Green C Math Class 10

EXAMPLE 10 In the given figure, O is the centre of the circle, AB the diameter
and arc BC = arc CD. Prove that AD||OC.

Solution : Given : (i) O is the centre of circle. D C

(ii) AB is the diameter. AB
(iii) BC = CD O

To prove : AD||OC

Proof:

Statements Reasons

1. ∠BAD 1 BCD 1. Inscribed angle is equal in degree to the
2 half of its corresponding arc

2. ∠BAD 1 ( BC + CD ) 2. Whole part axiom from statement 1
2

3. ∠BAD 1 ( BC + BC ) 3. BC = CD (given)
2

4. ∠BAD 1 ×2 BC 4. Form statement (3)
2

i.e. ∠BAD BC

5. ∠BOC BC 6. Central angle of a circle is equal in
degree to its corresponding arc
6. ∠BAD = ∠BOC
7. From statements (4) and (5)

7. ∴ AD||OC 8. Being corresponding angles equal from
statement (6)

Remember :

Angle at the centre is A Angles in the same C D
twice the angle at the x segment are equal. x x
cirumference. O
2x B
B
CA

C A

The angle in a semicircle A O Opposite angles in a a dD
is 90°. cyclic quadrilateral add
B upto 180°. b
C
c

a + b = 180°, c + d = 180° B

GREEN Mathematics Book-10 195

The perpendicular A O A straight line drawn B
drawn from the centre to M from the centre to O
a chord bisects the chord. B tangent at point of
contact is perpendicular. P
AM = BM
OP⊥AB A

An external point A The angles in the D
where two tangents B
meet is equidistant alternate segments c
from the points of
contact. P of a circle are equal C a bB

AP = BP drawn on the d

tangent. i.e. a = b, A
c=d

Equal arcs subtend Prallel chords
equal angles at the subtend equal
center of circle or at the arcs.
point of circmference.

EXERCISE 11.1

A. Very Short Questions C
A

1. a. If AB and CD are equal, write the relation of chord AB

and CD. D
PB

R

b. If chord PQ and chord RS are equal, write the relation of

PQ and RS . Q
S
C
A

c. If AB||CD, write the relation of AC and BD .

B D
M P

d. If MP = NQ , write the relation of MN and PQ.

D
NQ

69

e. What is the size of ∠ABC? A C

B

196 GREEN Mathematics Book-10

A

2. a. In the figure alongside, O is the centre of the circle. O B
What is the relation of ∠AOC and ∠ABC. C D
b.
In the figure alongside, PQRS is cyclic quadrilateral. A

Write the relation of P B
c. C
i. ∠QPS and ∠QRS Q S
ii. ∠PQR and ∠RSP R

In the figure alongside, A, B, C and D are concyclic
points. Write the relation of:

i. ∠BAC and ∠BDC

ii. ∠ABD and ∠ACD. AB C
D
d. In the figure alongside BCDE is cyclic quadrilateral.

Write the relation of ∠ABE and ∠EDC. E

P
R

e. In the figure alongside, O is the centre of circle. What

is the measure of ∠QPT and ∠QRT? Q T
O

3. a. Define segment angles with an example. BC D
b. Define central angles with an example. A
c. Define concyclic points with an example.
d. Define cycle quadrilateral with an example.
4. In the given figure A, B, C, D, Q and P are concyclic points,

a. If AB = CD , write the relation of ∠APB and ∠CQD.
b.
5. a. If ∠APB = ∠CQD, write the relation of AB and CD . PQ

O is the centre of circle. If A, B and C are concyclic points and AC is diam-
eter, write the value of ∠ABC.

b. If P, Q, R and S are concyclic points, write the relation of :

i. ∠QPR and ∠QSR ii. ∠PQS and ∠PRS
iii. ∠QPS and ∠QRS iv. ∠PQR and ∠QPS

GREEN Mathematics Book-10 197

B. Short Questions V W
6. Find the value of x in each figure.
c.
CX

a. b. x 72°
x
xB O
O O
98° Z
26°
38°
A
YX
Y

7. Calculate the values of x and y from the following figures where O is the cen-
ter of circle.

A P

a. b. c.

40° x
O
OP OS
x
60° y 100° y
BC x R
Q
RQ

8. Calculate the values of x and y from the following figures where O is the cen-

ter of circle.

X

a. P b. c. x

M x 30° 230°
y
O

O x N O
40° YZ
N Q L 130°

M

9. Calculate the values of x and y from the following figures where O is the cen-

ter of circle. WZ

a. b. x c. A D

A O D X 30° O Y 35° x
x y
20°
B
O

By

CC

10. Calculate the values of x, y and z from the following figures where O is the

center of circle. T

N Py A
a. y
b. c. x

OO O

110° L 130° S z D
Ny x y

T QR B

198 GREEN Mathematics Book-10 50° C
N

11. Calculate the values of x and y from the following figures where O is the cen-
ter of circle.

M A

a. b. N y c. y

PO 140°x O 20°
100°O
50° O 25° x C
70°
y° B
x°

Q R P

12. Calculate the values of x and y from the following figures where O is the cen-

ter of circle. P W

a. b. c.

x

O OO

y Z X 30° y
110°
QxR x Z
X

S YY

13. Calculate the values of x, y and z from the following figures where O is the
center circle.

a. b. P A
14. a.
c.

70° 63°

OC O BO E
y xS
x D C
x Qy C
AB zR

In the given figure, O is tThe centre of circle and AB is AO

the diameter. If ∠ABC = 70°, find ∠BDC. B

AE

D
b. In the given figure, AB = AC, ∠ABC
O

= 70° and AC|| ED. Find ∠CAE. p
B CD

y
60° s

x
c. From the given adjoining figure, find the values q
85°
of x and y.
R

tA

15. a. In the given figure, O is the centre of the circle in O z
which x = 150° and y = 45°. Find the value of z.
x D
C
By

GREEN Mathematics Book-10 199

b. In the given figure, O is the centre of the circle, O

BC||AD and ∠AOC = 90°. Find ∠BAD. B 90°
A
C
D

c. In a cyclic quadrilateral PQRS, if ∠P = 3 ∠Q = 2 ∠R, find the measures of ∠P

and ∠S. A

d. In the given figure, AB is a DO
diameter of the circle with
center O. If ∠DOE = 40°, find 60°
the measure of ∠DCE. 40°

C EB C
B
C. Long Questions

16. a. In the adjoining figure, prove that : ∠PDB = ∠PAC. P D
A



b. In the adjoining figure, AB = L A
AC, ∠BAL = 60° and ∠ALM
= 30°. Prove that AMBL is a 30°
cyclic quadrilateral.
B
M C

DC

17. a. In the adjoining figure, O is the centre of circle, A O B
AB is the diameter and BC = CD . Prove that X
b. AD||OC. A Q
D S
If two chords AB and CD of a circle intersect
c. at the point M, prove the ∠AMC = ∠BMD = M
1 B
2 ( AC + BD ).
C

P

Two chords PQ and RS of a circle intersect at an

1
external point X. Prove that ∠PXR = 2 ( PR – QS ).

R

200 GREEN Mathematics Book-10

A

18. a. In the figure, BE = DE. Prove that AB = CD. B
E
b. A
C D

C

In the given diagram, OD
AOB is diameter of
the circle. If AC = CD,
prove that ∆BAD is an
isosceles triangle.

B

c. ABC is an isosceles triangle and XY is parallel to the base BC. If XY cuts AB

at X and AC at Y, prove that four points B, C, Y and X are concyclic.

YX

19. a. In the figure, AD||BC. Prove that ∠AYC = ∠BXD.



X A A D
b. Two circles intersect at A P Q C

and B. Through A, two B
straight lines PAQ ans XAY Y

are drawn terminated by

the circumferences. Prove

that ∠PBX = ∠QBY. B

c. AB is a diameter of circle with the centre O. Chord A OB
CD is equal to radius OC. Chords BD and AC, when
20. a. produced, meet at point P. Prove that ∠APB = 60°. CD
p
P Q
T E
b. In the figure, QR is a bisector
of ∠SQT and PQRS is a cyclic RF D
quadrilateral. Prove that DPSR is S C
c. an isosceles triangle.

In the figure ∠AFC = ∠BED. Prove that AD||BC. A

B
AP B

In the given diagram, two circles

intersect at points P and Q. Prove

that: AC||BD. CQ D

GREEN Mathematics Book-10 201

21. a. In the given figure, PQRS is a cyclic quadrilateral P

and RS is produced to T. If ∠PRQ = ∠PRS, prove Q S
D
that DPRT is an isosceles triangle. S
R
T
F
R C

b. In the figure MN is the diameter N
of a circle with centre O. If PR = M
OP
PQ, prove that ∠OSP = ∠ORP.
Q

c. In the figure, DM⊥EF and FN⊥ED. Prove that ∠EMN = N

22. a. ∠EDF. A
E
M
b. In the adjoining circle M and N are M PQ N R
the mid points of arcs AB and AC C

c. respectively. Prove that : AP = AQ. B C

A

In the given diagram P, Q and R are the mid P
points of side AB, BC and AC of triangle ABC

respectively. If AD⊥BC, then prove that :PDQR is a

cyclic quadrilateral. B DQ
L

In the adjoining diagram AB = AC, 30° A
∠ALM = 30° and ∠LMB = 60°. Prove B 60°
that AMBL is a cyclic quadrilateral.
M

11.4 Tangents to a circle

1. Tangent and secant

A line which is drawn from an external point of a N O B
circle and meets the circle at one and only point on the A E
circumference of the circle is called a tangent. TD

The point A where the tangent touches the circle is called
the point of contact.

In the given figure, TAN is a line which meets the circle
at a point A. So, TAN is a tangent and the point A is point
of contact.

In the same figure, a ling TB cuts the circle at two points
D and E . So, the line TB is called a secant line.

202 GREEN Mathematics Book-10

2. Common tangent to two circles

If a line touches two given circles, then it is called a common tangent to the circles.
Different conditions of common tangent

When two circles touch When two circles
internally, as in the figure, touch externally, as
only one common tangent P in the figure, three
can be drawn at the point common tangents
of contact. can be drawn.

When two circles Common tangent When two circles do
intersect in two distinct not intersect each other
points, as in the figure, as in the figure, then
two common tangents four common tangents
can be drawn can be drawn.

3. Properties of tangent

(i) The tangent at any point of a circle is perpendicular to O
the radius through the point of contact. In the figure,
OA⊥TN. TA N

(ii) The length of tangents drawn from an external point to a A
circle are equal.
O
(iii) Alternate segment and its angles : The segment opposite T
to the angle formed by the chord of a circle with the
tangent at a point is called the alternate segment for that D B
angle. T
OB
In the figure, TAN is a tangent to the circle at the point A C
and AB is a chord. The chord divides the circle into two
segment ACB and ADB. These segments are alternate to AN
each other.
DB
For ∠BAT, the alternate segment is ∠ACB.
N A C
For ∠BAN, the alternate segment is ∠BDA. Fig. I T

(iv) Angle in the alternate segment : Here, in the figure (i) D B
∠BAT and ∠BDA are said to be angles in the alternate
segment. Similarly, in figure (ii), ∠BAN and ∠BCA are N A C
said to be another pair of angles in the alternate segment. Fig. II T

The angles in the alternate segments are always equal. B C
A O
∴ ∠DAN = ∠DBA and ∠BDA = ∠BAT
T
(v) Relation between a tangent and secant : If AT is the
tangent drawn from point A to a circle and AC is a secant
then AT² = AC × AB

GREEN Mathematics Book-10 203

Theorem – 11 P

Tangents drawn from an external point on a circle are equal in length.
Experimental Verification

A T1

O BA
P

O

B T2 T2 T1

Fig. I Fig. II

Step: I With the help of a compass and a pencil two circles of different radii with
Step: II centre 'O' are drawn. In each circle from an external point P two tangents
PAT1 and PBT2 are drawn.

With the help of a divider and a ruler, the length PA and PB of tangents
from the external point P to the points of contact A and B of each figure are
measured and the results are tabulated below :

To verify : PA = PB

Table of observation:

Fig PA PB Result
I PA = PB
II

Conclusion: The above experiment shows that the tangents drawn from an external point
on a circle are equal in length.

Theorem – 12

Line segment joining center of the circle and point of contact is perpendicular to the
tangent.

Or,

A tangent is perpendicular to the radius of the circle at point of contact.

Experimental Verification

O O

T N
A
N A
Fig. a T

204 GREEN Mathematics Book-10 Fig. b

Step: I With the help of a compass and a pencil two circles of different radii with
centre at 'O' are drawn. In each figure, radii OA and tangent TAN are drawn.

Step: II With the help of protractor, ∠OAT and ∠OAN of each figure are measured
and the results are tabulated as:

To verify : ∠OAN = ∠OAT = 90°

Table of observation:

Fig. ∠OAN ∠OAT Result
a
b ∠OAN = ∠OAT = 90
i.e. OA⊥TN

Conclusion: The above experiment shows that the tangent is perpendicular to the radius
of the circle at point of contact.

Worked Out EXAMPLES

EXAMPLE 1 In the given figure, O is the centre of circle. B
Solution :
TAN is a tangent to the circle where A is the x

o

point of contact. If ∠BAN = 40° and ∠ABO =

x°, find the value of x. 40°

∠BAN = 40° T A N

∠OAN = 90° [∵ radius (OA) ⊥ tangent (TAN) at A]
∠OAN = ∠OAB + ∠BAN [∵ Whole part axiom]
∴ ∠OAB = 90° – 40° = 50°

∠ABO = ∠OAB [∵ OA = OB]

∴ x = 50

EXAMPLE 2 In the given figure, O is the centre of
Solution :
circle. TAN is a tangent to the circle at A o

where A is the point of contact. If OA = B
8cm and AT = 15cm, find the length of BT.

Here, T AN

AT = 15cm, OA = 8cm, BT = ?

∠OAT = 90° [∵ radius (OA) ⊥ tangent (TAN) at A]

∴ ∆OAT is a right angled triangle.

OT² = OA² + AT²

GREEN Mathematics Book-10 205

= (8)² + (15)² ∵ OA + OB (radii of the same circle)

= 64 + 225 = 289
∴OT = 289 = 17 cm

Now, BT = OT – OB

= OT – OA
= 17cm – 8cm
= 9cm

EXAMPLE 3 In the given figure, O is the centre of a o B
Solution : circle. TAN is a tangent to the circle at A. N
If ∠ACB = 65° and ∠CAT = x°, find the C 65°
EXAMPLE 4 value of x.
Solution : x
Here, T
A

A line TCB passes through the centre O.

So, CB is a diameter of the circle.

∴ ∠CAB = 90° ∵ ∠CAB lies on the semi-circle

∠CAB + ∠ACB + ∠ABC = 180° ∵ Sum of all three interior angles of ∆ABC
is equal to 180°
or, 90° + 65° + ∠ABC = 180°

∴ ∠ABC = 180° – 155° = 25° ∵ Angles in the alternate segments
∠CAT = ∠ABC

∴ x = 25° A

In the adjoining figure two circles BD

intersect each other at A and D. Their T C
common tangent TP meets the two
P
circles at B and C. Prove that, ∠BAC +

∠BDC = two right angles.

Given : Two circles intersect each other at points A and D. TP is a
common tangent at B and C.

To prove : ∠BAC + ∠BDC = 180°
Construction : Points A and D are joined.
Proof:

Statements Reasons
Being angles in the
1. ∠DAB = ∠DBC 1. alternate segment of the
circle.
2. ∠DAC= ∠DCB 2.
By reason similar to 1.
3. ∠DAB + ∠DAC = ∠DBC + ∠DCB 3.
By adding (1) and (2).

206 GREEN Mathematics Book-10

4. ∠BAC = ∠BAD + ∠DAC 4. By whole parts axiom
5. ∠BAC = ∠DBC + ∠DCB
6. ∠DBC + ∠DBC + ∠BDC = 180° 5. From statements (3) and
7. ∠BAC + ∠BDC = 180° (4)

6. Sum of all three interior
angles of ∆DBC

7. From statements (5) and
(6)

Proved

EXAMPLE 5 In the adjoining figure PQ is a tangent to the B
Solution : circle at A. C is the middle-point of arc AB. If
CM⊥PQ and CN⊥AB, prove that CM = CN. o
N

Given : A circle with the centre O. PQ is a C
tangent to the circle at A. C is the mid
P A MQ

- point of ACB .

i.e. AC = BC , CM⊥PQ and CN⊥AB.

To prove : CM = CN

Construction : A and B joined with C.

Proof:

Statements Reasons

1. Chord AC = Chord BC 1. Equal arcs intercept equal chords
( AC = BC )

2. ∠BAC= ∠ABC 2. Base angles of isosceles ∆ABC are
equal as AC = BC from statement 1.

Angle made by the chord AC with

3. ∠CAM = ∠ABC 3. the tangent PQ is equal to the angle
made by the same chord in the

alternate segment of the circle

4. ∴∠BAC = ∠CAM 4. From statement 2 and 3

5. In ∆ACM and ∆ACN 5.

i. ∠AMC = ∠ANC (A) i. Both being 90° (CM⊥PQ and
CN⊥AB)

ii. ∠CAM = ∠CAN (A) ii. From statement (4)

iii. AC = AC iii. Common side

6. ∆ACM ≅ ∆ACN 6. By A.A.S. congruency axiom

7. ∴ CM = CN 7. Corresponding sides of congruent
triangle are equal

Proved

GREEN Mathematics Book-10 207

EXERCISE 11.2

A. Very Short Questions

1. Calculate the values of unknown angles from each of the following circles

where O is the center of the circle. N

a. b. C c. 60°

B x x A
Ox O BO

T 60° 116° T
NT A BN

A

2. Calculate the values of unknown angles from each of the following circles
where O is the centre of the circle.

B B N

C b. 65° c. C
40° O
a.

x O O A
50° x N
T A
C x
A
T
N

3. Calculate the values of unknown angles from each of the following circles:

BC

a. x b. c. 50°x y
A
Cy 80° C xy BD z
22° NT
T NT B N
A A 30°



4. Calculate the value of unknown angles from each of the following circles
where O is the center of the circle.

AC C B
Ox A
a. b. c.
70°
Px O 120° A Ox 65° P

BD B DC

208 GREEN Mathematics Book-10

5. a. In the figure, the length of the tangent PT is 16cm. O
If OP = 20cm, find the radius of circle.

P

T



b. In the given diagram, AB = 4cm, BC = 2cm and BA is O
the tangent. Find the length of radius of the circle. C

B
A

O
17cm
c. In the given figure 'O' is the center and AB is a tan-
gent of the circle. Find the radius of the circle.

A
B 15cm

6. Calculate the values of unknown angles from each of the following circles where O
is the centre of the circle.

a. b. z y

O 50° O 110° D
y
xz x
60°

c. In the figure, TE is a tangent to the circle with the O
centre O. Find the ∠OAE = 132°. Find the value
of x. C

T x E
BA

B. Short Questions O
A
7. a. In the figure, PT is a tangent
of the circle with the centre O.
If ∠OTP = 40°, find the ∠PAT.

T C
P

b. In the figure, PQ is a tangent to the circle at A. BD 30° O D
is a diameter and O is the centre. If ∠OAD = 60°, B

and ∠DBC = 30°, find (i) ∠CDB, (ii) ∠QAB and (iii) 60° P

∠PAD. A

Q

GREEN Mathematics Book-10 209

C

c. In the figure, AB is a diameter and AC is a chord 30°
of a circle such that ∠BAC = 30°. The tangent at C A
BD
intersects AB produced to D. Prove that BC = BD.

8. a. O is the centre of the circle X ?
and BX is the tangent at X. C 2cm B
If BC = 2cm and AC = 16cm, A O
find the length of BX. 16cm

E

C

b. In the given figure, O is the centre of a circle in Ox

which AD = DC. If AC is the tangent to the circle at A D B
C, find the value of x.



c. In the figure alongside, O is centre of the O
circle. PQ is a tangent and P is the point C
of contact. If PQ = 5cm and OQ = 13cm,
find the length of CQ. Q
P

A

9. a. In the given figure, O is the centre of the cir-

cle. If TBD is a tangent with B as the point of Ox
C
contact and ∠ODB = 38°, find the values of x
y 38°
and y. T B D

G

D A

b. In the adjoining figure EF is

a diameter of the circle. GD E 65°
is a tangent with the point of OF
contact at D. If ∠DFE = 65°,

find the ∠DGF.



c. In the given figure, if TCN is a tangent at C, B 70° D
∠DCN = 40° and ∠ABC = 70°, what must be the

value of ∠ACD? 40°

T CN

210 GREEN Mathematics Book-10

10. a. In the given figure, RN is the tangent to the M
circle and N the point of contact. If ∠PNM =
2∠PNR and ∠PRN = 20°, calculate the value of P
∠RMN. R 20°

N

A

b. In the adjoining figure, PR is a 30°

c. tangent at T. ∠BAT = 30°, PT = Bx
TA. Find the value of x.
11. a. P R
T

In the given figure AB is a tangent and R is a point P Q
of contact. If PR = QR and ∠PRQ = 50°, find the B
value of ∠BRQ. R
T T
A
C
In the adjoining figure two P AC
circles touch externally at X.
PT is a common tangent to B
the circles at A and B. Prove
that ∠AXB = 90°. X

b. In the adjoining figure, O is the centre of the circle. TAN is B A

a tangent to the circle at the point A. If BC||OT, prove that O

∆ABC ∼ ∆ OAT.

PAT N

c. In the figure, M is any point on the M NQ
chord AB and MN||PT. Prove that C

MNCB is a cyclic quadrilateral. B A



d. In the adjoining figure three circles

touch externally at P and Q points.

If AB and AC are tangent, prove that

AB = AC. B C

PQ

GREEN Mathematics Book-10 211

12

Trigonometry

Estimated Teaching Periods : 10

Hipparchus was a Greek astronomer who lived between 190-120 B.C. He
is considered the father of trigonometry, a branch of mathematics which
studies the angles of sides of triangles. Hipparchus also developed the first
accurate star map.

Contents

12.1 Area of triangles and quadrilaterals
12.2 Height and distance
Test questions

Objectives

At the end of this unit, students will be able to:
find the area of a triangle and quadrilateral by using trigonometry formulae.
solve the problems of height and distance.
solve problems related to the height and distance of houses, poles, etc. in a day to day
life.

Materials

Poles, sticks, etc.

212 GREEN Mathematics Book-10

12.1 Area of triangles and quadrilaterals

Area: The space occupied by an object is called an area of the object.

Triangle: The plane figure bounded by three sides is called triangle.

Similarly, we can say that the space occupied by a triangle is known as area of triangle.
There are different triangles and different formulae to find area of different triangles. But
in this chapter we are going to learn new formula to find area of any triangle with its two
given sides and angle included by them.

At first,

We know, Area of a triangle = 1 × base × height
2 A
∴ Area of ∆ABC = 1 × a × AE ................. (i) cb

Here, in ∆ ABE, 2

AB = hypotenuse (h) Height

AE = perpendicular (p) BC
∴ Sin B = AE E

AB a = base
By cross multiplication, we get,

AE = Sin B. AB A

= AB . Sin B

= c. Sin B cb

Putting the value of AE in equation (i)

Area of ∆ABC = 1 × a × c × Sin B BC
Hence, 2 a

Area of ∆ABC = 1 × a × c × Sin B
2
In this way, we can get,

Area of ∆ABC = 1 × a × b × Sin C
2
and Area of ∆ABC = 1 × b × c × Sin A
2
Conclusion : To find the area of a triangle, we need any angle and sides beside the given

angle.

For example. In ∆MNO M

Area of ∆MNO, (i) with reference to angle 40°, 4cm 80° 5.5cm
Area of ∆MNO = 1 × ON × OM × Sin 40°
N 60° 40° O
2

6cm

GREEN Mathematics Book-10 213

(ii) With reference to angle 60°.
Area of ∆MNO = 1 ON × MN × Sin 60°

2
(iii) With reference to angle 80°
Area of ∆MNO = 1 MN × MO × Sin 80°

2

Some formulae and relations:

a. If a median of a triangle is given it bisects the triangle into two equal parts. A median
in an isosceles triangle and in equilateral triangle is perpendicular to the side. The
median of a triangle passes through the mid point of the side.

A

Figure I : Here, in figure 2. AD is the median of ∆ABC;
which passes through D. D is mid point of
BC.

Hence, Area of ∆ABD = Area of ∆ADC B C
C
Figure II : Here, in figure 2. ∆ABC is an isosceles triangle. B D
Since AB = AC and D is mid point of BC. So, BD
= DC. And AD is a median of ∆ABC. It passes Fig. I
through mid point of BC. In this figure AD is
median as well as perpendicular to the side BC. A

Hence, ∆ABD = ∆ADC D

Fig. II

Note :

Every equilateral triangle is an isosceles triangle. Hence, all the relation of figure 2 can
be applied to the equilateral triangle.

Scalene Triangle : Here, ∆ABC is a scalene triangle. A

So, Area of ∆ABC = s(s – a) (s – b) (s – c)

∴ s = a + b + c , where, cb
2

BC= a, AC = b and AB = c. BC
a

214 GREEN Mathematics Book-10

Parallelograms : CD C CD C
D
D

A BA BA BA B

(i) (ii) (iii) (iv)

In above figures No. (i) is a square, (ii) is a rhombus, No. (iii) is a rectangle and
No. (iv) is a parallelogram. But these all are parallelograms too. And we know
diagonal of a parallelogram bisects the parallelogram into two equal triangles.

Hence, ∆ABD = ∆BCD.

S.N. Name Figure Area

1. Equilateral triangle a 3 a²
4

2. Isosceles triangle a b
b 4 4a² – b²

3. Right angle triangle p h 1 × b × p
b 2

4. Square d l² or 1 × d²
2

l

5. Rhombus h 1 × d1 × d2 or b × h
2

b

6. Parallelogram h b×h

b

GREEN Mathematics Book-10 215

Worked Out EXAMPLES

EXAMPLE 1 In ∆ABC, AC = 4.5 cm, AB = 6.5cm ∠B = 40° and A

∠C = 60°, find the area of the DABC. 4.5cm 6.5cm

Solution : We have, C 60° 40° B

AC = 4.5cm

AB = 6.5cm,

∠C = 60° and ∠B = 40°

Here, our required angle is ∠A

\ ∠BAC + 60° + 40° = 180° [∵ Sum of interior angles of a ]triangle

\ ∠BAC = 80°

Now, we know, Area of DABC = 1 × AC × AB × Sin ∠BAC
2

= 1 ×4 .5 × 6.5 × Sin 80°cm²
2

= 14.40cm²

C

EXAMPLE 2 In the triangle ABC, AD is the median, D
∠BAD = 30°, AB = 8cm, and AD = 7cm. Find
the area of DABC. 7cm

Solution : We have, A 30° B
8cm
A B = 8cm, AD = 7cm and AD is median of
DABC

Now, A rea of DABD = 1 AB × AD × Sin 30°
2

= 1 × 8 × 7 × Sin 30°
2

= 1 × 8 × 7 × 1
2 2

= 8×7 = 14cm²
4

Again, As we know,

Median bisects the triangle,

DABC = 2 (Area of DABD)

= 2 × 14cm²

= 28cm² .

216 GREEN Mathematics Book-10

EXAMPLE 3 Find the area of the quadrilateral PQRS.

Solution : S 5cm R
120°
6cm 6cm

P 60° 9cm Q

In ∆PQS, PQ = 9cm

PS = 6cm

Area of ∆PQS = 1 PS × PQ × Sin60°
2

= 1 × 6cm × 9cm × 3
2 2

= 27 3 cm²
Again, 2

Area of ∆RQS = 1 SR × RQ × Sin120°
2

= 1 × 5cm × 6cm × 3
2 2

= 15 3 cm²
2

Area of the quadrilateral PQRS = Area of ∆PQS + Area of ∆SQR

= 27 3 cm² + 15 3 cm²
22

= 42 2 3 cm²

= 21 3 cm²

= 36.372 cm²

GREEN Mathematics Book-10 217

EXAMPLE 4 If area of ∆XYZ is 20.78cm², XY = 8cm and ∠XYZ = 60°, find the length
Solution : of YZ.

In ∆XYZ X

Area of ∆XYZ = 20.78cm² 8cm

XY = 8cm

∠XYZ = 60° Y 60° Z

YZ = ?

Now,

We know,

Area of ∆XYZ = 1 X Y × YZ × Sin 60°
2

or, 20.78 = 1 × 8 × YZ × 3
2 2

or, YZ = 20.78
2× 3
\ The length of YZ = 6cm.

EXAMPLE 5 In the parallelogram PQRS, PQ = 8cm , S R
Solution : ∠PSQ = 80°, QS = 6cm and ∠QRS =60°. 80° 60°
Find the area of PQRS.
6cm

In the parallelogram PQRS, SQ is a P Q
diagonal.
8cm

\ ∠SRQ = 60°

∠PSQ = 80°

SQ = 6cm and

PQ = 8cm

We know, opposite angles are equal in parallelogram

\ ∠QPS = ∠QRS = 60°

Again, In DPQS

∠QPS + ∠QSP + ∠PQS = 180°

or, 60° + 80° + ∠PQS = 180° Sum of interior angles of a triangle.

or, ∠PQS = 40°
Here,
Area of DPQS = 1 × PQ × QS × Sin 40°
2

= 1 × 8cm × 6cm × Sin 40°
2

= 15.42cm²

218 GREEN Mathematics Book-10

Hence,
Area of parm PQRS = 2 (Area of DPQS)

= 2 × 15.42cm²

= 30.84cm²

EXAMPLE 6 If the area of the trapezium AECD D C
is 46.77cm², find the length of AE,
where, ∠CBE = 60°.

Solution : We have, AE
A rea of trap. AECD = 46.77cm² B
∠CBE = 60°
Each angle of an equilateral triangle.

\ ∠DAB = 60° ∵ Corresponding angle of ∠CBE

AE = ?

Now, Let,

AB = BE = EC = CD = AD = BC = a

AD = a and BC = a
Here,
A rea of DBEC + Area of parm ABCD

= Area of trap. AECD

or, 46.77cm² = (a × a × Sin 60°) + (a × a × Sin 60°)
2

or, 46.77cm² = oa ² × 3p + oa² × 3 p
42

or, 46.77cm² = 3 × a ² + 2 3 ×a2
4

or, 187.08cm² = 3 (a² + 2a²)
or, 187.08cm² = 3a²

3
or, 1083cm² = a²
or, a = 36
∴ a = 6
∴ AE = 2 × 6 cm
= 12cm.

GREEN Mathematics Book-10 219

EXAMPLE 7 Find the area of the polygon ABCDEA. A E
Solution : Here,
2 cm
AB =2 cm, BC = 1cm 3cm
∠ABC = ∠ACD = ∠ADE = 90°
CD = 2cm, DE = 3cm B D
We know, 1 cm C 2 cm
AC = AB² + BC²
= 2² + 1²
= 5 cm

Again, in DACD. ∵ Being DABC is a right angled triangle.

AD = AC² + CD²

= ( 5 )2 + 2²
= 5 + 4

= 9 cm = 3cm

Now, In DADE,

Area of DADE = 1 × AD × DE sin90°
2

= 1 × 3 × 3 cm² × 1
2

= 4.5 cm²

Area of DACD = 1 × CD × AC × sin90°
2

= 1 × 2 × 5 cm² × 1
2

= 5 cm2

Area of DABC = 1 × BC × AB × sin90°
2

= 1 × × 1 × 2 cm² × 1
2

= 1cm2

Here, Area of polygon ABCDE = Area of (DACD + DADE + DABC)

= 4.5cm² + 5 cm² + 1cm²

= 7.73cm²

220 GREEN Mathematics Book-10

EXERCISE 12.1

A. Very Short Questions

1. Find the area of the triangles given below:

a. A b. A c. A

5cm 60° 4cm 8cm 68° + x 8cm

30° Bx 2x C

B 4cm C B C 9cm

d. A e. X f. X

80° 6.5cm Y ZY Z
C
B 70° Perimeter = 30cm
7.5cm
h. P 6cm A
X 7cm x
Z 2x 3x i. 2x
g. 6cm Q
RB 30° C
Y 30° 3x
6cm
AB + BC = 20cm

B. Short Questions
2. Find the area of the quadrilaterals given below:

a. D 4cm C b. S 4cm R c. P 5cm O
150°
Q M
6cm

3cm
3cm
A 60° B P 60° N
8cm6cm
4 cmT

d. D 5cm10cmCe.6 cmC f.D 6cm C
8cm60°D150°
A
30° 120°
5cm B
BA 30° B A

GREEN Mathematics Book-10 221

A D

3. a. In the given figure, ABCD is a rhombus. If ∠BED = E
C
∠ADE = 90°, ∠CDE = 30°, CE = 2cm and CD = 4cm, D

find the area of trapezium ABED. B C

b. Find the area of kite ABCD where AD = 5cm, DB = 15cm, A
∠ADB = 30°.

12cm B
PO

c. Find the area of parallelogram MNOP. 230° N
7
M 10cm

d. Area of the parallelogram ABCD is 48 cm². If D C
∠DAB = 30° and AD = 8cm, find the length of R
AB. A 30° C
B
P

e. The given figure PQR is a triangle where Q
PR = 7 2 cm, QR = 6cm and the area of the
DPQR = 21cm². Find the measurement of ∠PRQ.

f. In the figure, AD is the median of the ∆ABC. If B 6cm
∠CAD = 30°, AD = 6cm and AC = 8cm, find the area A
of DABC. 30°

D

g. In the figure, ED = 10cm, AC = 13cm and BD = 5cm. 10cm 5cm D
Find the area of trapezium ACDE. E C

A 13cm B

222 GREEN Mathematics Book-10

h. The given figure is a rhombus. If MN = 12cm and the area M

of the rhombus = 72 3 sq.cm, find the measurement of P
∠NOP.
6cm O N
P

i. Find the area of quadrilateral MNOP. 30°1 cm4cm O
5cm
M N
1cm 8cmAE

j. Find the area of the polygon.

D D C
B 1 cm C 1 cm

k. In the given figure, ABCD is a parallelogram and

BCE is an equilateral triangle. If DC = 10cm, BE =

6 3 cm, find the area of the trapezium. A

D E
B

C

l. In the figure given alongside, BCE is an equilateral B E
D
triangle and ABCD is a rhombus. If the area
of the trapezium AECD is 27 3 cm², find the A
measurement of AB.

m. Find the area of quadrilateral ABCD in which ∠DAC = 75° and 5cm 7cmC

∠BAC = 60°. A

B
M

n. Find the area of kite MNOP in which MO = 30cm, 10cm
MP = 10cm, ∠NOM = 30° and ∠OPM = 105°. 30cm

105°
NP

30°

O

PQ T

o. In the given figure, area of trapezium 120° S
PRST is 12 3 cm², find the length of PQ
R
where PQ = QR = RS = ST = QT.

GREEN Mathematics Book-10 223

12.2 Height and Distance

Introduction

Trigonometric ratios are used to solve the problems related to the height and distance.
Measuring length, breadth and height of a room is not difficult. But height of pole, tower,
building, tree and so on is a bit difficult to find. Similarly, to find the width of river is
also a bit difficult task. Therefore, from long time trigonometric ratio and Pythagoras
formula have been used to make this task easy and simple. The skill and knowledge
about trigonometry helps us to solve above problems. Also, only right angled triangle is
required. Trigonometric ratio is useful to take the measurement of height of the following
objects.

Height Height
Height

Line of sight:

Suppose we are looking at an object standing on the ground. Clearly, the line of sight (or
line of vision) to the object is the line from our eyes to the object which we are looking at.

Angle of elevation: Line of sight P

Let P be the position of an object above the horizontal Angle of elevation
line OX, where O is the position of the eye of an observer OX
looking at the object. Then ∠XOP is known as angle of
elevation. The angle XOP is formed by the line of sight Horizontal line
OP with horizontal line OX.

Angle of depression: O Horizontal line
Let P be the position of an object below the horizontal line X
OX, where O is the position of the eye of an observer looking Angle of depression
downward the object. Then ∠XOP is known as angle of
depression. The angle XOP is formed by the line of sight OP Line of sight
with the horizontal line OX
Here, ∠XOP = ∠MPO MP

224 GREEN Mathematics Book-10

Basic trigonometric ratios are

Sin q = Perpendicular (P)
Hypotenuse (h)

Cos q = Base (b)
Hypotenuse (h)

Tan q = Perpendicular (P)
Base (b)

From the above three trigonometric ratios, we use p, b, h and q. And among four items
three are related. But among three if any two are given, we can find the remaining item.

For example: A
Here, in the right angled triangle ABC,

∠ACB = 90°, ∠ABC = q = 30° ?

BC = base = 50m B θ =30°
C
AC = Perpendicular = ?

Therefore,

We found the relation between q, base (b) and perpendicular (p). But among three, we
have two values.

i.e. q = 30° Remember • Slop
b = 50m • Slant
P = ? • Perpendicular • Ladder
• ree • String
So, easily we can apply, Tan q = p • Pole • Hypotenuse
b • Pilar
• Height
• House

or, Tan 30° = AC m • Base
50 • Shadow
• A part
or, AC = Tan 30° × 50 m • A way
• Distance
AC = 28.87 m

∴ The length of perpendicular is 28.87m.

GREEN Mathematics Book-10 225

Worked Out EXAMPLES

EXAMPLE 1 A boy observed a bird sitting on the pole of height 20m and found
Solution :
angle of elevation 30°. Find the distance A

between the boy and the pole. 20m = P

Let AB be the height of the pole, BC be 30° B
the distance between boy and pole and C
∠ACB be 30°, the angle of elevation.
b = ?

In right angled triangle ABC,

We have, Tan q = p
\ b

or, Tan 30° = 20bm

or, b = 20m
Tan 30°

\ b = 34.64m

Hence, the distance between the boy and pole is 34.64m.

EXAMPLE 2 Inclination of a ladder leaning against a wall is 30° and the foot of the
ladder is 2m away from the wall. Find the length of the ladder.

C

Solution : Let AB be the distance between the foot of the
ladder and the wall, BC be the height of the
wall, AC be the length of the ladder and angle A 30° B
of elevation be 30°. 2m

Now, in right angled DABC,

q = 30°

b = 2m

h = ?

Now, Cos q = AB [∵ q, b, h are correlated, cos q = b ]
or, AC h
or,
\ Cos 30° = 2m
h

h = 2
Cos 30°

h = 2.31m.

Hence, the length of the ladder is 2.31.m.

226 GREEN Mathematics Book-10

EXAMPLE 3 Find the angle of elevation of the top of a tower of height 100
3 m from a point at a distance of 100 m from the foot of the tower.

R

Solution : 100 3 m

Let the height of the tower RQ be 100 3 m

and PQ = 100m. Let the ∠RPQ = θ P θ Q
I n right angled D PQR, 100m

\ Tan q = RQ [∴ Tanq = p ]
PQ b

or, Tan q = 100 3 m
100m

or, Tan q = 3
\ θ = Tan – 1 ( 3 ) = 60°
Hence, angle of elevation = 60°.

EXAMPLE 4 A boy of height 1.5m is flying a kite with string 200m and he made
Solution :
angle of elevation at his eye 60°. Find the A
height of kite from the ground.

Let AB be the length of string, BC be the 200m

height of boy, AD be the height of kite from 60° E
Rectangle 1.5m
the ground and angle of elevation be 60°. B
D
C

So, ∠ABE = 60°, AB = 200m, BC = ED = 1.5m

Now, in right angled triangle ABE,

\ Sin ABE = AE
AB

AE
or, Sin 60° = 200m

or, AE = Sin 60° × 200m

or, AE = 173.20m

\The height of kite (AD) = AE + DE
= 173.20m + 1.5m
= 174.70m

Hence, the kite is 174.70m high from the ground.

GREEN Mathematics Book-10 227

EXAMPLE 5 The vertical stick 10cm long casts a shadow 8cm long. At the same

time, a tower casts a shadow 30m long.

Determine the height of the tower. D

Solution :

Let, AB be the stick, BC be the shadow of x A
stick, DE be the height of tower and EF be
the shadow of tower and the angle of eleva-
tion is the same, i.e. q.

Now, in right angled triangle ABC 10cm

\ Tan q = 10 θ
8 B 8cm C
E 30m
or, Tan q = 45

\ q = Tan–1 5 = (51.34)°
4

Again, In right angled triangle DEC,

\ Tan q = DE
30m

or, Tan(51.34)° × 30 m = DE

\ DE = 37.5m

Hence, the height of the tower is 37.5m.

EXAMPLE 6 A tree is broken due to the storm and the broken part bends so that
the top of the tree touches the ground making an angle of 30° with the
ground. The distance from the foot of the tree to the point where the
top touches the ground is 10 meters. Find the height of the tree before
being broken.

Solution : Let AB be a tree before being broken. The broken part AC touches

the ground at D, making an angle of 30° with the ground. A

Let, AC = xm

\ CD = xm and CB = ym, DB = 10m x

Now, in right angled DCBD, C

\ Tan 30° = y xy
10

\ y = Tan 30° × 10m D 30°

\ y = 5.8m 10m B

228 GREEN Mathematics Book-10

Again,
x = BD² + BC²
= 10² + (5.8)²
= 100 + 33.64
= 133.64
\ x = 11.56m
Hence, h eight of tree = AC + BC = x + y
= (11.56 + 5.8)m
= 17.36m

EXAMPLE 7 A tree of height 60m standing on the bank of a river is observed from
the opposite bank of the river and found the angle of elevation to be
30°. What is the width of the river?

Solution : Let AB be the height of tree and A
its height is 60m. BC is the width B
of the river and angle of elevation 60m
is 30°.
30°
Now, C

In right angled DABC,

AB = P = 60m

BC = b = ?

θ = 30°
Now,
Tan θ = p
b
or,
AB
or, Tan 30° = BC
\
1 60m
3 = BC

BC = 60 3 m

Hence, the width of river is 60 3 m

GREEN Mathematics Book-10 229

EXAMPLE 8 A player sitting on the top of a tower of height 20m observed the an-

gle of depression of a ball lying on the

ground as 60°. Find the distance be- A
tween the foot of the tower and the ball. D 60°

Solution :

Let AB be the tower and its height is 20m

20m. Angle of depression ∠DAC = 60°.

The distance between tower and ball is 60° B
BC = x = ? C
x
∠ACB =∠ DAC = 60° being alternate
angles equal.

Now, In right angled triangle ABC,

20
Tan 60° = x

or, 3 = 20
x

or, x = 20
3

or, x = 11.55m

Hence, the distance between tower and ball is 11.55m.

EXERCISE 12.2

A. Long Questions

1. a. Angle of elevation from the roof of a house to the top of a tree is found to be
30°. If the heights of the house and tree are 6m and 18m respectively, find the
distance between the house and the three.

b. A person of height 2m observes the angle of elevation of the top of a pole 62m
high which is in front of him and finds it to be 45°. Find the distance between the
person and the pole.

c. A man 1.6 m tall observes the angle of elevation of top of house and found to be
60°. If the distance of man from the foot of the house is 100m, find the height of
the house.

d. From the roof of a house 6m high, the angle of elevation of the top of a temple
66m high is observed to be 60°. Find the distance between the house and the
temple.

e. A man observes the top of a pole of 52m height, situated infront of him and finds
the angle of elevation to be 30°. If the distance between the man and the pole is
86m, find the height of the man.

f. An observer from the top of a house of 8m high and finds the angle of elevation
of the top of the tower to be 60°. If the distance between the tower and the house

230 GREEN Mathematics Book-10

is 20 3m, find the height of the tower.

2. a. From the roof of a house of 6m tall, the angle of elevation of the top of a tower
was observed and found to be 30°. If the distance between the house and the
tower was 19 3m, determine the height of the tower.

b. A man of 2m. observes the angle of elevation of the top of a tower and finds to
be 60°. If the distance between the man and the tower is 45 m, find the height of
the tower.

c. A man 1.7m tall observes the angle of elevation at the top of the column and
finds to be 60°. If the distance between the man and the column is 25m, find the
height of the column.

d. A man 1.6m tall stands 30m. far from the foot of the pole. He finds the angle of
elevation to be 45° while observing the top of the pole, find the height of the
pole.

3. a. A boy of height 1.5 m is flying a kite. When the length of string is 200m, it makes
an angle of 30° with the horizon, find the height of the from the ground.

b. An observer finds the angle of elevation of the top of a pole to be 60°. If the
height of the pole and the observer are 25.3 m and 1.3m respectively, find the
distance between the observer and the pole.

c. The thread of a kite makes an angle of 60° with the horizon while a boy of height
1.8 m is flying a kite. If the length of the thread is 300 m, find the height of the
kite form the ground.

d. A man of height 1.68m observes a pillar of height 51.68m and finds the angle of
elevation of the top of the pillar to be 60°. How far is the man from the base of
the pillar? Find it.

e. A man of height 1.7m observes the angle of elevation of the top of a pole situat-
ed in front him and finds it to be 60°. If the height of the pole is 61.7m, find the
distance between the pole and the man.

f. A boy of 1.7m. tall, is flying a kite. When the length of the string of the kite is 1
80m., it makes an angle of 30° with the horizon. What is the height of the kite
from the ground?

g. The horizontal distance between two towers is 70m. The angles of depression of
the top of the first tower, when seen from the top of the second tower, is 30°. If
the height of the second tower is 120m, find the height of the first tower.

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

b. From the top of a hill 51.6m high, the angle of depression of the top of the tree
of height 6.6m, situated in front of the hill, was observed and found to be 30°.
Find the distance between the hill and tree.

c. Form the top of a tower of 80m high, the angle of depression of the top of a pole

GREEN Mathematics Book-10 231

of 20m high lying in front of the tower is observed and found to be 45°. Find the
distance between the tower and the pole.

d. From the top of a tower, the angle of depression of the roof of a house 20m high
which is 60m. away from the tower was found to be 60°. Find the height of the
tower.

e. From the top of a tower, the angle of depression of the top of a tree 6.6m high
and 45 3 m away from the tower was found to be 30°. Find the height of the
tower.

f. From the top of a tower 91 m high, the angle of depression of the top of a house
12m high on the same level of the ground was observed and found to be 30°.
Find the distance between the tower and the house.

5. a. A circular pond has a pole standing vertically at its centre. The top of the pole is
30m above the water surface and the angle of elevation of it from a point on the
circumference is 60°. Find the length of diameter of the pond.

b. A pole on the bank of a river is 20m. high and the angle of elevation of the top of
the pole form the opposite bank is 30°. Find the width of the river.

c. The circumference of a circular pond is 176 m and a pillar is fixed at the centre
of the pond. If a person finds the angle of elevation of 60° of the top of the pillar
from any point on the bank of the pond, find the height of the pillar above the
water level.

d. A pillar of 60 3 m height was fixed in the center of a circular meadow. The angle
of elevation of its top was found to be 30° when observed from a point of the
circumference of the meadow. Find the diameter of the meadow.

6. a. The angle of elevation from the roof of a house to the top of a tree is found to
be 30°. If the height of the house ad tree are 8m and 20m respectively, find the
distance between the house and the tree.

b. The angle of elevation of the top of a tree as observed from the roof of a house
is found to be 60°. The height of the house and the tree are 5m and 17m respec-
tively. Find the distance between the house and the tree.

7. a. A man observes the top of 80 3 height from 240m far from the foot of the tower.
Find the angle of elevation.

b. What is the angle of elevation when a lamp at the top of a lamp post 80 3 m
high is observed from a distance of 240m?

c. A girl 1.75 m tall is 50m away from the tower 51.75m high. What is the angle of
elevation of the top of the tower from her eye?

8. a. Atree of 14 meters height is broken by the wind. The tree’s top touches the ground
making an angle of 60°. Find the length of the broken part of the tree.

b. If the top of a tree which is broken by the wind makes an angle of 60° with the
ground at a distance of 15 3 m from the foot of the tree, find the height of the
tree before it was broken.

232 GREEN Mathematics Book-10

9. a. Two vertical poles are fixed 60m apart. The angle of depression of the top of the
first poles as seen from the top of the second pole which is 150m high is 30° find
the height of the first pole.

b. A man 1.73 m tall is standing few meters away from the tower of height 91.8 m
on the same level of the ground. If he finds the angle of elevation to top of the
tower to be 60°, how far is he from the foot of the tower? Find it.

c. A flag staff of height 7 meters stands on the top of a tower. The angle subtend-
ed by top of the flag staff at a point 48m away on the ground is 45°. Find the
height of the tower.

Unit Test Time : 1 hour
FM. 32
Each question carries 2 marks.
C

1. a. If the area of ∆ABC is 27 3 m², AC = 12 3 m and

∠BAC = 30°, find the length of AB. A 30° B
P

b. Find the area of DPQR, in which ∠P = 70°, ∠R = 50°, PQ 5cm 6cm 70° R
= 6cm and QR = 8cm. Z
50°
Q 8cm
W

8cm

2. a. Find the area of parallelogram WXYZ, where, XW = 5cm, X Y
7cm
XY = 7cm and WY = 8cm.

b. Find the area of the DABC, if the perimeter and the ratios of the sides are 56cm

and 3:5:6 respectively. P

3. a. In the DPQR, PQ = 6cm, and ∠PQR = 75°. Find the area of

DPQR. Q R

GREEN Mathematics Book-10 233

b. If the area of DXYZ is 20 3 cm², XY = 8cm and X
∠YXZ = 60°, find the length of XZ. 60°

P Y Z

4. a. Find the area of DPQR, in which R R
PR = 6cm, QR = 10cm and Q
S
∠QRP = 30°. Q D
30° C
b. Find the area of the quadrilateral PQRS, where

PR = 3 SR = 2 PQ = 24cm. P

5. a. Find the area of kite ABCD, in which BD = 70cm,

∠DCB = 105°, ∠ADB = 45° and AB = 40cm.

W 30° Z A

b. Find the area of rhombus WXYZ. 8cm

XY B

Each question carries 4 marks.

6. Solve the following:

a. A vertical straight tree, 15m high, is broken by the wind and its top touches the
ground and makes an angle of elevation 60° with the ground. Find at what height
from the ground was the tree broken?

b. A boy 1.6m tall is standing on the ground at a distance of 50m from a tree and finds
the angle of elevation of the top of the tree to be 30°. What is the height of the tree?

c. A straight wooden pole is in the centre of a circular pond. A boy of height 1.5m
observes the top of the pole and finds 30° angle of elevation. Find the height of the
pole above the water level, if the diameter of the pond is 80m.

234 GREEN Mathematics Book-10

13

Statistics

Estimated Teaching Periods : 17

Pafnuty Chebyshev (1821-1894) is a Russian mathematician who is well
known for Chebyshev’s Theorem, which extends the properties of normal
distributions to other, non-normal distributions.

Contents

13.1 Introduction
13.2 Quartiles-Introduction
13.3 Cumulative frequency curves or ogives

Objectives

At the end of this unit, students will be able to:
introduce statistics; represent the statistical data by means of statiscal diagram such as
histogram, line graph, pie chart, frequency polygon and cummulative frequency curve.
calculate the value of mean, median, mode and quartiles of ungrouped data.

Materials

Scale, signpen, colour pencils, card board, chart paper, graph paper, etc.

GREEN Mathematics Book-10 235

13.1 Introduction

Statistics :

The word "Statistics" has been derived from the Latin word "Status" or Italian word
"statista" meaning thereby a political state. At the beginning in society statistics was
considered as science for fulfilling the needs of state administration. Also it was known
as science of kings. During the time the chiefs or the kings needed the information about
the population, strength of military, etc. for forming ancillary and fiscal policies. But now
a days, it is very essential in modern society. It is not only important for administrative
work to run the state. Its area is broad; we are using it in Economics, Science, Business,
Research, Bank, Insurance field, etc. So, the use of statistics is rapidly growing universally.
According to Mr. A.L. Boweley, statistics is the "science of counting" or the "science of
averages".

The four broad aspects of statistics are as follows:

1. Collection of data: It is the initial step in the statistical investigation.

2. Presentation of data: After collecting data it should be presented in :

a. Tables b. Diagrams c. Graphs.

3. Analysis of data: After the presentation of data, analysis of data takes place. We need
the concept of average dispersion, analysis of variance, experimental designs, etc.
The choice of appropriate techniques of analysis would depend upon the type of
data available and the purpose of enquiry.

4. Interpretation of data : Interpretation of data is very important task and requires high
degree of skill and experience. If the data are not properly interpreted, the whole
objects of enquiry may be defeated and fallacious conclusions be drawn.

Importance of Statistics:

• It is useful in the administrative works of state.

• It is useful to the business sector.

• It is useful to the economists.

• It is useful to the researchers.

• It is useful in plans making.

• It is useful to the bankers, insurance companies, etc.

• It is useful in supervision.

• It helps in the development of other science.

The meaning of the statistical terms:

Data: The pieces of information are called data. It is for representing the facts or ideas in
a formalized manner.

236 GREEN Mathematics Book-10

Primary data: The data which is collected by the investigator for a certain objective is
called primary data. Thus, it is original in character. For example: the census of statistics
collected by the census department of the government of Nepal provides first hand
information. It is collected on the spot by the investigator involving himself/herself with
the target group. Methods to collect primary data are: interviews, information from
correspondents, questionnaire method, etc.

Secondary data: When the primary data is used for next purpose by any other person
or department, then it is called secondary data. It is collected through offices from the
primary data collection.

For example : When the data collected by the census department of the government of
Nepal is used by any other, then it is considered as secondary data.

Mode

The value of a set of data repeated more than other is called mode.

For example: Mode in individual series

In the data 5, 6, 6, 9, 8, 7, 6, 5, 4, 5, 6, find mode.

Arranging above data in ascending order.

We have, 4, 5, 5, 5, 6, 6, 6, 6, 7, 8, 9

The number 6 is repeated maximum times. Therefore, mode is 6.

If two or more than two value are found repeated at maximum times then we use
another formula.

Mode = 3 median – 2 mean i.e. mode = 3 (median of data) –2 (mean of data)

For example :

10, 11, 11, 12, 12, 12, 13, 13, 13, 14, 15

Here, the data 12 and 13 are repeated maxium times.

So, Mean = 10+11+11+12+12+12+13+13+13+14+15

= 136 = 12.37 11
11
Again,

Position of median = N+1 th
2
items

= 11+1 th ∴ mode = 3 median – 2 mean
2 = 3 × 12 – 2 × 12.37
items

= 6th items = 36 – 24.73

Median = 12 = 11.27

GREEN Mathematics Book-10 237

Mode in descrete series

Find the mode value from the following data.

Size of shoes (X) Number of pairs (f) Here, the size of shoes numbered size
9 is repeated 20 times. So, mode value
6 10 is 9.
7 14
8 15
9 20
10 11

N = 70

Mean (x) or Arithmetic mean or Average:

A distinct value For example:
obtained from the Marks in maths (x) = 40, 50, 20, 60, 80, 90
given data dividing
the sum of data by Number of students (N) = 6
the number of data is
called mean or average Sum of marks (∑x) = 40 + 50 + 20 + 60 + 80 + 90
or arithmetic mean.
= 340
Mean (x) = Σx

N
= 340 = 56.66

6

Individual Series: For Example:

The data are given Marks in maths (x) = 10, 80, 70, 90, 60
individually in this
series. Σx
N
Mean (x) =

= 10+80+70+90+60 = 310 = 62
5 5

Descrete Series: For Example:

The data which are Mean (x) = Σfx
given with their Data (x) N
corresponding frequency
are called descrete series. 10 20 30 40 50
4
f 1325 200

fx 10 60 60 200

238 GREEN Mathematics Book-10

Continuous Series: For Example: 0–10 10–20 20–30 30–40
Class 2 124
The series in which f
the data are given in
grouped form with ∵ Mean (x) = Σfm , where m is mid value of each class.
the corresponding N
frequency are called
continuous series.

Like, 0 + 10 = 5 is the mid value of first class interval.
2

i.e. Mid value (m) = Lower limit + Upper limit
2

∴ Mean (x) = Σfm
N

Grouped data:

There are two types of grouped frequency distribution.
i. Exclusive type ii. Inclusive type

For example:

Exclusive type: In this type, the upper limit of the class interval is same as the lower
limit of the succeeding class interval. Upper limit is not required to count in the
particular class.
In 0 – 10, all the numbers from 0 to 9 are included but 10 will be excluded.
Inclusive type: The classes in which both the upper and lower limit are included in
the particular class are called inclusive type.
In 0 – 10, all the numbers from 0 to 10 are included.

Correction Factor (CF) = Lower limit of 2nd class – upper limit of 1st class
2

or,
Correction Factor (CF) = Lower limit of a class – upper limit of previous class

2

GREEN Mathematics Book-10 239

Worked Out EXAMPLES

EXAMPLE 1 Use tally marks for the same marks obtained in maths :
5, 5, 10, 12, 18, 20, 18, 12, 20, 15, 5, 5, 20, 5
Solution :
Marks Tally marks Frequency (f)

5 |||| 5
| 1
10 || 2
12 | 1
|| 2
15 ||| 3

18 N = ∑f = 14

20

EXAMPLE 2 Find the mean (x) of the given data : 10, 12, 18, 20, 30.
x = 10, 12, 18, 20, 30
Solution :
N = 5

Σx = 10 + 12 + 18 + 20 + 30 = 90
Mean (x) = Σx

N
= 90

5
= 18

EXAMPLE 3 Observe the data given below and answer the following questions:

Solution : 5, 10, 6, 8, 10, 5, 6, 12, 8, 10, 5, 10
i. How many students got 5 marks?
ii. Which marks is repeated more than other?
iii. Which mark is not repeated?
i . Three students got 5 marks.

ii. 10 is repeated more than other.

iii. 12 is not repeated.

EXAMPLE 4 Write the frequency for the given tally marks.
(|||| ||), (||||, ||||, |||), (||||, ||||)
Solution : The frequency for the 1st, 2nd and 3rd are, 7, 13, 9 respectively.

240 GREEN Mathema tics Book-10

EXAMPLE 5 Find the mean (x) from the given observation.

Marks 30 40 50 60 70 80

No. of Students 2 5 4 12 8 9

Solution : Marks (x) f f×x
30 2 60
5 200
40 4 200
50 12 720
60 8 560
70 9 720
80 N = 40 Σfx = 2460

Mean (x) = Σfx
N

= 2460
40

= 61.5

EXAMPLE 6 If mean (x) of the data given is 12, find the value of x.

8, 10, x, 12, 14, 16

Solution : Here,

N = 6

Σx = 8 + 10 + x + 12 + 14, 16 = 60 + x

Mean (x) = Σx
N

∴ Mean (x) = 60 + x
6

or, 12 = 60 + x
6

or, 72 = 60 + x

∴ x = 12

GREEN Mathematics Book-10 241

EXAMPLE 7 Find the mean (x) from the given data:

Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
f 21354

Solution : Marks (x) f Mid value (x) f × x

0 – 10 2 5 10

10 – 20 1 15 15

20 – 30 3 25 75

30 – 40 5 35 175

40 – 50 4 45 180

Σf = 15 Σfx = 455

We know,
Mean (x) = Σfx

N

= 455 = 30.33
15

EXAMPLE 8 The mean of the data given below is 36. Determine the value of p.

Age (in year) 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60

No. of students 3 8 15 p 4

Solution : Age (class) Mid value (x) f f×x

10 – 20 15 3 45
20 – 30 25 8 200
30 – 40 35 15 525
40 – 50 45 p 45p
50 – 60 55 4 220

N = 30 + p Σfx = 990 + 45p

Here, we know,
Mean (x) = 36
N = 30 + p
Σfx = 990 + 45p
Mean (x) = Σfx

N
or, 36 = 990 + 45p

30 + p
By cross multiplication, we get
or, 1080 + 36p = 990 + 45p
or, 9p = 90
∴ p = 10

242 GREEN Mathematics Book-10

EXAMPLE 9 Find the average mark from the following frequency distribution

by deviation method.

Marks 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100

No. of students 6 8 12 8 6

Solution : Marks Mid-value (x) No. of students (f) d = x – 75 fd
50 – 60 55 6 – 20 – 120
60 – 70 65 8 – 10 – 80
70 – 80 12 0
80 – 90 a = 75 8 10 0
90 – 100 85 6 20 80
95 120
Here, N = 40 Σfd = 0

a = 75

Σfd = 0
N = 40

x = ?

We, know,
Mean (x) = a + Σfd

N
= 75 + 0

40

= 75

EXAMPLE 10 If ∑fx = 50 + 8a, ∑f = N = 11 + a and mean (x) = 6, find the value of a.

Solution : We, have

Σfx = 50 + 8a

Σf = N = 11 + a
Mean x = 6

a = ?

Now, we know,

Mean (x) = Σfx
Σf

50 + 8a
or, 6 = 11 + a

By cross multiplication, we get,

or, 66 + 6a = 50 + 8a

or, 16 = 2a

∴ a = 8

GREEN Mathematics Book-10 243

EXAMPLE 11 If ∑fx = 36 + 48a and N = 9 + 12a, find the mean (x).

Solution : We, have

Σfx = 36 + 48a

Σf = N = 9 + 12a

Mean (x) = ?

Now, Mean (x) = Σfx
Σf

Mean (x) = 36 + 48a
9 + 12a

= 132((33++44aa))
= 132 = 4

EXAMPLE 12 If the mean (x) = 10, ∑fx = 500, find the number of frequency.

Solution : We, have

Mean (x) = 10

Σfx = 500

Number of frequency (N) = ?

We know, Σfx
N
Mean (x) =

or, N = MeΣafnx(x)

N = 51000
∴ N = 50

EXAMPLE 13 If the data are given in ascending order, find the mode.

10, 12, 20, 20, 30, 30, 30, 40

Solution : Number No. of repeatation

10 1
1
12 2
20 3
1
30

40

Here, 30 is repeated three times, which is the highest frequency.
Hence, mode = 30.

244 GREEN Mathematics Book-10