Math 10

8. In the given figure, two circles intersect at D and E, A
prove that: AFEG is a cyclic quadrilateral.
FG
9. In the given figure, AB = BC = CD = AD and ∆AEF is an E
equilateral triangle, prove that: DB//EF.
B DC
10. In the given figure, O is the center of the circle, prove
that: ∠COE – ∠BOD = 2∠CAE. B
AF

C

DE

B C
A O

D E

11. In the given figure, PQ//RS and QT = RS, prove that PR
∠PQR = ∠RQT. T

Tangent QS

1. Find the value of unknown angles.

a) b) B c) T A

A C D Cx N x O
600
x y 300 y
400 S
E B
A B
T
[Ans: x = 600, y = 600] [Ans: x = 700, y = 700] N
[Ans: x = 300]

d) D e) f) R

T C yB T 360 P
450
x0
O

x x x0
T 600 z N
A2 B 300 C A Q

[Ans: x = 800] [Ans: x = 300, y = 150, z = 450] S [Ans: x = 720]

g) A h) A i) A

350

600 C B OC xD yT

Tx x0 700

C B

B [Ans: x = 200] [Ans: x = 700, y = 400]
[Ans: x = 600]

Oasis School Mathematics-10 295

2. In the adjoining figure, AB, BC and AC are tangents A
and AB = BC. Prove that AR = CR.
PR
O

B QC

3. The circle touches the sides of ∆ABC, AB, BC and B QC

CA at points P, Q and R respectively. Prove that:

AP + BQ + CR = PB + QC + RA = 1 (AB + BC + AC.) PR
2
A
4. In the given figure, from any external point P, PS and
PT are tangents drawn to the circle with the center at S
O. Prove that OP and QR are perpendicular to each Q P
other.
OM
R

B

5. In the given figure, FA and FB are tangents from F to F CA
a circle having center O. CED is a tangent drawn at EO
point E. Prove that:
DB
(i) CA + DB = CD. (ii) FA + FB = FC + FD + CD.

6. In the given figure, TA and TB are two tangents A
drawn from an external point T. TO

Prove that: (i) ∠ATB = ∠OAB +∠OBA. B

(ii) ∠ATB = 2∠OAB. A
B
7. In the given figure, two circles meet externally
at point C. AB is a common tangent at A and B. C

Prove that: ∠ACB = 90°.

296 Oasis School Mathematics-10

Full Marks: 24

Attempt all the questions:

Group A [3 × 1 = 3]

1. (a) ∆ABC and ∆DBC are standing on the same base BC and between the parallel lines
AD and BC. If ∆ABC = 40 cm2, find the area of ∆DBC.

(b) ABCD is a cyclic quadrilateral, then what is the sum of ∠A and ∠C.?

(c) In the given figure, TAN is a tangent at A and
∠ATO = 400, find the value of ∠AOT.

Group B [4×2=8] P S
Q RT
2. (a) In the adjoining figure, ∆PST and square PQRS both
lie on the same base PS and between the parallel lines
PS and QT where the area of ∆PST = 36 cm2. Find the
length of SQ.

(b) In the given figure, OPQR is a parallelogram. Find O P
the values of x and y.
y
3. (a) In the given figure, AB is a diameter and ABCD is
a cyclic quadrilateral. If ∠ABD = 300, find the value x
of x. RQ

(b) In the given figure, TA and TB are two tangents DC
drawn from the point T. If ∠ATB = 500, find the
value of x. x

A 300 B

O

A 500
Ox
T
B

Oasis School Mathematics-10 297

Group C [4×4=16]

4. Parallelogram ABCD and ∆EBC both lie on the same base BC and between the parallel
lines ED and BC, prove that ∆EBC = 1/2 parallelogram ABCD.

AC

5. In the given figure, AB//CD, ∠ABC = ∠CBE. E
Prove that BE = CD. BP

D

6. Verify experiementally that opposite angles of a cyclic quadrilateral are supplementary.

7. Construct a parallelogram ABCD in which AB = 5.9 cm, ∠ABC = 750 and BC = 6.8 cm.
Also construct a triangle ABF equal in area to parallelogram ABCD.

AB

Group D [5 × 1 = 5]

8. In the given figure, ABCD is a parallelogram, AF and BC DF C
are produced to meet at E.

(i) Prove that ∆DEF = ∆BFC E

(ii) If the area of parallelogram ABCD = 40 cm2 and area of ∆ADF = 12 cm2,
find the area of ∆DFE.



298 Oasis School Mathematics-10

Trigonometry

10Estimated Teaching Hours

Contents

• Area of Triangle
• Height and Distance

Expected Learning Outcomes

At the end of this unit, students will be able to develop the following
competencies:
• To find the area of a triangle and parallelogram using the formula

of trigonometry
• To solve the problems of height and distance using the formula

of trigonometry

Materials Required

• Protractor, field tape, chart paper, A4 size paper.

Oasis School Mathematics-10 299

Unit

16 Trigonometry

16.1 Warm-up Activities A

Discuss the following in your class and draw a conclusion.

In the given right-angled triangle, 5 3cm

• find the length of AC. P B q C
• find the value of 'q'. 5cm

In the given right-angled triangle,

• find the length of PQ and PR. Q 600 R

15cm

• How to calculate a side and two angles of a right-angled triangle if two sides are given?
Explain with an example.

• How to calculate the two sides and an angle if an angle and a side of a right-angled
triangle are given.

16.2 Area of Triangle A

i. When the base and height of a triangle are given: DC
A
In this case, we can calculate the area of a triangle by using,
b
Area of triangle = 1 base × height.
2

∴ In ∆ABC, if AD⊥BC.

Area of ∆ ABC = 1 BC ×AD. B
2

ii. When the three sides of a triangle are given: (Heron's formula)

i.e. BC = a, AC = b, AB = c,

Then, area of ∆ABC c
= s(s − a)(s − b)(s − c)

Where, s = Semi-perimeter of ∆ABC B aC

= a+b+c
2

iii. When two sides and the angle contained by them are given: (Heron's formula)

In ∆ABC, BC = a, AC = b, AB = c,

If ∠B is the given angle, AD is drawn perpendicular to BC.

300 Oasis School Mathematics-10

In ∆ABD, Sin ∠ABD = AD A
Sin B = AB
AD = AD cb
or, AB
or, BDC
AB Sin B
a
= c Sin B.
A
Now, area of ∆ABC = 1 BC. AD D
2 C

= 1 ac. Sin B.
2

Area of ∆ABC = 1 ac Sin B …………….. (i)
2

Again, if ∠C is the given angle,

Sin ∠ACD = AD
or, Sin C = AC

AD
AC

or, AD = AC. Sin C

= b. Sin C.

Now, area of ∆ABC = 1 . BC. AD.
2

= 1 . a. b. Sin C
2

∴ Area of ∆ABC = 1 ab Sin C ………… (ii)
2

If ∠A is the given angle, BD is drawn perpendicular to AC.

In ∆ABD, Sin A = BD
Sin A = AB

or, BD = BD
or, c
c Sin A.

Now, A rea of ∆ABC = 1 AC × BD B
2

= 1 bc. Sin A.
2

= 1 bc Sin A.
2

∴ Area of ∆ABC = 1 bc Sin A ………… (iii)
2

From equations (i), (ii) and (iii)

1 ac Sin B = 1 ab Sin C = 1 bc Sin A.
2 2 2

Dividing each by 1 abc
2

Oasis School Mathematics-10 301

SinB = SinC = SinA
Taking reciprocals, b c a

a = b = c
SinA SinB SinC

The sides of a triangle are proportional to the Sine of the opposite angles.
This relation is known as the Sine law.

Note: Equations (i), (ii) and (iii) are also valid if the given angle is obtuse-angled.

Area of Parallelogram

Area of parallelogram ABCD = 2 (Area of ∆ABC) [∵ Diagonal bisects the parallelogram]

= 2 × 1 AB × BC. Sin B.
2

= AB × BC Sin B. AD
Therefore,

∴ Area of parallelogram ABCD = AB × BC. Sin B.

Similarly, area of parallelogram ABCD = BC × CD. Sin C.

= AB × AD. Sin A. B C

= AD × CD Sin D.

Worked Out Examples

Example: 1

Find the area of ∆ABC in the given figure.

Solution: Here, A

AB = 5 cm.

BC = 8 cm. 5 cm 5cm

Now, using formula ∠B = 30°

Area of ∆ABC = 1 AB × BC Sin 30° B 300 C
2 8cm

= 1 × 5 × 8 × 1 cm²
2 2

= 10 cm² P

Example: 2

Find the area of ∆PQR in the given figure.

Solution: Here,

Q 750 R

302 Oasis School Mathematics-10

PQ = PR = 5 cm.

∠Q = ∠R = 75° [Base angles of an isosceles triangle]

or, ∠P + ∠Q + ∠R = 180° [Sum of three angles of a triangle]
or, ∠P = 180° – 150°
or, ∠P = 30°

Now, 1
2
Area of ∆PQR = × PQ × PR. Sin P.

= 1 5 × 5 × Sin 30°
2

= 1 × 5 × 5 × 1 cm2
2 2

= 25 cm2
4

∴ Area of ∆ PQR = 6.25 cm2

Example: 3

In the given figure, AD is a median of ∆ABC, AB = 8 cm and AD = 6 cm. Find the area of

∆ABC. A
Solution: Here,

AB = 8 cm, ∠BAD = 30°, AD = 6 cm, 300
8 cm
Area of ∆ABD = 1 AB × AD. Sin ∠BAD 6 cm
2
B DC
= 1 8 × 6 × Sin 30°
2

= 1 8 × 6 × 1
2 2

= 12 cm2

Since AD is the median,

Area of ∆ABC = 2 ∆ABD [Median divides a triangle into two triangles

of equal area]

= 2 × 12 cm2
= 24 cm2
∴ Area of ∆ABC = 24 cm2

Example: 4

In the given figure, AB = 10 cm, BC = 12 cm and area of ∆ABC = 30 3 cm2. Find the value of ∠ABC.

Solution: A

Here, AB = 10 cm, BC = 12 cm, area of ∆ABC = 30 3 cm2 10 cm

We have, B?

Area of ∆ABC = 1 × AB × BC × Sin ∠ABC. 12 cm C
2

Oasis School Mathematics-10 303

or, 30 3 = 1 × 10 × 12 × Sin ∠ABC
2

or, 30 3 = 60 × Sin ∠ABC

or, Sin ∠ABC = 30 3
60

or, Sin ∠ABC = 3

2

or, Sin ∠ABC = Sin 60°

∴ ∠ABC = 60°

Example: 5

In the parallelogram ABCD, AD = 6 cm, AB = 4 cm, ∠BCD = 60°. Find the area of the

parallelogram ABCD. DC
Solution:
600

Here, AB = 6cm, AD = 4cm, 6cm

∠A = ∠C = 60° [Opposite angles of a parallelogram]

Area of parallelogram ABCD = AB × AD . Sin ∠BAD A 4cm B

= 6 × 4 × Sin 60°

= 6 × 4 × 3 cm2

2

= 12 3 cm2

∴ Area of parallelogram ABCD = 12 3 cm2

Example: 6

In the given figure, ABCD is a rhombus whose area is 50 cm2. If ∠ADC = 300, find the length

of each side. AB
Solution:

Let AD = DC = AB = BC = a

Area of rhombus ABCD = AD × DC Sin 300­ 300 C

1 D
2
50 = a × a ×

or a2 = 100

or a = 10 cm.

∴ Length of each side of rhombus = 10 cm

Example: 7

In the given figure, ABCD is a rhombus and ∆DPC is an AD
equilateral triangle. Find the area of the trapezium ABPD.
Solution: 600

Here, ABCD is a rhombus and ∆DPC is an equilateral triangle. B C P

∠DCP = ∠ABC = 60°, BC = AB = AD = CD = CP = DP = 6 cm.

304 Oasis School Mathematics-10

Now,

Area of rhombus ABCD = AB × BC × Sin ∠ABC.

= 6 × 6 × Sin 60°

= 6×6× 3 cm2

2

= 18 3 cm2

Again,

Area of equilateral triangle DPC = 1 CD × CP × Sin ∠ DCP
2
1
= 2 × 6 × 6 × Sin 60° [∴ ∠DCP = ∠ABC ]

= 1 × 6 × 6 × 3 cm2
2
2

= 9 3 cm2

∴ Area of trapezium ABPD = Area of rhombus ABCD + Area of ∆ DPC

= 18 3 cm2 + 9 3 cm2

= 27 3 cm2

Example: 8

In the given figure, from the given information, find the area of parallelogram ABCD.

Solution:

Here, ABCD is a parallelogram

AB = 8 cm, AE = 9 cm, ∠AED = 60°.

Now, ∠AED +∠EAB = 180° [Sum of the co–interior angles]

or, 60° + ∠EAB = 180° E D C

or, ∠EAB = 180° – 60° 600

∴ ∠EAB = 120° 9cm

Now, Join EB.

Area of ∆ EAB = 1 AB × AE× Sin ∠EAB A 8 cm B
2

= 1 8 × 9 × Sin 120°
2

= 1 × 8 × 9 × 3 cm2
2
2

= 18 3 cm2

Again,

area of parallelogram ABCD = 2 ∆EAB. [∴ Triangle and parallelogram on the same base and
between the same parallel lines]

= 2 × 18 3 cm2
= 36 3 cm2

Oasis School Mathematics-10 305

Example: 9

In the given figure, AB = BC = 6 cm, AD = CD, BD = 12 cm, ∠ABD = 60°. Find the area of

quadrilateral ABCD. A
Solution:

Here, AB = BC, AD = CD, ∠ABD = 60°

∴ ∆ABD ≅ ∆BCD. B 600 12cm D

Area of ∆ABD = Area of ∆ BCD. 6cm

Now, Area of ∆ ABD = 1 AB × BD × Sin ∠ABD. C
2

= 1 × 6 ×12 × sin 600
2

= 1 × 6 ×12 × 3 cm2. = 18 3 cm2
2
2

∴ Area of quadrilateral (kite) (∵ longest diagonal bisects the kite)

= 2 ∆ABD

= 2 × 18 3 cm2

= 36 3 cm2
Example: 10

In the adjoining trapezium ABCD, BC = 2 AD = 18 cm, BD = 12 cm ∠DBC = 30°, AD||BC.

Find the area of the trapezium ABCD.

Solution: AD

Here, BC = 18 cm, BD = 12cm, ∠DBC = 30° 12 cm

Area of ∆DBC = 1 × BD × BC. Sin ∠DBC B 300 C
2 18 cm

= 1 × 12 × 18 × Sin 30°
2

= 1 × 12× 18 × 1 cm2
2 2

= 54 cm2

Again, BC = 2AD

AD = 1 BC [Given]
2

= 1 × 18 cm
2

= 9 cm

∠DBC = ∠ADB = 30° [Being AD||BC]

Area of ∆ABD = 1 AD × BD Sin ∠ADB
2

= 1 × 9 × 12 × Sin 30°
2

= 1 × 9 × 12 × 1 cm2 = 27 cm2
2 2

306 Oasis School Mathematics-10

∴ Area of trapezium ABCD = Area of ∆ABC + Area of ∆ABD
= 54 cm2 + 27 cm2
= 81 cm2
∴ Area of trapezium ABCD = 81 cm2

Exercise 16.1

1. Find the area of each of the following triangles.

(a) A (b) P (c) A
300
5 cm 6 cm 450

B 300 CQ R B C
6 cm
4 cm

(d) (e) P (f) A

P 5 2cm Q
450 10 2cm

6 cmR 450 6 3cm

6 cmR Q B 450 C
3 cm
2. Find the area of each of the following triangles.

(a) A A (c) A

(b)

D
300 D
5 cm 3 cm
5 cm 600
C
BDC B 5 cm
300
B 12 cm C

3. Find the area of each of the following parallelograms.

(a) D C (b) D C (c) D C

10 cm 10cm

A 600 B A 450 B A 600 B
8 cm
8 cm

(d) S R (e) S R (f) A E B
C
12 cm 300 1000 5 3cm

6 cm 12 cm

P Q P 500 Q D 600 10 3cm

8 cm



Oasis School Mathematics-10 307

4. (a) In the given triangle ABC if AC = 8 cm, C
∠BAC = 30° and the area of ∆ABC = 24 cm2, find
the length of AB. 8cm

(b) In ∆ABC if BC = 9 cm, ∠C = 60° and the area of B 300 A
∆ABC = 27 3 cm2, find the length of the side AC. B

18cm 5cm 7cm 8cm 9cmC 600 A
(c) In the given figure, area of ∆PQR = 75 cm2, P

∠PRQ = 300. Find the length of PR. Q 300 R

5. (a) If the area of ∆PQR = 8 cm2, PQ = 4 cm, QR = 8 cm, 20cm
calculate the magnitude of ∠PQR.
P

4cm

R 8cm Q

C

(b) If the area of ∆ABC = 20 2cm2,
AB = 10 cm, AC = 8 cm, find the value of ∠BAC.

A 10cm B
P

(c) In the adjoining figure, PR = 7 cm, QR = 16cm and
area of ∆PQR = 28 3 cm2. Find the value of PRQ.

R 16cm Q
A

(d) In the given figure, AB=5cm, BC = 8cm and area B 400 C
of ∆ABC = 10 3cm2, ∠ACB = 400. Find the value of
∠BAC. 8cm

AB

6. (a) The adjoining figure is a parallelogram ABCD. If D 600 C
its area is 180 cm2, AD = 18 cm and ∠ADC = 60°, D C
calculate the length of DC.
600 B
(b) In the given figure, the area of the rhombus ABCD is
18 3 cm2. Find the length of the sides of the A
rhombus.

308 Oasis School Mathematics-10

(c) In the given parallelogram ABCD, DC = 10 cm, AB
∠BDC = 30°, area of parallelogram ABCD is
D 300 C
90 cm2. Find the length of BD. A B

(d) In the given figure, ABCD is a parallelogram, D 600 C
DC = 10 cm, ∠ACD = 600 and area of parallelogram D
ABCD = 60 3 cm2. Find the length of AC. 10cm

7. (a) In the given figure, ABCD is a rhombus. Area of A
the rhombus = 18 3 cm2. Find the value of ∠ABC.
B C
(b) In a rhombus ABCD, if AB = 12 cm and the area of D 6cm C
the rhombus ABCD = 72 cm2, find the measure of
∠BAD. AB

12cm

(c) The area of a parallelogram ABCD is 48 square cm. D C
If AB = 12 cm, BC = 8 cm, find the value of ∠ABC.
8cm
12cmA 12cm B
P Q
(d) In the adjoining figure, PQRS is a parallelogram.
SR = 5 cm, QS = 8cm and area of parallelogram 8cm
PQRS = 20 3 cm2. Find the value of ∠PQS.
S 5cm R

(e) In the given figure, PQRS is a parallelogram. P TQ
Area of parallelogram PQRS = 48 cm2, TS = 12cm,
SR = 8cm and ∠PST = 500. Find the value of ∠PQR. 500

S 8cm R
A 10cm

8. (a) In the given figure AD = DC, AB = BC = 10 cm,
BD = 16 cm, ∠ABD = 30°. Find the area of the kite
ABCD. D 300 B

C D

A

14cm
300

(b) From the given figure, find the area of the

quadrilateral ABCD. B 450
12 cm
C

Oasis School Mathematics-10 309

9. (a) In the given quadrilateral ABCD, AB = 2 3 cm, A 8cm D

AC = 6 cm, AD = 8 cm and the area of the quadri- 2 3cm 600 C
lateral ABCD is 15 3 cm2. Find the value of ∠BAC. 6cm A

(b) In the given figure, AC = 16 cm CD = 12 3 cm , B 300

3 B 16cm
4
∠ACD = 60°. If the area of ∆ABC = ∆ACD, find

the length of AB. 600 D

C 12 3cm

A

10. (a) In the given figure, AD ⊥ BC, AD = DC. 9 2cm
AC = 9 2 cm. Find the area of ∆ABC.

600 D C

B

(b) In the given figure, ∆OQP is an equilateral triangle ON

in which OQ = 6cm. MNOQ is a rhombus. Find the

area of figure MNOP. P QM

(c) In the given figure, ABCD is a rhombus. BE⊥DE . If AD
∠ABC = 600, DC = 30 cm and BE = 54 cm, find the
area of the trapezium ABED. 600 900

B CE

(d) In the given figure, ABC is an equilateral trian- A

gle and ACD is a right angled triangle where

∠D=900. If∠CAD=300, AC=6cm,findtheareaofthe B D
quadrilateral ABCD. C

Answer

1. (a) 7.5 cm2 (b) 12.73 cm2 (c) 6.93 cm2 (d) 7.5 cm2 (e) 50cm2 (f) 54 cm2 2. (a) 15 cm2 (b) 30 cm2
(c) 12.99 cm2 3. (a) 69.28 cm2 (b) 33.94 cm2 (c) 86.60 cm2 (d) 48 cm2 (e) 36 cm2 (f) 129.90cm2
4. (a) 12 cm (b) 12 cm (c) 15 cm 5. (a) 300 (b) 450 (c) 600 (d) 800 6. (a) 11.55 cm (b) 6 cm (c) 18 cm
(d) 12 cm 7. (a) 600 (b) 300 (c) 300 (d) 600 (e) 800 8. (a) 80 cm2 (b) 108.4 cm2 9. (a) 300 (b) 27 cm
10. (a) 63.88 cm2 (b) 46.77 cm2 (c) 630 3 cm2 (d) 23.38 cm2

Do you know !
Greek mathematician Hipparchus is known as the father of trigonometry.

310 Oasis School Mathematics-10

16.3 Height and Distance

Height and distance is the application of trigonometry. In height and distance, we can find the
height of an object or a distance of an object from a point without measuring its length.

In height and distance, we have to solve right-angled triangle making height and distance as

two sides of a triangle.

Here are some basic terms which are used to solve the problem of height and distance.

Angle of elevation

If an object A is observed from the position O, O is the position of the eye A

of the observer. Line OA is called the line of sight. OB is the horizontal Line of sight
line. Then ∠AOB is the angle of elevation of A from O.

Therefore, when the position of the object lies above the eye of an observer, angle of B
the angle made by the line of sight with the horizontal line through the O elevation
Horizontal line

position of the eye is called the angle of elevation.

Angle of depression horizontal line B
angle
If an object ‘A’ is observed from position O, O is the position of the O Line of sight of depression
eye of the observer. Line OA is the line of sight, OB is the horizontal
line through O. Then ∠AOB is the angle of depression of ‘A’ from O. A

When the position of the object lies below the eye of the observer,

the angle made by the line of sight with the horizontal line through the position of the eye is
called the angle of depression.

Worked Out Examples A

Example: 1

The angle of elevation of the top of a house at a distance of 20 m from its
foot on a horizontal plane is found to be 60°. Find the height of the house.

Solution:

Let, AB be a house, C the position of the observer, BC the distance between C 600 B
20 m
the foot of the house and the observer.

Here, angle of elevation, ∠ACB = 60°

BC = 20 m
AB = ?
In right angled triangle ABC
AB
Tan 60° = BC

or, 3 = AB
BC

Oasis School Mathematics-10 311

or, AB = 20 3m
or, AB = 20 × 1.732 m
or, AB = 34.64 m
∴ Height of the house = 34.64 m.

Example: 2
From the top of the tower 100 m high, the angle of depression of an object on the ground is

30°. Find the distance between the object and the foot of the tower. A D
Solution:
300

Let AB be a tower and C the position of the object. BC is the distance 100m 300 C
between the object and the foot of the tower.

Here, ∠DAC = angle of depression of C from A B

Since, DA||CB. ∠DAC = ∠ACB = 30°


AB = 100 m

BC = ?

In the right-angled triangle ABC, AB
BC
Tan 30° =
100
1 BC
or, 3 =

or, BC = 100 3m

or, BC = 100 × 1.732

or, BC = 173.2 m

Hence, distance between the tower and
the object = 173.2 m.

Example: 3

The length of the shadow of a pole 3 m high is 3 3 m. What is the altitude of the sun? At the
same time, if the length of the shadow of a house is 30 3 m, find the height of the house.

Solution:

Let AB be a pole and BC its shadow. ∠ACB is the altitude of the sun. At the same time, QC is
the length of the shadow of the house PQ

Here, AB = 3m, BC = 3 3m, QC = 30 3m A P

In ∆ ABC, 3m ?
θ
Q
C 3 3m B

30 3m

312 Oasis School Mathematics-10

Tan q = AB Again
or, Tan q BC
or, Tan q In∆PQC
or, Tan q
=3 tan θ = APQB
33 tan 300 = QBCC

=1 PQ
3
30 3

= Tan 300 1 = PQ
3 30 3

∴ PQ = 30m

∴ Altitude of the sun = 30° ∴Height of the house = 30m.

Example: 4 C

A man 1.6 m tall observes the angle of elevation of building to be 60°. If E
the distance between the foot of the house from the observer is 60 m, find D
the height of the building.

Solution: 600
60m
Let, AB be the man, CD a building, BD the distance between observer and A

the building. 1.6m

Here, ∠CAE = Angle of elevation of C from A = B

60°

BD = AE = 60 m

ED = AB = 1.6 m

In right angled triangle AEC, CE
AE
Tan 60° =

or, 3 = CE
60

or, CE = 60 3 m.

= 60 ×1.732 m

= 103.92 m.

∴ Height of the building = CD

= CE + ED = 103.92 +1.6 m = 105.52 m.

Example: 5 A

A kite is flying in the sky. If a 300 m string is let out and if it makes 300m ?
an angle of 45° to the horizontal line, find the height of the kite from 450 B
the ground.

Solution: C
Let A be the kite. AC is the length of the string of the kite.

AB is the vertical height of the kite. ∠ACB is the angle made by the string with the horizontal
line.

Oasis School Mathematics-10 313

Here, AC = 300 m

∠ACB = 45°

AB = ?

In ∆ ABC, AB
AC
Sin 45° =
AB
or, 1 = 300

2

or, AB = 300 m = 300 × 2 m = 150 2m = 212.13m
2 2

∴ Height of the kite is 212.13 m.

Example: 6

The top of a tree, which is broken by the wind, makes an angle of 60° with the ground at
a distance of 3 3m away from the foot of the tree. Find the height of the tree before it was
broken.

Solution:

Let BC be the height of the tree before it was broken, AC the broken part of the tree, D a point

on the ground where the top of the broken part touches. C

Here, BD = 3 3m

∠ADB = 60°

In ∆ right angled ADB, A

Cos60° = BD
or, AD
1 =
2 3
600
AD
D 3 3m
or, AD = 6 3 m. B

∴ Length of the broken part of the tree = AC = AD

= 6 3 m.

Again, In ∆ABD Tan 60° = AB
BD

or, 3 = AB
33

or, AB = 9 m.

∴ Height of the tree before it was broken.

= AB + AC

= (9 + 6 3 )m

= (9 + 6 × 1.732) m

= 9 + 10.392

= 19.392

= 19.39 m.

314 Oasis School Mathematics-10

Exercise 16.2

1. (a) The angle of elevation of the top of a building from a point 40 m away from the
bottom of the building is found to be 60°. Find the height of the building.

(b) From the top of a tower 90 m high, the angle of depression of an object on the
ground is observed to be 30°. Find the distance between the object and the foot of
the tower.

(c) A tree 60 m high on the bank of a river subtends an angle of 60° on the opposite
bank. Find the breadth of the river.

(d) A man observes the top of a tower 80 3 m in height from a distance 240 m from the
foot of the tower. Find the angle of elevation.

2. (a) Find the length of the shadow of a vertical pole of height 40 ft, if the sun's rays are
inclined at an angle of 45° with the ground.

(b) When the sun's altitude is 30°, the length of the shadow of a pillar is 5 m. Find the
height of the pillar.

(c) The shadow cast by a tree of height 8 3 m is 8 m long. Find the altitude of the sun.

(d) The length of the shadow of a pole 3 m is 3 m. At the same time, find the length
of the shadow of a tower 30 m high.

3. (a) A ladder leans against a wall, with its foot 20 m from the bottom of the wall. If it
makes an angle 60° with the ground, find the length of the ladder.

(b) A ladder of length 20 m is leaning against a vertical wall. The ladder makes an
angle of 30° with the wall. Calculate the height up to which the ladder reaches on
the wall.

(c) A boy is flying a kite. The stretched part of a string, which is 60 m long makes an
angle of 60° with the horizontal line. Find the vertical height of the kite.

4. (a) A man of height 1.5 m observes the angle of elevation of the top of a pole situated in
front of him and finds to be 60°. If the height of the pole is 121.5 m, find the distance
between the pole and the man.

(b) A man stands 50 m away from the foot of a pole. He finds the angle of elevation to
be 30° while observing the top of the pole. If his eyes are 1.5 m above the ground,
find the height of the pole.

(c) A man of height 1.5 m, standing 48 m away from the foot of a building, find the
angle of elevation of the top of the building to be 30°. Find the height of the building.

(d) A boy of height 1.1 m was flying his kite. When the length of a string of the kite was
33 m, it made an angle of 30° with the horizon. What is the height of the kite from the
ground?

Oasis School Mathematics-10 315

(e) A man observes the top of a pole of height 52 m, situated in front of him and finds
the angle of elevation to be 300. If the distance between the man and pole is 87 m,
find the height of the man.

(f) A girl having height 1.54 m is 30 m away from the tower whose height is 53.5m.
Find the angle of elevation of the top of the tower from her eyes.

5. (a) From the top of a building, the angle of elevation of the top of a tower is found to
be 30°. If the distance between the building and the tower is 120 m, find the height
of the building, if the height of the tower is 100 m.

(b) The angle of elevation from the roof of a house to the top of a tree is found to be 300.
If the height of the house and tree are 8 m and 20 m respectively, find the distance
between the house and the tree.

(c) From the top of a tower the angle of depression of the roof of a house 20 m high and
60 m away from the tower was observed and found to be 600. Find the height of the
tower.

(d) Two pillars of height 115.36 m and 150 m are at a distance of 60m. Find the angle of
elevation of the top of the second from the top of the first.

(e) A boy was flying his kite from the roof of a house having height 12.5 m. The kite
was flying at a height of 42.5 m from the ground. If the string of the kite makes an
angle of 30° with the horizon, find the length of the string.

6. (a) The upper part of a tree broken by the wind makes an angle of 60° with the
horizontal line at a distance of 15 m from the foot of the tree. Find the height of the
tree before it was broken.

(b) A tree of height 24 m is broken by the wind such that the broken part of the tree
makes an angle of 30° with the ground. Find the length of the broken part of the
tree.

(c) The upper part of a tree broken by the wind is 18 m long. If it touches the ground
at a distance of 9 m from the bottom of the tree, find the angle made by the broken
part of the tree with the ground and the height of the tree before it was broken.

7. (a) The diameter of a circular pond is 120 m. The angle of elevation of the top of a pillar
situated in the middle of the pond, observed from the edge of the pond, is found to
be 60°. Find the height of the pillar above the surface of water.

(b) A pillar is fixed at the center of a circular meadow of diameter 60 m. The angle
of elevation of the top was found to be 60°, when observed from a point on the
circumference of the circular meadow. Find the height of the pillar from the ground.

(c) A pole of height 80 3 m is fixed at the centre of a circular pond. The angle of
elevation of the top of the pole from a point on the circumference of the pond is
found to be 300. Find the diameter of the pond.

316 Oasis School Mathematics-10

(d) At the centre of a circular pond, there is a pole of 11.62 m height above the surface
of the water. From a point on the edge of the pond, a man of 1.62 m height observed
the angle of elevation of the top of the pole and found it to be 30°. Find the diameter
of the pond.

(e) The diameter of a circular pond is 100m and a pillar is fixed on the centre of the
pond. A person finds the angle of elevation of the top of the pillar is Q from the
bank of the pond. If the depth of the pond is 1.5m and total height of the pillar is
51.5m, find the value of Q.

8. (a) A flagstaff stands on the top of a tower. The angles of elevation of the top and the
bottom of a flagstaff as observed from a point 20 m away from the foot of the tower
are found to be 60° and 45° respectively. Find the height of the flagstaff.

(b) A flagstaff stands on the top of a tower. The angle subtended by the flagstaff and
the tower at the point 40 m away from the bottom of the tower are 150 and 300
respectively. Find the height of the flagstaff.

(c) The angle of elevation of the top of the incomplete house of height 60 m is found
to be 45°. By how much should the height of the house be increased so that the
elevation of the top from the same point is 60°.





Answer (b) 155.88 m (c) 34.64 m (d) 300
1. (a) 69.28 m (b) 2.89 m (c) 600 (d) 17.32 m
2. (a) 40 ft. (b) 17.32m (c) 51.96 m 4. (a) 69.28 m
3. (a) 40m (c) 29.21 m (d) 17.6 m (e) 1.77 m (f) 600
(b) 30.37 m (b) 20.78m (c) 123.92 m
5. (a) 30.72 m (e) 60 m 6. (a) 55.98 m (b) 16 m
(d) 300 7. (a) 103.92 m (b) 51.96 m (c) 480 m
(c) 600, 33.59 m 8. (a) 14.64 m (b) 16.9 m (c) 43.92 m
(d) 34.64 m (e) 450

Project Work

Find the indication of the sun's rays at different times of the day by using your height and

length of your shadow.

Time length of shadow tan q = your height q

your shadow

Oasis School Mathematics-10 317

Miscellaneous Exercise X

Area of Triangle and Parallelogram 450
1. In the given figure, XY = YZ, ∠YXZ = 450 and area of
Y Z
∆XYZ = 36 2 cm2. Find the length of XY. (Ans: 12 cm) P
6 2cm
2. In the given figure, PQ = 6 2 cm, QR = 10 cm, ∠RPQ = 750 and 750
area of ∆PQR = 30cm2. Find the value of ∠PRQ . (Ans: 600)

R 10cm Q

3. If the area of a parallelogram ABCD is 40 square 8cm AD
cm, AB = 8cm and diagonal AC is 10 cm, find the
10cm

measure of ∠BAC. (Ans: 300) BC

4. In the figure given, ∆ABC is an equilateral triangle A E
and ACDE is a parallelogram. If AB = 6 cm, C 10cm D
6cm

CD = 10 cm, find the area of the trapezium ABDE. B

(Ans: 67.54cm2)

5. In the given figure alongside, ABCD is a parallelogram and A D

∆DCE is an equilateral triangle. If BC = 2AB and area of

trapezium ABED is 125 3 cm2, find the length of AB. B C E
4

(Ans: 5 cm)

Height and Distance

1. The angle of elevation of a tower from the two places 20 3 m apart are 45°and

30°respectively. Find the height of the tower. [Ans: 47.32 m]

2. The shadow of the pole is 7 m longer when the sun's altitude is 45° than when it is 60°.

Find the height of the tower. [Ans: 16.56 m]

3. From the top of a tower 100 m high, the angle of depression of two objects on the same

side of the horizontal plane at the foot of the tower is 30° and 45°. Find the distance

between the objects. [Ans: 73.2 m]

4. A pillar is fixed in the centre of a circular meadow whose circumference is 84π cm. If the

angle of elevation of the top of the pillar from a point on the circumference be 300, find

the height of the pillar. [Ans: 24.25m]

318 Oasis School Mathematics-10

5. A pole is fixed at a corner of a rectangular meadow having dimension 16 m × 12 m. If the
top of the pole is observed from the opposite corner, the angle of elevation is 600. Find

the height of the pole. [Ans: 34.64 m]

6. The height of a tower is three times the height of a house. The angle of elevation of the
top of the tower from the top of the house be 450. If the distance between the house and

tower be 60m, find the height of the house and the tower. [Ans: 30 m, 90 m]

Attempt all the questions: A Full Marks :18
Group A [3×2= 6]
600 B
1. (a) Find the area of given rhombus ABCD.
D 15 3m

C

A

(b) If the area of ∆ABC in the given figure is 27cm2, find 5 3cm 9cm
the value of ∠ABC.

BC

6 2m

A Q
P

(c) In the given figure, PQRS is a parallelogram. Its area 200 R
is 60 cm2. If AS = 5 3 cm and SR = 8 cm, find the value
of ∠PSR. S

8cm

Group B [3×4=12]

2. A boy who is 1.75m tall is flying a kite. When the length of the string of the kite is 150m,
it makes an angle of 300 with the horizon. What is the height of the kite from the ground?

3. From the top of a tower 80m high, the angle of depression of the top of a pole 20m high lying
in front of the tower is found to be 450. Find the distance between the tower and the pole.

4. If the top of a tree which is broken by the wind, makes an angle of 600 with the ground
at a distance of 15 3 m away from the foot of the tree, find the height of the tree before it
is broken by the wind.

Oasis School Mathematics-10 319

Statistics

17Estimated Teaching Hours

Contents

• Measures of central tendency
• Mean
• Median and Quartiles
• Cumulative Frequency Curve

Expected Learning Outcomes

At the end of this unit, students will be able to develop the following
competencies:
• To convert raw data into grouped data
• To find the mean from grouped data
• To find the median and quartile from grouped data
• To draw a less than type ogive curve
• To read the less than type ogive curve
• To calculate median class and quartile classes with their respective

frequency from the given less than type ogive curve

Materials Required

• A4 size paper, chart paper.

320 Oasis School Mathematics-10

Unit

17 Statistics

17.1 Warm-up Activities

Discuss the following in your class and draw a conclusion.
• Identify the type of given distribution
(i) 25, 27, 28, 30, 35, 40
(ii)

x 10 12 14 16 18
f 52 7 8 5
(iii)

Class 0-10 10-20 20-30 30-40 40-50
Frequency
57 38 6

In number (ii), frequency of 12 is 2, what does it mean?

In number (iii), frequency of (30-40) is 8, what does it mean?

Marks obtained by 20 students in a class are given below.

25, 27, 30, 35, 20, 26, 39, 43, 42, 45,

48, 50, 38, 37, 29, 28, 33, 37, 41, 43.

How do you prepare the frequency distribution table from the above data?

17. 2 Measures of Central Tendency

Statistics deals with the study of estimation. If we are working out with different
variables of the same kind, there can be a single value representing all of them. The
value so chosen stands for all of them and is known as the central tendency.

The following are the measures of central tendency:

i) Mean

ii) Median

iii) Mode

Mean (X)

The mean or average of the given distribution is denoted by X . It is calculated by using
the following mathematical formulae.

Oasis School Mathematics-10 321

I. Direct Method:
(a) For individual distribution,

Mean (X) = ∑x
n

Where, ∑x = sum of the variables

(b) For discrete distribution,

Mean (X) = ∑fx
N

(c) For continuous distribution,

(X) = ∑fm
N

Where, m is mid–value of variable class and N

is the number of variables

II. Short-cut Method:

(a) For individual distribution,

Mean (x) = a + ∑d
n

where, a = assumed mean

d=x–a

n = number of variables

(b) For discrete distribution,

Mean (x ) = a + ∑fd
N

Where, a = assumed mean

d = x –a, N = ∑f

(c) For continuous distribution,

Mean (x) =a+ ∑fd
N

Where, a = assumed mean

d=m–a
m = mid-value

In continuous distribution, we can also use the step deviation method to calculate
the mean.

According to this method,

Mean ( X ) = a+ ∑fd' × i
N

where, a = assumed mean

d' = m–a
i

m = mid-value

i = size of the class interval.

322 Oasis School Mathematics-10

Worked Out Examples

Example: 1
Calculate the arithmetic mean from the given data.

Marks 0–20 20–40 40–60 60–80 80–100
No. of students
5 10 12 18 5

Direct method:
Solution:

Marks Mid value (m) Frequency (f) fm
50
0 – 20 10 5 300
600
20–40 30 10 1260
450
40–60 50 12
∑fm = 2660
60–80 70 18

80–100 90 5

∑f = (N) = 50

Now, we know, Mean ( X ) = ∑fm
∑f

= 2660 = 53.20
50

Hence, the mean marks is 53.2

Short–cut method
Solution:
Let assumed mean (a) = 50

Class Mid value (m) Frequency (f) d = m–a f × d

0 – 20 10 5 –40 –200

20 – 40 30 10 –20 –200

40 – 60 50 12 00

60 – 80 70 18 20 360

80 – 100 90 5 40 200

∑f = 50 ∑fd = 160
a = assumed mean = 50
Where,

d = deviation = difference between mid–value and assumed mean

∑f = frequency of the variable class

We know, ∑fd
N
Mean (X)=a+ = 50 + 160 = 50 + 3.2 = 53.2
50

Oasis School Mathematics-10 323

Step–deviation method:

Solution:
Let, assumed mean (a) = 50

Class Mid value (m) Frequency (f) d' = m–a fd'
0 – 20 i –10

10 5 –2

20 – 40 30 10 –1 –10

40 – 60 50 12 0 0

60 – 80 70 18 1 18

80 – 100 90 52 10
∑fd' = 8
∑f = 50

Where,
a = assumed mean = 50
d = deviation of assumed mean from the mid-value
i = size of the class interval

f = frequency of the variable class

Now, we know,

Mean ( X ) = a + ∑fd' × i
N

= 50 + 8 × 20
50

= 50 + 3.2 = 53.2

Exercise 17.1

1. (a) Write the formula to calculate means in the continuous data using direct method.
(b) Write the formula to calculate mean in continous data using short-cut method. and

step deviation method.

2. (a) If ∑fm = 400 and N = 20, find x. (b) If ∑fm = 540 and ∑f = 27, find x.

(c) If a = 20, ∑fd = 80 and N = 20, find x. (d) If a = 40 and ∑fd' = 15, i = 10, N = 30, find x.

3. (a) In a series if x = 10, ∑fm = 700 + 5m and ∑f = 40+3m, find the value of m and total
number of terms.

(b) If the number of terms (N) = 12 + a, ∑fm = 168 + 14a, find the mean.

(c) In a continuous series, mean (x) = 32, ∑fx = K and ∑f = 20, find the value of k.
a
(d) If ∑fm = 4000, x = 80, and N = 40 + 10 , find the value of a.

(e) If ∑fm = 2000 + 200a, ∑f = 50 and x = 80, find the value of a.

324 Oasis School Mathematics-10

4. Find the value of mean by using: (iii) Step deviation method
(i) Direct method (ii) Short-cut method

(a)

Marks 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30
Frequency 20 10 30 50 40

(b)

Classes 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
Frequency 3 5 842
(c)
0 – 2 2 – 4 4 – 6 6 – 8 8 – 10
Size: 5 72 3 4
Frequency

5. (a) Find the value of 'a' if the mean of the given data is 28.

Class 17.5–22.5 22.5–27.5 27.5–32.5 32.5–37.5
Frequency
12 18 12 a

(b) Find the value of f if x is 54.

Class 0–20 20–40 40–60 60–80 80–100
Frequency 7 f 10 9 13

(c) If the mean of given data is 25, find the value of x.

Marks obtained 0–10 10–20 20–30 30–40 40–50
Number of students 3 5 45 x

Answer 2. (a) 20 (b) 20 (c) 24 (d) 45
1. Consult your teacher (c) 640 (d) 100 (e) 10 (c) 3
3. (a) 12, 76 (b) 14 (c) 4.43 5. (a) 18 (b) f = 11
4. (a) 20.17 (b) 23.64

17.3 Median

In the given set of variables when arranged either in ascending or in descending order of magnitude,

the value which divides the whole arrangement into two equal halves is called the "median". It is
also called positional average. It is denoted by Md. e.g. median of 1, 5, 10, 11, 20 is 10.

Calculation for median:

a) Individual distribution: The positional value of median for individual distribution is
n+1 (th
given by ( 2 item, where n is No. of observations.

Oasis School Mathematics-10 325

(b) Discrete distribution: Let x1, x2, x3, ............ xn be the values of the variables and their

respective frequencies be f1, f2, f3 .......... fn. Then the value of the median is given by
N is the number of variables, i.e., total number of frequencies.
(N2+1(thposition, where

(c) Continuous series, or grouped frequency distribution: ( N (th
2
Median position for grouped data is obtained by the formula items, where N is

number of variables. But here we will be able to find the class variable in which median

falls. Now, in order to calculate the exact value of the median from its class, we use the

following formula.



Median = l+ N – c.f.
2
×i
f

Where, l = lower limit of median class

N = number of variables

i = size of the median class

c.f. = cumulative frequency preceding the median class

f = frequency of the median class

Worked Out Examples

Example: 1
Find the median from the given distribution:

x: 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
4
f: 2 5 7 12

Solution:

Computation for the median:

x f c.f.

0 – 10 22
10 – 20 5 2+5=7
20 – 30 7 7 + 7 = 14
30 – 40 12 14 + 12 = 26
40 – 50 4 26 + 4 = 30

∑f = 30

Here, number of variables (N) = 30
For median =N2 3=20 15

326 Oasis School Mathematics-10

The c.f. which is just greater than 15 is 26, whose corresponding class interval is (30 – 40).
Hence, the median that lies in the class is (30 – 40).

l

Example: 2

Find the value of x from the given data, if the median (md) = 52.

Class interval 15–30 30–45 45–60 60–75 75–90
10
No. of students 10 8 15 p

Solution:

Computation for the median:

class No. of students (f) c.f.

15–30 10 2
30–45 8 2+5=7
45–60 15 7 + 7 = 14
60–75 p 14 + 12 = 26
75–90 10 26 + 4 = 30

N = 43+p

Since, Md = 52, it lies as the class (45 – 60).
∴ l = 45, f = 15, c.f = 18, i = 15

We have, N – c.f.
Md = l + 2
×i
f

52 = 45 + 43+p – 18
2
× 15
15

52–45 = 43 + 8 – 36 × 1 × 15
2 15

7 = 7+8
2

or, 14 = 7 + 8

or, 14 = 14 – 7 = 7

∴ p=7

Oasis School Mathematics-10 327

Exercise 17.2

1. (a) What is the position of median in continuous series? Write the formula to
calculate exact median in continuous series.

(b) In a continuous series, the median lies in the class (20-30), whose corresponding
frequency is 10. If N = 50, c.f. proceeding the median class is 20, find the exact median.

(c) In a continuous series, N = 60, the median lies in the class (30–40), whose
corresponding frequency is 10, c.f. proceeding the median class is 27, find the
exact median.

2. Compute the median from the following data:
(a)

x: 0–5 5–10 10–15 15–20 20– 25

f: 10 15 12 18 5
(b)

Marks 0–10 10–20 20–30 30–40 40–50 50–60

No. of students 1 5 8 6 8 2
(c)

Age (in years) 0–2 2–4 4–6 6–8 8–10
85734
No. of persons

(d)

Class 0 – 30 0 – 60 0 – 90 0 – 120 0 – 150 0 – 180
5 15 37 62 76 80
Frequency
(e)

Marks 0–10 0–20 0–30 0–40 0–50
No. of Students 4 12 24 44 62

3. (a) The following are the marks obtained by 35 students of class X in Mathematics.

28, 47, 22, 56, 69, 70, 82, 36, 40, 50, 55, 62, 31, 37, 42, 57

60, 72, 81, 90, 91, 56, 34, 57, 49, 70, 90, 61, 65, 80, 81, 87, 63, 28, 34
(b)
(i) make a frequency table of class interval 10.

(ii) find the median marks.

The following are the daily wages of 30 workers in a factory.

95, 80, 76, 91, 70, 52, 50, 45, 67, 70, 85, 96, 98, 72, 65, 55, 48, 73

86, 82, 49, 55, 72, 79, 78, 62, 63, 82, 89, 92

(i) make the frequency table of class interval 10.

(ii) find the median.

328 Oasis School Mathematics-10

4. (a) If the median of the given distribution is 17.5, find the value of f.

x: 5–10 10–15 15–20 20–25 25–30
f: 20 30 40 f 10

(b) If the median of the given distribution is 25.625, find the value of p.

x: 0–10 10–20 20–30 30–40 40–50

f: 3 p 8 6 3

(c) If the median of the given data is 24, calculate the value of x.

Marks 0–10 10–20 20–30 30–40 40–50
21 x 15 10
No. of students 9

5. If the median of the given distribution is 30, find the values of 'x' and 'y'.

Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 –60 Total
100
Number of students 10 x 25 30 y 10

Answer

1. (a) Consult your teacher (b) 25 (c) 33
(c) 4.14 (d) 93.6
2. (a) 12.08 (b) 31.67 4. (a) f = 40 (b) p = 5 (e) 33.5
(c) x = 25
3. (a) 59.16 (b) 75

5. x = 15, y = 10

17.4 Quartiles

Quartiles are the values which divide the whole arrangement of data into four equal parts.
Hence, there are three quartiles in a distribution, which are denoted by Q1, Q2 and Q3.

Note: Q2 represents the value of median.

The lower quartile is denoted by Q1 and upper quartile is denoted by Q3. Calculations for Q1
and Q3 are as follows:

a) Positional value of Q1

(i) In individual distribution

nN + 1 th
4 item.
( ) Q1 =

(ii) In discrete distribution

( ) Q1 = N+1 th
4
item.

Oasis School Mathematics-10 329

(iii) In continuous distribution, position of Q1 = ( N (th.
4

N – c.f.
4
Q1 = l + ×i
f

Where, l = lower limit of Q1 class

f = frequency of Q1 class

c.f. = c.f. preceding the Q1 class

i = size of the class interval

b) Positional value of Q3

(i) In individual distribution
3(n + 1) th
Q3 item = 4 item

(ii) In discrete distribution,

Q3 item = 3(N+ 1) th item
4

(iii) In continuous distribution,

( ) Q3 item = 3N th
4
item

Q3 =l+ 3N − c.f. ×i
4 f

Where, l = lower limit of Q3 class

f = frequency of Q3class

c.f. = c.f. preceding the Q3 class
i = size of the class interval

Worked Out Examples

Example: 1
Compute the lower and upper quartiles:
x: 50 – 60 60 – 60 70 – 80 80 – 90 90 – 100
f: 5 7 8 10 3
Solution:

330 Oasis School Mathematics-10

We have,

Class (x) Frequency (f) c.f.
50 – 60 5 5
60 – 70 7
70 – 80 8 5 + 7 = 12
80 – 90 10 12 + 8 = 20
90 – 100 3 20 + 10 = 30
30 + 3 = 33
∑f = 33

For lower quartile (Q1)

N=4 34=3 8.25.
c.f. just greater than 8.25 is 12, which belongs to the class (60 – 70). Hence, Q1 class is

(60 – 70).

Here, l = 60, i = 70 – 60 = 10 c.f. = 5, f = 7

We have,  N − c.f. 
∴ Q1 = 4 f
l +   × i = 60 + 8.25 − 5 × 10
  7
 

= 60 + 3.25 × 10
7

= 60 + 4.64 = 64.64

For upper quartile (Q3):

3N = 3 × 33 = 24.75.
4 4

c.f. just greater than 24.75 is 30, which corresponds to the variable class (80 – 90)

Q3 class is (80 – 90).

Now, Lower limit (l) of Q3 class = 80

Class height (i) = (90 – 80) = 10
=
c.f. = 20

f 10

( )l + 3N − c.f.
4
∴ Q3 = ×i
= f
=
80 + 24.75 − 20 × 10
10

80 + 4.75 = 84.75

Oasis School Mathematics-10 331

Example: 2 40-60 60-80 80-100
Calculate the value of 'a' if Q1 = 33. a 2 10

Marks obtained 0-20 20-40
Frequency 3 5
Solution:

Marks f c.f.
0 - 20 3 3
20 - 40 5 8
40 - 60 a 8+a
60 - 80 2 10 + a
80 - 100 10 20 + a
N = 20 + a

Given, Q1 = 33

Since Q1 = 33, it lies on the class (20 - 40)

Where, l = 20, f = 5, c.f. = 3, i = 20

We have, Q1 = ll + N − c.f. i
4 f
×

or, 33 = 20 + 20 + a −3 × 20
4
5

or, 33 − 20 = 20 +a − 12 × 20
4× 5

or, 13 = 8+a × 20
20

or, 13 = 8 + a

or, a =13 − 8

∴ a=5

332 Oasis School Mathematics-10

Exercise 17.3

1. (a) What is the position of Q1 in a continuous series? Write the formula to calculate the
exact value of Q1.

(b) What is the position of Q3 in a continuous series. Write the formula to calculate the
exact Q3.

2. (a) In a continuous series, N = 40, Q1 lies in the class (10-20), whose corresponding fre-
quency is 5, c.f. proceeding the Q1 class is 8, find Q1.

(b) In a continuous series, N = 60, Q3 lies in the class (20–25), whose corresponding fre-
quency is 8, c.f. proceeding Q3 class is 40, find the exact Q3.

3. Calculate the lower and upper quartile from the following data:

(a)

Marks 10 – 20 20 – 30 30 – 40 40 – 50

No. of students 20 12 8 30

(b) 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30
Age in years 54 3 6 7
No. of people

(c) 0–9 10 – 19 20 – 29 30 – 39 40 – 49
Marks 5 43 21
No. of students

(d) 5–10 5–15 5–20 5–25 5–30
Marks obtained 20 50 100 140 150
No. of students

4. (a) The following are the marks obtained by students in Maths in the SEE Test examination.
50, 52, 53, 29, 46, 85, 90, 84, 64, 39, 32, 77, 65, 43, 80, 95, 85, 09, 12, 01.
(i) Make a frequency table of class interval 10.
(ii) Calculate the first and third quartiles.

(b) The following are the marks obtained by students in English in an examination.
51, 22, 63, 46, 57, 79, 21, 39, 51, 32, 43, 52, 60, 38, 45, 40, 32, 60, 63.
Make a frequency table of class interval 10 and find its first quartile (Q1).

(c) 15, 12, 23, 35, 46, 57, 18, 12, 39, 51, 32, 43, 25, 59, 18, 38, 45, 40, 32, 33.
Make a frequency table of class interval 10 and find its third quartile (Q3)

Oasis School Mathematics-10 333

5. (a) If first quartile of the given data is 31, find the value of 'k'.

Class 10–20 20–30 30–40 40–50 50–60 60–70
Frequency 8 7 6
45 k

(b) If Q3 of the given data is 60, find the value of f1

Class 10–20 20–30 30–40 40–50 50–60 60–70 70–80
Frequency 354 5 4 f1 3

(c) If Q3 of the given data is 22.5, find the value of m.

Class 0-6 6-12 12-18 18-24 24-30 30-36
Frequency 5 645m 2

(d) If Q1 of the given data is 27.5, find the value of 'p'.

Class 10–20 20–30 30–40 40–50 50–60 60–70
Frequency 578 p8 6

6. (a) If Q1 of the given data is 15.625, find the value of x and y.

Class 0-10 10-20 20-30 30-40 40-50 50-60 Total

Frequency 6 x 12 10 y 2 42

(b) If the Q3 of the given data is 56.25, find the value of 'a' and 'b'.

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Total

No. of students 3 5a 5 4 b 3 26

Answer

1. Consult your teacher

2. (a) 14 (b) 23.12 3. (a) Q1 = 18.75, Q3 = 44.17 (b) Q1 = 11.56, Q3 = 25.54

(c) Q1 = 7.5, Q3 = 27 (d) Q1 = 12.92, Q3 = 21.56

4. (a) (i) Consult your teacher. (ii) Q1 = 35, Q3 = 82.5

(b) Q1 = 36.875 (c) Q3 = 45 5. (a) 10 (b) 4 (c) 3 (d) 7 6. (a) x = 8, y = 4 (b) a= 4, b = 2

17.5 Ogive (Cumulative Frequency Curve)

The sum of the frequencies of all the values up to a given value is known as cumulative
frequency. It is denoted by c.f.

The cumulative frequency curve is also known as ogive. It is a graphical representation
of the c.f. distribution of a continuous variable, where the upper limit (or lower limit) of
each class interval is taken as x–coordinate and its corresponding c.f. as y–coordinate,
and the points are plotted on a graph. We can obtain the curve with freehand.

334 Oasis School Mathematics-10

(i) More than ogive (or more than cumulative frequency curve)

If the lower limit of each class interval is taken as x–coordinate and its
corresponding cumulative frequency as y–coordinate, the curve so obtained
by joining the points by freehand is called more than ogive (or more than
cumulative frequency curve). It is a decreasing curve from left to right.

(ii) Less than ogive (or more than cumulative frequency curve)

If the upper limit of each class is taken as x–coordinate and its corresponding
cumulative frequency as y–coordinate, the curve so obtained by joining the
points by freehand is called less than ogive (or less than cumulative frequency
curve). The less than ogive is an increasing curve from left to right.

The following steps are necessary to construct a less than (or more than) ogive.

• Make a less than (or more than) cumulative frequency table.

• Choose a suitable scale and mark the upper class limits (or lower class limit)
of each class interval along x–axis and c.f. along y–axis.

• Plot the coordinates on a graph.

• Then, join the points by freehand to obtain a curve less than ogive.

Use of cumulative frequency curve to obtain the values of quartiles.
From the cumulative frequency curve, we can find the median and quartiles.

Steps:

(i) Draw the less than ogive curve.

(ii) Find N for the median and N and 3N for the lower and upper quartiles, respectively.
2 4 4

(iii) Locate these values on y–axis. From these points, draw lines parallel to the x–axis

to meet the curve.

(iv) From the points on the curve, draw perpendiculars to the x–axis. The points where
these perpendiculars meet the x–axis, give the values of the median and quartiles.

Note:
(i) Medians and quartiles can also be obtained from a more than the cumulative frequency

curve.
(ii) Corresponding x–coordinates of the meeting point of less than type ogive and more

than type ogive is the median.

Oasis School Mathematics-10 335

Worked Out Examples

Example: 1

Draw a less than cumulative frequency curve and find the median, more than cumulative
frequency curve and find the median class.

Class 0–10 10–20 20–30 30–40 40–50 50–60

Frequency 6 4 8 12 6 4

Solution:
For less than ogive curve.

Class Upper limit Frequency Less than c.f.
0–10 10 6 6
10–20 20 4
20–30 30 8 6+ 4 = 10
30–40 40 12 10 + 8 = 18
40–50 50 6 18 + 12 = 30
50–60 60 8 30 + 6 = 36
36 + 8 = 44
N = 44

For more than ogive curve.

Class Lower limit Frequency More than c.f.
38 + 6 = 44
0–10 0 6 34 + 4=38
10–20 10 4 26 + 8 = 34
20–30 20 8
30–40 30 12 14 + 12 = 26
40–50 40 6 8+ 6 = 14
50–60 50 8 8

N = 44
Y

Cumulative Frequency 50 Less than ogive
40

30

20

10 10 20 MdMore than ogive X
30 40 50 60
X' O

Y' Class Interval

336 Oasis School Mathematics-10

From the graph, the point of intersection of more than and less than ogive lies in the
class (30 – 40) .
Therefore median lies in the class (30 – 40)

Example: 4
Find Q1 and Q3 class from the given ogive curve. Also find their respective frequency.

Y

Cumulative Frequency 50
40

30
20
10

X' O 10 20 30 40 50 60 X
Less than ogive
Y' Class Interval

Solution:

From the graph, N = 50

For Q1 class,

Position of Q1 = N th = 50 th item = 12.5th item
4 4

From the graph,

12.5th item lies on the class (10 − 20)

∴ Q1 class = (10 − 20)

Again, for Q3 class,

Position of Q3 = 3N th = 3 × 50 th item = 37.5th item
4 4

From the graph,

37.5th item lies on the class (40 − 50) Y

∴ Q3 class = (40 − 50). 50
From the given ogive curve, frequency of the
Cumulative Frequency
class (10-20 is (20-5) = 15 40

From the given ogive curve, 30
frequency of the class (40-50) is 20
(40-30) = 10. 10

X' O 10 20 30 40 Q3 50 60 X

Y' Q1 Class Interval

Oasis School Mathematics-10 337

Exercise 17.4

1. Draw a less than ogive and more than ogive from the data given below. Compute the
median class.

(a)

Marks 10–20 20–30 30–40 40–50 50–60 60–70
No. of students 5
(b) 8 17 10 5 3

Wages (in Rs.) 20–30 30–40 40–50 50–60 60–70 70–80

No. of students 8 12 6 79 8

2. Find the first and third quartile classes and their respective frequency from the given
less than ogive curve.

Cumulative Frequency (a) (b) Cumulative Frequency 60Y X
(d) 50
Y 40
30
50 20
40 10
30
20 O 10 20 30 40 50 60
10 Class Interval

O 10 20 30 40 50 60 X
Class Interval

(c)

Cumulative frequency Y Cumulative frequency Y X

50 X 50
45 45
40 40
35 35
30 30
25 25
20 20

15 15
10 10
5 5

O 10 20 30 40 50 60 70 80 90 100 O 10 20 30 40 50 60 70 80 90 100
Marks Marks

338 Oasis School Mathematics-10

Y

3. (a) Look at the given ogive and find the median class. Frequency 200
160
120 X
80
40

O 4 8 12 16 20
Class Interval

(b) From the cumulative frequency (c) Find the median and modal class

curve, find the median class and from the given ogive curve.

its frequency.
Y
50 Y

45 Cumulative Frequency 50
Number of people40
40 30
20
35 10

Frequency 30 O 10 20 30 40 50 60
Class Interval
25 Y

20

15

10

5

O 10 20 30 40 50 60 70 80 90 100 X

C lass Interval

(d) Study the given ogive curve and give answers to 14
the following questions. 12
10
i) Which class has the greatest frequency? 8

ii) How many people of age more than 40 years are 6
there? 4
2

O 10 20 30 40 50 60 70 80 X
Age

(e) Study the given ogive curve and answer the questions 60Y
given below: 50
Number of workers
(i) Prepare a frequency distribution table. 40
30

(ii) Find the median class. (iii) Find the modal class. 20

(iv) Find the number of workers whose wages are less 10 X
than Rs. 40. O 10 20 30 40 50 60

(v) Find the number of workers whose wages are more Wages (in Rs.)
than Rs. 30.

Answer

1. Consult with your teacher. 2.(a)Q1class=(20-30),f=5,Q3class=(40-50),f=20(b) Q1class=(20-30), f=5,
Q3class=(40-50),f=20(c)Q1class=(20-30),f=10,Q3class=(40-50),f=15 (d)Q1class=(30-40),f=5,Q3class=
(70-80), f = 10 3. (a) (12-16) (b) (50–60), f = 5 (c) Median class = (30-40), Modal class = (10-20)

(d) (i) (10-20), (ii) 5(e) (ii) (40-50) (iii) (40-50) (iv) 30 (v) 35

Oasis School Mathematics-10 339

Miscellaneous Exercise

Statistics

1. Find the median from the given data.
a)

Class 35-45 65-75 75-85 45-55 55-65
Frequency 7 9 6 8 10

b) (Ans: 60)

Marks Below 20 Below 40 Below 60 Below 80 Below 100

Number of students 2 10 21 27 30

2. (a) Calculate the lower quartile from the given data. (Ans: 49.09)

Class 0≤ x< 20 20≤ x< 40 40≤ x< 60 60≤ x< 80 80 ≤ x< 100
Frequency 5 7 6 4 3
(Ans: 23.57)
(b)

Class 0–10 0–20 0–30 0–40 0–50 0–60

Number of workers 6 9 14 19 25 30

(Ans: 32)
3. The following are the marks obtained by 20 students in Mathematics in an examination.

15 12 23 35 46 57 18 12 39 51

32 43 25 59 18 38 45 40 32 33

i) make a frequency table of class interval 10. ii) find the third quartile (Ans 15.75)

4. a) If the median of the given data is 24, find the value of x.

Class 0-10 10-20 20-30 30-40 40-50
Frequency 4 12 x 9 5

b) If the third quartile of the given data is 35.5, find the value of y. (Ans: 10)

Class 0-10 10-20 20-30 30-40 40-50 50-60

Frequency 6 8 12 y 4 2

c) If the first quartile of the data is 31, find the value of x. (Ans:10)

Class 10-20 20-30 30-40 40-50 50-60 60-70

Frequency 4 5 x 8 76

(Ans: 10)

Project Work

Collect the marks obtained by the students of your class in your terminal examination.
Prepare the necessary class and the suitable continuous data and find subject-wise mean,
median and quartiles. Present the results in your class.

340 Oasis School Mathematics-10

Full Marks : 23

Attempt all the questions.

Group A [2 × 1 = 2]

1. (a) If 'a' is the assumed mean and 'n' the number of terms in a continuous series,

write the formula to calculate the mean (x).

(b) In a continuous series: N = 40, l = 20, f = 5, c.f. = 20, i = 10, find Md.
Group B [4×2=8]

2. (a) In a continuous series, ∑fm = 400 + 20a, x = 50+a and N = 10. Find the value of a.

(b) Average mark obtained by 20 boys in a class is 30 and that of 15 girls is 35. Find the

3. (a) average mark of the class. Y
From the given less than ogive curve,
Cumulative frequency
find the median class and its frequency. 70
60

50
40
30

20

10 X
O 10 20 30 40 50 60
Class

Y

(b) Find the Qgi1vcelnasosgainvde modal class Cumulative frequency 160
from the curve. 140
120
100 X
80
Group C [3×4=12] 60
4. Find the median from the given data: 40
20

O 10 20 30 40 50 60 70 80 90

Class

Class 0-10 10-20 20-30 30-40 40-50 50-60

Frequency 4 6 20 10 7 3

5. Marks obtained by 40 students of class X are given below. Prepare the frequency
distribution table of class interval 10 and hence find the first quartile.

27, 31, 33, 48, 49, 50, 62, 79, 68, 70, 60, 58, 67, 39, 28, 30, 56, 29, 22, 37, 42, 68, 51, 76, 78, 65,
43, 33, 51, 20, 31, 64, 35, 45, 65, 75, 54, 43, 34, 25.

6. If the Q3 of the given data is 84.75, find the value of 'a'.

Class 50-60 60-70 70-80 80-90 90-100

Frequency 5 78 a 3

Oasis School Mathematics-10 341

Probability

11Estimated Teaching Hours

Contents

• Mutually exclusive and mutually non-exclusive events
• Independent event
• Tree diagram

Expected Learning Outcomes

At the end of this unit, students will be able to develop the following
competencies:
• To identify whether the given events are mutually exclusive or not
• To solve the problem related to mutually exclusive and mutually

non- exclusive events
• To find the probability of independent events
• To draw a tree diagram and find the probability of all possible

outcomes

Materials Required

• Flash card, chart paper, playing cards, dice, coin

342 Oasis School Mathematics-10

Unit

18 Probability

18.1 Warm-up Activities

Discuss the following in your class and draw a conclusion.
Write the set of all possible outcomes in the given random experiment.
(i) A coin is tossed once.
(ii) A dice is rolled once.
(iii) A coin is tossed twice.
• On whose range lies the probability of an event?
• What is the probability of a sure event?
• What is the probability of an impossible event?

18.2 Probability

Random experiment: If an experiment is repeated under homogeneous conditions,
the result of which is not the same but may be one of the various possible outcomes,
then such an experiment is called a random experiment. In a random experiment, the
outcome of each trial depends on chance, which cannot be predicted with certainty.
For example, if we toss a coin, we may get either a head or a tail. If we throw a dice
which has six faces marked with 1, 2, 3, 4, 5 and 6, then any one of the faces may come
up. It is not possible to say which face will come up. These are examples of a random
experiment.

Trial and event: Performing a random experiment is called a trial, and the outcome or
combination of outcomes is called events. For example, throwing a dice is a trial and
getting any one of the faces 1, 2, …, 6 is an event.

Sample space: The set of all possible outcomes of a random experiment is called sample
space. The sample space is usually denoted by the letter 'S'. Some examples of sample
space are given below.

For example:
i. In tossing a single coin, there are two possible outcomes, namely a head (H) or a tail (T).

So, the sample space for a coin is given by S = {H, T}.
ii. If two coins are tossed simultaneously, then the sample space is given by:

S = {H, T} which means head on the first coin and tail on the second coin. Similarly,

Oasis School Mathematics-10 343

{T, H} means tail on the first coin and head on the second coin and so on.Coin 2

Coin 1
HT
H HH HT
T TH TT

Exhaustive outcomes: The total number of all possible outcomes of a random experiment
is called the exhaustive outcome for the experiment.

For examples:
In the case of a fair coin, the total number of exhaustive outcomes = 2.
i.e., {H, T}. If two coins are tossed simultaneously, the possible outcomes are {HH,
HT, TH, TT}, i.e.,

• heads on both tosses
• head on the first toss and tail on the second toss
• tail on the first toss and head on the second toss
• tails on the second toss

∴ Total number of exhaustive events = 22 = 4.

Similarly, if three coins are tossed, then the total number of exhaustive
events = 23 = 8.
If a dice is rolled and a coin is tossed, then the possible results are-

(1, H), (2, H), (3, H), (4, H), (5, H), (6, H)

(1, T), (2, T), (3, T), (4, T), (5, T), (6, T).

∴ Total number of exhaustive events = 2 × 6 = 12.

Coin → H T
Dice ↓ (1, H) (1, T)

1

2 (2, H) (2, T)

3 (3, H) (3, T)

4 (4, H) (4, T)

5 (5, H) (5, T)

6 (6, H) (6, T)

344 Oasis School Mathematics-10