chart

• To draw the histogram from the given data and get the
information from it

• To draw the more than and less than cumulative frequency
curve and get the information from the given curve

• To calculate the central values, Mean, Median and Mode
from the given individual and discrete data

• To calculate the partition values Q1 and Q3 from the given
individual and discrete data

Teaching Materials

• A4 size paper, Graph sheet, Chart paper, Models of pie chart,
Histogram, etc.

284 Oasis School Mathematics-9

Unit

16 Statistics

16.1 Warm-up Activities

Discuss the following in the class and draw the conclusion.
• Collect the marks obtained by the students of your class in

Mathematics in an examination.

• What type of data is this?

• What is the average marks of the class?
• How to calculate the average marks?

16.2 Collection of Data

A set of numerical facts collected for the purpose of investigation in order to fulfill certain
aims and targets is called "statistical data". There are mainly two types of data:

(a) Primary data

Data collected by the investigator himself or herself for the first time is called primary
data. The data collection by this method is more relevant and reliable but more time
consuming. Primary data can be collected by the investigator by direct personal
interview, by oral interviews, by correspondence, questionnaires through the post, etc.

(b) Secondary data

The data collected by the secondary sources are called secondary data. For example, the
weather forecast by radio, television or newspaper, the number of population of different
localities by the statistical department etc., all are secondary sources of collection of
data. Since we have to depend upon other sources, the secondary data are not so reliable
and may not be much relevant.

Frequency table

After the collection of data by primary or secondary sources is over, they should be
represented properly for the presentation so that the investigator can work out with
the set of collected data. The data collected at the very first time are called raw data. In
order to represent them systematically, we can take the help of frequency table where tally
marks ( / ) are used to indicate the number of variables of certain type like age, mark, height.
The presentation of data in such a table is called frequency table. For example, consider the
marks of 15 different students that are collected as below in unit test of full marks 10.

1, 5, 6, 9, 2, 8, 10, 4, 7, 6, 4, 2, 8, 7, 8,

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To represent the above raw data, the following frequency table can be used where
number of repeated items are represented by the number of tally mark.

Frequency table

Variable Mark (X) Tally mark Frequency (f)
/
1 // 1
2 2
3 // 5
4 / 2
5 / / 1
6 / / 2
7 / / / 2
8 / 3
9 / 1
10 1

Note:

If the number of any variable repeats 5 times, we use cross tally over the
first four tally as (////).

Frequency distribution:

When the data are arranged in a table sharing their respective frequencies in the
frequency column, that tabular arrangement of data is called frequency distribution.
Frequency distribution can be prepared in three following ways:

(i) Individual series or ungrouped data.

(ii) Discrete series or ungrouped frequency distribution

(iii) Continuous series or grouped frequency distributions

(a) Individual Series

In the individual series, items are listed individually one by one. No frequency of the
variables is mentioned. So, it is just an ungrouped data e.g. marks of 10 students are
given as.

1, 2, 5, 6, 4, 3, 1, 7, 3 and 8.

(b) Discrete or Ungrouped frequency distribution

In this type of distribution, the variables can be only the exact and specific value
with their corresponding frequency. For example, marks of 15 students are tabulated
as below.

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Marks 123 4 5

No. of Students 3 4 5 2 1

(c) Continuous series or grouped frequency distribution

The data which are represented by continuous variables is called continuous series.
For, e.g. ,

Class Interval 0–10 10–20 20–30 30–40 40–50
No. of Students 2 3 5 1 4

The above table shows the marks of 15 students in terms of different classes. We
can observe that there are two students whose marks lie between the range 0 to 10
(exclusive). Here the lower and upper limits of the first class are 0 and 10. The difference
between them is called class height or width or size or magnitude.

∴ Class size = Upper limit – Lower limit

The mid–value of the class is obtained by the average value of upper and lower limits.

( )
∴ Mid–value = Upper Limit + Lower Limit
2

In above case, the lower limit of the class is also considered as its value while the upper

limit will be excluded and will be included for the next class. But this is not applied for

the upper most class interval.

If the distribution is inclusive (discontinuous) i.e. the upper limit of the class is not equal

to the lower limit of another class, first we have to adjust it in the from of continuous

series by the selection of correct correction factor.
( ) Here, Correction factor = Difference of lower class of any group and the upper class of previous group
2
For example:

Variables 1–4 5–9 10 – 14 15 – 19
No. of items 2 5 6 3

In above table, the lower class limit of 2nd class is 5 and upper class limit of 1st class is

4. So the correct factor

c.f. = 5 – 4 = 1 = 0.5
2 2

Hence, the above discontinuous data is changed into continuous by subtracting the

correction factor from the lower class limit and adding it to the upper class limit. Now

the distribution becomes.

Variables Class Corrected Class Frequency
1–4
0.5 – 4.5 2

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5–9 4.5 – 9. 5 5
10 – 14 9.5 – 14.5 6
15 – 19 14.5 – 19.5 3

Cumulative frequency distribution

The frequency distribution table in which frequencies are cumulated either from top
to bottom or bottom to top is called cumulative frequency distributions. There are two
types of cumulative frequency distributions.

(a) L ess than cumulative frequency distribution

(b) More than cumulative frequency distribution

(a) Less than cumulative frequency distribution

The frequency distribution in which the frequencies are cumulated from top to
bottom is called less than cumulative frequency distribution. Let us see the table given
below.

Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
7
No of Students 2 4 35
(Frequency )

Less than cumulative frequency table of the given data.

Marks No of Students Upper limit Less than cumulative
(frequency) frequency
0 – 10 2 10
10 – 20 4 20 2
20 – 30 3 30 2+4=6
30 – 40 5 40 6+3=9
40 – 50 7 50 9 + 5 = 14
14 + 7 = 21

Remember !

• Marks of 2 students are less than 10.

• Marks of 6 students are less than 20.

• Marks of 9 students are less than 30 and so on.
MarkNs ootfe9: students are less than 30 and so on.

Less than c. f. represents the frequency of less than the upper limit of
corresponding class.

288 Oasis School Mathematics-9

(b) More than cumulative frequency distribution :

The frequency distribution in which frequencies are cumulated from bottom to top is
called more than cumulative frequency distribution. The table below shows the more
than cumulative frequency distribution.

Marks No of Students Lower More than cumulative lfSirnteaeqrutbeotonttcoaymddf r othme
(frequency) Limit frequency
0 – 10
10 – 20 2 0 19 + 2 = 21
20 – 30
30 – 40 4 10 15 + 4 = 19
40 – 50
3 20 12 + 3 =15

5 30 7+ 5 = 12

7 40 7

Note:
More than c. f represents the frequency of more than the lower limit of
corresponding class.

Remember !
• Marks of 21 students are more than 0.

• Marks of 19 students are more than 10.

• Marks of 15 students are more than 20 and so on.

Worked Out Examples

Example: 1

Prepare a discrete frequency table of the marks of 21 students of grade IX in a school.

1, 5, 6, 8, 5, 3, 4, 5, 9, 2, 8, 2, 5, 10, 6, 9, 7, 6, 5, 2, 10

Solution: Marks Tally Marks Frequency (f)

1 / 1
2 /// 3
3 / 1
4 / 1
5 //// 5
6 //// 3
7 / 1
8 // 2
9 // 2
10 / 1

Σf = 21

Oasis School Mathematics-9 289

Example: 2

Daily expenses by different persons (in Rs.) are given below. Construct a continu-
ous frequency table of class interval as 0–20, 20–40, etc.
5, 12, 25, 10, 40, 60, 85, 55, 32, 15, 37, 56, 80, 95, 14,

36, 72, 65, 45, 90, 60, 33, 8, 19, 12, 35, 65, 52, 72, 91
Solution:

Class expenses Tally Marks Frequency (f)
0 – 20 //// /// 8
20 – 40 //// 5
40 – 60 //// 5
60 – 80 //// // 7
80 – 100 //// 5

Σf = 30

Example: 3

Form a (i) less than cumulative frequency table and (ii) more than cumulative
frequency table from the following data.

Wages (Rs) 0 – 100 100 – 200 200 – 300 300 – 400 400 – 500
30 10 5
No. of workers 10 15

Solution:

Less than cumulative frequency table:

Wages (Rs) No. of workers (f) Less than c.f.
10
0 – 100 15 10
100 – 200 30 10 + 15 = 25
200 – 300 10 25 + 30 = 55
300 – 400 5 55 + 10 = 65
400 – 500 65 + 5 = 70
∑f = 70

290 Oasis School Mathematics-9

More than cumulative frequency table:

Wages (Rs.) No. of workers (f) More than c.f. Start to add frequency
from the bottom.
0 – 100 10 60 + 10 = 70
100 – 200 15 45 + 15 = 60
200 – 300 30 15 + 30 = 45
300 – 400 10 5 + 10 = 15
400 – 500 5 5

Exercise 16.1

1. Construct a frequency distribution table from the marks obtained by 25
students using tally marks.

24 12 23 13 1 24 16 23 23 12 16 23

16 23 12 13 23 12 16 13 23 16 23 24

2. Following are the wages earned by 30 workers in a week. Construct frequency
distribution table using tally marks.

75 65 85 80 95 90 65 65 75 80

95 95 65 75 85 85 95 95 95 75

85 85 80 80 80 85 85 90 80 80
3. Construct grouped frequency table from the following data of class interval 5.
25 11 20 35 16 15 31 30 33 29

39 19 23 38 24 12 17 18 16 10

15 19 27 36 38 27 31 22 19 22

4. Construct a cumulative frequency table (less than and more than) from the
following grouped data:

(a) Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
No. of students 7 6458

(b) Earning in (Rs) per day 50 – 100 100–150 150–200 200–250 250–300 300–350

No. of workers 12 13 10 8 6 12

Oasis School Mathematics-9 291

5. Construct an ordinary frequency table for the following data:

(a) Age (less than) 5 10 15 20 25 30 35
No. of persons 7 12 15 22 30 35 50

(b) Height (less than) in cm. 70 80 90 100 110
No. of plants 37 10 12 15

6. Construct an ordinary frequency table from the following data:

(a) Weight (in kg) No. of persons
Above 40 100
Above 50 85
Above 60 65
Above 70 40
Above 80 15

(b) Marks (more than) 0 10 20 30 40 50
No. of students 25 22 17 15 12 5

7. Construct an ordinary frequency table from the given data:

(a) Class 10–19 20–29 30–39 40–49 50–59 60–69

Frequency 8 7 9 12 4 7

(b) Wage 10 – 14 15 – 19 20 – 24 25 – 29 30 – 34
Number of workers 56 12 8 4

Answer
Consult your teacher.

16.3 Graphical Representation of Data

The collected data in raw form are often difficult to understand. To make the data more
meaningful and more understandable the collected data are presented in the form of
graph or geometrical diagram such as line graph, pie–chart, bar–diagram, histogram,
frequency polygon, Ogive etc. In this part, we will discuss line graph, pie–chart, histo-
gram and Ogive curve.

292 Oasis School Mathematics-9

Line Graph or Time Series Graph

The graph obtained by plotting the variables along x-axis and corresponding
frequencies along y–axis and by joining all those points on the cardinal plane is
called a line graph. Let's see an example.

Marks 20 25 30 35 40 45 50
No. of Students 10 20
10 5 8 10 15

Number of students Y Steps:
• Take variable along X-axis and
50
45 frequency along Y-axis.
40 • Plot all the points.
35 • Join all the points by a straight
30
25 line.
20
15 Note: On the questions related to time,
10 take time along x–axis and values
5 of the variable along y-axis.

X' 20 25 30 35 40 45 50 55 60 65 X

Y' Marks

Example: 2
Draw a pie–chart for the following data of expenditure in a family.

Items Food Cloths Rent Education Medicine Others
10 10 15 5
Expenditure (in %) 40 20

Solution:

Angular measures of

(i) Food = 40 × 360° = 144°
100

(ii) Clothing = 20 × 360° = 72°
100

(iii) Rent = 10 × 360° = 36°
100

(iv) Education = 10 × 360° = 36°
100

(v) Medicine = 15 × 360° = 54°
100

(vi) Others = 5 × 360° = 18°
100

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Rent Education Index
36º 36º Food
Cloth
Fig:Pie-Chart Cloth Medicine Rent
72º 54º Education
Medicine
Others Others
18º

Food 144º

Histogram

It is the diagram representing the data in terms of adjacent bar diagram. The width of
bar represents the class interval while the height represents its corresponding frequency.
The following histogram shows the number of employees in an organization earning
different ranges of basic salary per month.

Salary (In Rupees) 1500 — 2000 2000 — 2500 2500 — 3000 3000 —3500
No. of workers 100 50 70 20

Solution:

Number of workers 100

90
80

70
60
50
40
30
20
10

0 1500 2000 2500 3000 3500

Salary

Histogram is constructed on continuous data. If the data are not continuous
(ungrounded), first of all we must change them into continuous distribution by the help
of correction factor which is defined earlier. The width of each bar of histogram gives
the class range, so it depends upon the range of group. Histogram is one of the best
ways to represent the grouped data for further analysis and conclusion.

Cumulative frequency curve or Ogive

The curve obtained by plotting the variable values along x–axis and cumulative
frequency along y–axis is called "cumulative frequency curve" or "Ogive". We can plot
two types of Ogive curve.

(a) Less than Ogive/ cumulative frequency curve:

(b) More than Ogive/cumulative frequency Curve:

294 Oasis School Mathematics-9