Green C Math Class 10

e. Three years ago, the sum of ages of the father and his son was 54 years. 3 years
later the father's age will be double of his son's age after 9 years. Find their
present ages.

f. The ages of two girls are in the ratio of 2:3. 6 years hence, the ratio of their ages
will be 11:15. Find their present ages.

g. If 3 is added to the numerator of a fraction, the value of the fraction becomes 1.
When 2 is subracted from the denominator, the value of fraction becomes 1 ,
2
find the fraction.

h. A number consisting two digits exceeds 4 times of the sum of the digits by 3.
If 27 is added to the number, the places of the digits of the number are inter
changed. Find the number.

i. There is a number between 10 and 100. The number is 8 times the sum of its
digits. If 45 is subtracted from it, the result is the number formed by inter
changing its digits. Find the number.

j. Three years ago, the ratio of the ages of A and B was 4:3. Three years hence, the
ratio of their ages will be 11:9. Find their present ages.

k. A year later, a mother's age will be 5 times of her daughter’s age. Two years
ago if the mother's age was three times of her daughter's age of four years
later, find their present ages.

l. In a two digit natural number, the digits of unit place is 4 more than that of the
digit in ten's place. If the product of their digit is 21. Find the numbers.

m. In a number of two digits, one of the digit is greater than other by 5. When the
digits are reversed, the number becomes 3 of the original number, find the
8
original number.

n. The sum of the digit of two digit number is 6. The number with the digit in
reverse order is 4 times the original number. Find the original number.
7

o. The number consists of two digits. The sum of its digits is 16. If 18 is added to
the number, the digits interchange their place. Find the number.

p. A number consists of two digits. The sum of its digits is 16. If 18 is subtracted
from the number, the digits interchange their place. Find the number.

q. A number lies between 10 and 100. If the number formed by reversing its digits
is added to it, the sum is 143 and if the same number is subracted from it, the
remainder is 9. Find the number.

r. The present age of the father is thrice as old as the age of the daughter 5 years
hence. If 12 years ago, the age of the father was twice as old as the age of the
daughter 8 years hence, find their present ages.

s. The ages of two girls are in the ratio 5:7. Eight years ago, their ages were in the
ratio 7:13. Find their present ages.

t. 2 years ago, father's age was 9 times of the son's age but 3 years later, it will be

GREEN Mathematics Book-10 145

five times only. Find the present ages of father and son.

u. 5 years ago, Ramesh was 2 times as old as Sabin. Ten years hence, Ramesh
3
will be 5 times as old as Sabin. Find their present ages.
6

v. Find the fraction which has reduced value 2 , if the numerator is increased by
3
4 and the denominator is decreased by 2, then resulting fraction has the value

2.

w. If the numerator of a fraction is multiplied by 4 and the denominator is reduced
by 2, the result is 2. If the numerator of the fraction is increased by 15 and 2

is subtracted from the double of the denominator, the result is 9 . Find the
7

fraction.

8.2 Word problems on quadratic equations

Worked Out EXAMPLES

EXAMPLE 1 Solve: x = 36
9 x

Solution : x = 36
9 x

or, x2 = (9 × 36)

or, x2 = (3 × 6)2

∴ x = ± 18

EXAMPLE 2 Solve : y2 – 3y – 10 = 0

Solution : y2 – 3y – 10 = 0

or, y2 – 5y + 2y – 10 = 0

or, y(y – 5) + 2(y – 5) = 0

or, (y – 5) (y + 2) = 0

Either, y – 5 = 0, or, y + 2 = 0

y = 5 or, y = – 2

∴ y = 5 or –2

146 GREEN Mathematics Book-10

EXAMPLE 3 If the product of two consecutive integers is 56, find the integers,

Solution : Let the smaller integer be x, then other is (x + 1)
Then, x (x + 1) = 56
or, x2 + x = 56
or, x2 + x – 56 = 0
or, x2 + 8x – 7x – 56 = 0
or, x(x + 8) – 7 (x + 8) = 0
or, (x + 8) (x – 7) = 0
Either,
x+8=0
or, x = – 8
or, x – 7 = 0
x=7
∴ x = – 8 or – 7
∴ The integers are –8 and –7 or 7 and 8.

EXAMPLE 4 Find an integer when it is added with its inverse, the sum will
26 .
be 5

Solution : Let the integer be x then its inverse = 1
x

Now, x + 1 = 26
x 5

or, x2 + 1 = 26
x 5

or, 5x2 + 5 = 26x

or, 5x2 – 26x + 5 =0

or, 5x(x – 5) – 1 (x – 5) = 0

or, (x – 5) (5x – 1) = 0

Either,

x–5 =0

or, x = 5

or, 5x – 1 = 0

or, 5x = 1

or, x = 1 (rejected)
5

∴ the required integer is 5

GREEN Mathematics Book-10 147

EXAMPLE 5 A rectangular garden 50m long and 34m wide is surrounded by
a uniform road. Find the width of the road if total area of the
road is 540m2.

Solution : Let the width of road is xm

Then length of garden with road = (50 + 2x)m x

Breadth of garden with road = (34 + 2x)m 34m
Now, (50 + 2x) (34 + 2x) – 50 × 34 = 540m2 50m

or, 1700 + 100x + 68x + 4x2 – 1700 = 540m2

or, 4x2 + 168x – 540 = 0

or, 4(x2 + 42x – 135) = 0

or, x2 + 45x – 3x – 135 = 0

or, x(x + 45) – 3(x + 45) = 0

or, (x + 45) (x – 3) = 0

Either, (x + 45) = 0

or, x = – 45 (Which is impossible)

or, x – 3 = 0

or, x = 3

∴ The required width of the road is 3m.

EXAMPLE 6 If 5 is added to the square of a number, the result is 14, find the
number.

Solution : Let the number be x.

By the question,

x² + 5 = 14

or, x² = 14 – 5

or, x² = 9

or, x² – 9 = 0

or, x² – 3² = 0

or, (x – 3) (x + 3) = 0

Either x – 3 = 0

x = 3

x + 3 = 0

x = – 3

∴ The required number is 3 or – 3.

148 GREEN Mathematics Book-10

EXAMPLE 7 The present ages of a father and his son are 42 years and 16 years
respectively. Find how many years ago, the product of their ages
Solution : was 272?

Let x years ago the product of their ages was 272.

So, (42 – x) (16 – x) = 272

or, 672 – 16x – 42x + x² = 272

or, x² – 58x + 672 – 272 = 0

or, x² – 58x + 400 = 0

or, x² – 50x – 8x + 400 = 0

or, x(x – 50) – 8 (x – 50) = 0

(x – 50) (x – 8) = 0

Either, x – 50 = 0

x = 50

50 years is not possible

x–8 = 0

x = 8

∴ 8 years ago the product of their ages was 272.

EXAMPLE 8 In 11 years, the age of Sanjay will be half the square of the age
13 years ago. Calculate the current age of Sanjay.
Solution : Let his present age be x years.
Age 13 years ago = x – 13
Age in 11 years = x + 11
According to question

x + 11 = (x – 13)2
2

or, 2x + 22 = x2 – 26x + 169

or, x2 – 28x + 147 = 0

or, x2 – 21x – 7x + 147 = 0

or, x(x – 21) – 7(x – 21) = 0

or, (x – 21) (x – 7) = 0

Either x – 21 = 0

x = 21
or, x – 7 = 0
x = 7 (Which is impossible)
∴ Sanjay's present age is 21 years.

GREEN Mathematics Book-10 149

EXAMPLE 9 250 oranges were divided equally among certain number of
students. If there were 25 more students, each would have
Solution : received half orange less. Find the number of students.

Let, number of students = x
Each will get = 250

x
If there are 25 more students,

new number of students = x + 25
Each will get = 250

x + 25
From the question,

250 – 250 = 1
x x + 25 2

or, 250x + 6250 – 250x = 1
x(x + 25) 2

or, x2 + 25x – 12500 = 0

or, x2 + 125x – 100x + 12500 = 0

or, x(x + 125) – 100 (x + 125) = 0

or, (x + 125) (x – 100) = 0

Either, x + 125 = 0

x = – 125

or, x – 100 = 0

x = 100
∴ Required number of students = 100.

EXERCISE 8.2

A. Very Short Questions b. x2 = 16 c. x(x – 15) = 0
1. Solve : f. 2a2 – 8 = 0
e. x2 – 1 = 35 i. b2 + 4 = 40
a. x2 – 1 = 0 99 l. 34p – 3p2 = 0
d. a(a + 1) = 0
h. x = 49
g. m2 = 7m 4x

j. y2 – 90 = 10 k. 2t2 – 15t = 0

150 GREEN Mathematics Book-10

B. Short Questions

2. Solve for unknown value :

a. x – 2 = x+4 b. (5x – 1)2 = 4 c. (2m – 1)2 = 49
x – 1 2x + 2 25

d. (9 – 2p)2 – 50 = 0 e. x+4 = x
x 2

C. Long Questions

3. a. If the sum of number and its reciprocal is 2, find the number.

b. Find two consecutive integers such that the square of the bigger number is
equal to the product of numbers.

c. If the square number is decreased by 15, the value is twice the original number.
Find the numbers.

4. a. The hypotenuse of a right angled triangle is 20cm. If the difference of lengths
of the two sides is 4 cm, find the length of each.

b. If 15 is subtracted from the square of a number, the result is 34. Find the
number.

c. If 41 is added to the square of a number, the result is 441. Find the number.

d. The sum of two numbers is 20 and their product is 99. Find the numbers.

e. If 17 is subtracted from the square of a natural number, the result is 152. Find
it.

f. If 3 is added to one-third of the square of a natural number, the sum is 30. Find
the number.

g. If 10 is added to 7 times the square of a positive number, the result is 353. Find

the number.

h. Find the dimension of the rectangle. 520m2 2x

x+7

5. a. If the sum of two numbers 28 and their product is 195, find the numbers.

b. Divide 24 into two parts, so that sum of the square of the parts is 360.

c. Find two consecutive integers such that the square of greater is 9 more than

square of the smaller.

d. The sum of two numbers is 60 and sum of their reciprocal is 3 . Find the
numbers. 40

e. If the perimeter of a rectangular plot of land is 104 metres and its area is 640
m², find the length and breadth.

f. Find two consecutive even integers, the sum of whose square is 164.

g. The present ages of two brothers are respectively 15 years and 22 years. After
how many years, the product of their ages will be 408?

h. If the perimeter of a rectangular plot of land is 108 ft. and its area is 720 sq. ft.,
find its length and breadth.

i. One side of right angled triangle exceeds the other by four centimeters. The

GREEN Mathematics Book-10 151

hypotenuse is 20cm. Find the length of the shorter side of right angled triangle.
j. The product of two consecutive negative integers is 1122. What are the

numbers?
k. Determine the value of 'k' so that the two roots of the equation x2 – kx + 36 = 0

are equal.
l. The length of the sides of a right angled triangle are measured as three

consecutive even numbers. Find the values of these sides if area of given
triangle is 24cm2.
m. Two consecutive numbers differ by two units and the sum of their squares is
580. What are these numbers?
n. Three sides of a right angled triangle are proportional to the numbers 3, 4 and
5. Find the length of each side if the area of triangle is 24cm2.
6. a. In a two digit number the product of the digits is 18 and their sum is 9. Find
the number.
b. The present age of father and his son is 30 years and 14 years respectively. Find
how many years ago, the product of their ages was 192.
c. In a two digit number, the product of the digits is 24. When 45 is added to the
number, the digits interchange their places. Find the numbers.
d. The difference of the ages of two brothers is 5 years and the product of their
ages is 84. Find their present ages.
e. In a picnic, the total bus fare of Rs. 360 is to be equally paid by a certain number
of students. If there are 3 more students each would have to pay Rs. 6 less.
Find the number of students and fare to be paid by a student.

Project Work:
a. Estimate the perimeter and area of your classroom. Calculate area and perimeter

by measuring the length and breadth of the room and find the difference in area,
length and breadth to estimate the measurement.

152 GREEN Mathematics Book-10

9

Area of Triangle and Quadrilateral

Estimated Teaching Periods : 13

Euclidean geometry is a mathematical system attributed to the Alexandrian
Greek mathematician Euclid, which he described in his textbook on
geometry: the Elements. Euclid's method consists in assuming a small set
of intuitively appealing axioms, and deducing many other propositions
(theorems) from these.

Contents

9.1 Review
9.2 Theorems
i. Area of parallelograms on the same base between same parallel lines are equal in area.

ii. Area of triangle is one half area of parallelogram on the same base between same parallel lines.
iii. Areas of triangles on the same base between same parallel lines are equal.

Objectives

At the ends of this unit, students will be able to:
prove the theorems related with triangle and quadrilateral.
calculate the area of different plane figures.
relate the area of different types of quadrilaterals and triangles.

Objectives

Chart about concept of congruent and similar triangles.
Graph papers.

GREEN Mathematics Book-10 153

9.1 Review

Area of triangles and quadrilaterals

In the daily life people have to find the area of rooms, area of 4 walls, area of land etc. The
shape of the land may be a square, a rectangle, a triangle, etc. In this unit, we will learn
to find the area of some geometrical figures and the relation between them if they are
formed on the same (or equal) base and between the same parallel lines.

Area of a triangle

In the given figure, ABC is a triangle. In DABC, AD is

perpendicular to the side BC. To find the area of the DABC,

we measure the side BC of the triangle and the length A

of perpendicular drawn on this side from the opposite

vertex. In the previous class, we have already discussed

about the area of the triangle. Area of the triangle is equal

to the half of the product of a side and the length of the

perpendicular from the opposite vertex to the side. BC
D
So, area of a triangle = 1 × side × perpendicular
2

= 12 × base × height (altitude)

∴ Area of D ABC = 1 × BC × AD
2

Area of a paralellogram

The given figure is a parallelogram ABCD. BD is its D C
diagonal which divides it into two equal triangles. From B
a vertex D, a perpendicular DE is drawn to the side AB.

Area of a parallelogram ABCD = 2 area of D ADB A
E
= 2 × 1 × AB × DE
2

= AB × DE

∴ The area of a parallelogram = base × height

154 GREEN Mathematics Book-10

Area of a quadrilateral D

In the given figure, ABCD is a quadrilateral and BD is its AF P2
diagonal. AE and CF are the length of the perpendicular P1 E
drawn from the vertices A and C respectively on the
diagonal BD.

BC

Area of the quadrilateral ABCD = area of D ABD + area of D BCD

= 1 × BD × AE + 1 × BD × CF
22

= 1 × BD (AE + CF)
2

∴ The area of a quadrilateral = 1 × diagonal × (sum of the perpendiculars)
2

= 1 × d × (P1 + P2) where d = BD, AE = P1 and CF = P2.
2

Area of a rhombus

In the given figure, ABCD is a rhombus. AC and BD are D C
two diagonals which intersect each other at a point O. We B
know that the diagonals of rhombus bisect each other at O
right angles.

A

∴ OA = OC, OB = OD and ∠COD = ∠BOC = ∠AOB = ∠AOD = 90°

Area of the rhombus ABCD = area of DABD + area of DBCD

= 1 × BD × AO + 1 × BD × OC
22

∴ Area of rhombus = 1 × BD (AO + OC)
2

= 1 × BD × AC
2

= 1 × product of two diagonals
2

= 1 × d1 × d2, where d1 and d2 are two diagonals
2

GREEN Mathematics Book-10 155

Area of a trapezium

In the given figure, ABCD is a trapezium having FA D C
AD||BC. DE⊥BC and BF⊥DF (Produced DA upto B E
F) are drawn. A diagonal BD divides the trapezium
ABCD into two triangles BCD and ABD. We know
that the perpendicular distance between two parallel
lines is the same measured at all points. Therefore, BF
= DE (since AD||BC). Suppose BF = DE = h.

Now,

Area of the trapezium ABCD = area of DBCD + area of DABD

= 1 × BC × DE + 1 × AD × BF
22

= 1 BC × h + 1 × AD × h
22

= 12 × h (BC + AD)

Hence, the area of a trapezium = 1 × height × sum of the parallel sides
2

Formula to find the area of different geometrical figures:

1. Area of the triangle = 1 × b × h
2

2. Area of a parallelogram = b × h

3. Area of a square = l² = 1 d²
4. Area of a rectangle 2

= l × b

5. Area of a rhombus = 1 × d1 × d2
2

6. Area of quadrilateral = 1 × d × (p1 + p2)
2

7. Area of a trapezium = 1 × h × (sum of parallel sides)
2

156 GREEN Mathematics Book-10

9.2 Theorems

Theorem – 1 DC

Diagonal of a parallelogram bisects the parallelogram.

OR The area of each triangle formed by a diagonal of parallelo-

gram is half of the area of the parallelogram. A B

Given : ABCD is a parallelogram in which BD is its diagonal.

To prove : Area of DBCD = Area of DADB = 1 area of ⎕ABCD.
2

Statement Reasons

1. In DABD and DBCD 1.
i. AB = DC (S) i. Being the opposite sides of ABCD
ii. AD = BC (S) ii. Same reasons as above
iii. BD = BD (S) iii. Being common side

2. DABD ≅ DBCD 2. By S.S.S. axiom

3. DABD = DBCD in area 3. Congruent triangles are equal in area

4. DABD + DBCD = ABCD 4. Whole part axiom

5. 2 DABD = ABCD 5. From statements (3) and (4)

6. DABD = 1 ABCD in area 6. From statement (5)
2

∴The area of DABD = area of DBCD =
1 area of ABCD.
2

Proved

Theorem – 2 D FC E

Parallelograms on the same base and between the same parallels

are equal in area.

Given : Parallelograms ABCD and ABEF are on the same

base AB between the same parallel lines AB and DE. A B
To prove : Area of ABCD = Area of ABEF

Statement Reasons
1. In DADF and DBCE 1.

i. AD = BC (S) i. Being opposite sides of ABCD

ii. ∠ADF = ∠BCE (A) ii. Being corresponding angles (since AD||BC)
iii. Same reason as (ii) but AF||BE.
iii. ∠AFD = ∠BEC (A) 2. By A.A.S. statement.
2. DAFD ≅ DBEC 3. Congruent triangles are equal in area
3. DADF = DBCE in area

GREEN Mathematics Book-10 157

4. DADF + trapazium ABCF = 4. Adding the same trapazium ABCF to both

DBCE + Trap. ABCF in area sides of st. (3) [By addition axiom]

5. ABCD = ABEF in area 5. Whole part axiom from st. (4)

Hence the area of the parallelograms are equal if they stand on the same base and between
the same parallel lines.

ALTERNATE METHOD: FC E

D

Parallelograms on the same base and between the same par-

allels are equal in area.

Given : Parallelogram ABCD and ABEF are on the same

base AB between the same parallel lines AB and XB
DE. A

To prove : Area of ABCD = area of ABEF

Construction : CX perpendicular to AB is drawn from the point C

Statement Reasons
1. Area of ABCD = AB × CX 1. Area of a parallelogram = base × height

2. Area of ABEF = AB × CX 2. Area of a parallelogram = base × height

3. Area of ABCD = Area of ABEF 3. From statment 1 and 2.

Proved.

Note : Motivate the students to prove following corollaries.
Corollary : A parallelogram and a rectangle on the same base and between the

Corollary same parallel lines are equal in area.
: Parallelograms on equal bases and between the same parallel lines are

equal in area.

Theorem – 3

The area of a triangle is equal to half of the area A FE D
of a parallelogram standing on the same base C
and between the same parallels.

Given : A triangle ABC and a parallelogram B

BCDE are on the same base BC and

between the same parallels BC and AD.

To prove : Area of DABC = 1 area of ⎕BCDE.
2

Construction : A line CF is drawn parallel to BA.

158 GREEN Mathematics Book-10

Statement Reasons
1. ABCF is a parallelogram 1. Being BC||AD (given) and BA||CF

2. DABC = 1 ABCF in area (construction)
2 2. Diagonal AC divides the

3. ABCF = BCDE in area parallelogram into two equal
triangles.
4. DABC = 1 BCDE in area 3. Being both parallelograms on the
2 same base BC and between the same
parallel lines BC and AD.
4. From st. (2) and st. (3)

Proved

Hence the area of a triangle is equal to half of the area of the parallelogram standing on
the same base and between the same parallels.

Alternative Method: AE D

Given : A triangle ABC and a parallelogram BCDE are on the
same base BC and between the same parallels BC and AD.

To prove : Area of DABC = 1 Area of BCDE. H FC
2 B

Construct : From E, draw EF⊥BC and from A draw AH⊥BC (produced CB to H).

Statement Reasons
1. Area of BCDE = BC × EF 1. Area of a parallelogram is base × height

2. Area of DABC = 1 × BC × AH 2. Area of a triangle = 1 × base × height
2 2

3. AH = EF 3. Being AD||BC and AH, EF are
perpendicular distance between
parallel lines AD and BC.

4. Area of DABC = 1 × BC × EF 4. From st. (2) and st. (3)
2

5. Area of DABC = 1 area of BCDE 5. From st. (1) and st. (4)
2

Corollary : Area of a triangle is equal to half of the area of a rectangle standing both
Corollary on the same base and between the same parallel lines.

: Area of a triangle is equal to half of the area of a parallelogram standing
both on the equal bases and between the same parallels.

GREEN Mathematics Book-10 159

Theorem – 4

Triangles on the same base and between the same A H D
parallels are equal in area.

Given: Triangles ABC and DBC are on the same base B C
BC and between the parallel lines BC and AD.

To prove : Area of DABC = Area of DBCD in area.

Construction: A line BH is drawn parallel to CD from B meeting AD at H.

Statement Reasons
1. BCDH is a parallelogram
1. BC||HD (given) and BH||CD (by
2. DBCD = 1 BCDH in area construction)
2
2. Diagonal BD bisects the parallelogram
3. DABC = 1 BCDH in area BCDH into two equal triangles.
2
3. Being DABC and BCDH standing
4. DABC = DBCD in area on the same base BC standing on the
same parallels BC and AD

4. From st. (2) and st. (3)

Proved
Hence, the triangles on the same base and between the same parallels are equal in area.

Alternative Method: AD

Given : DABC and DDBC are on the same base BC B G
To prove : and between the same parallels BC and C
AD. H

DABC = DDBC in area

Construction: AH⊥BC where CB produced up to H and DG⊥BC where BC produced
upto G are drawn from the points A and D respectively.

Proof

Statement Reasons

1. Area of DABC = 1 × BC × AH 1. Area of a triangle is equal to 1 × base ×
2 height. 2

2. Area of DDBC = 1 × BC × DG 2. Same reason as (1)
2
3. AH = DG 3. AD||BC and AH, DG are perpendicular
distances between two parallel lines AD
and BC.

4. Area of DABC = 1 × BC × DG 4. From st. (1) and st. (3)
2

5. Area of DABC = area of DDBC 5. From st. (2) and st. (4)

160 GREEN Mathematics Book-10

Corollary -1 : Triangles on equal bases and between the same parallels are equal in
area.

Corollary - 2 : A median divides a triangle into two triangles of equal area.

Corollary - 3 : Triangles of equal area standing on the same base or equal bases
between same parallel lines lie on the same side of the base.

Worked Out EXAMPLES

EXAMPLE 1 Find the area of a parallelogram whose base is 10cm and height is 6cm.

Solution : Here,

Base of the parallelogram (b) = 10cm

Height of the parallelogram (h) = 6cm

Now,

Area of the parallelogram (A) = base (b) × height (h)

= 10cm × 6cm

= 60sq. cm

EXAMPLE 2 In the given figure, ABCD is a parallelo- A D
Solution : gram. AE⊥BC and AN⊥CD. If BC = 12cm, E N
AE = 5cm and AN = 8cm, find the length of
CD. C

Here, B

BC = 12cm, AE = 5cm and AN = 8cm

Area of ABCD = BC × AE

= 12cm × 5cm

= 60sq. cm.

Again,

Area of ABCD = CD × AN

or, 60sq.cm = CD × 8cm

or, CD = 60sq. cm
8cm

∴ CD = 7.5cm

GREEN Mathematics Book-10 161

In the given figure area of a trapezium D C

EXAMPLE 3 AECD is 80 sq. cm and area of ∆DEC is
Solution :
27sq. cm. Find the area of ∆BCE.

Here, A BE

Area of ABCD = 2 times area of ∆ DEC ∴A B∆CDDEsCtaanndd parallelogram
on the same

= 2 × 27 cm² base DC and between the
same parallels DC and AE.

= 54 cm²

Area of trapezium AECD = Area of ABCD + Area of ∆ BCE
or, 80 cm² = 54 cm² + Area of ∆BCE

∴ Area of ∆BCE = 26 cm²

EXAMPLE 4 In the given figure, PQRS is a square. If P S T
Solution : N
QR||PT and QS = 6cm, find the area of
R
∆TNR where N is the mid-point of QT.

Q

Here,

PQRS is a square whose diagonal QS is 6cm.

So, area of the square PQRS = 1 (QS)²
2

= 1 (6cm)² = 18 sq.cm
2

∆ QRT and PQRS stand

Area of ∆QRT = 1 × area of square PQRS on the same base QR and
2 between the same parallels
QR and PT.
= 9 sq. cm.

Area of ∆QNR = 1 area of ∆QRT RN median bisects triangle.
2

= 1 × 9 sq. cm
2

= 4.5 sq. cm

EXAMPLE 5 In the adjoining figure, M and N are the mid- A

points of the sides AB and AC of the trian- M N
P
gle ABC respectively. Prove that the area of
C
∆BPC is equal to the area of the quadrilateral B
AMPN.

162 GREEN Mathematics Book-10

Solution : Given : ABC is a triangle. M and N are A
the mid-points of the sides AB
and AC respectively. M N
P
To prove: Area of ∆BPC = Area of quadrilat-
C
eral AMPN. B

Proof:

Statement Reasons

1. (DABN = 1 DABC) in area 1. BN median bisects DABC.
2 2. CM median bisects DABC
3. From statment 1 and 2.
2. (DMBC = 1 DABC) in area
2

3. (DABN = DMBC) in area

4. (DBMP + quad. AMPN = DBMP + 4. From statement 3 by whole part axiom.
DBPC) in area

5. Quad. AMPN = DBPC in area 5. From statement 4.

Proved

EXAMPLE 6 In the figure, PQRST is pentagon where QT||RS, QR||PS and TS||PR,

prove that area of ∆PQR = area of ∆PTS. P

Solution : Given : QT||RS, QR||PS and TS||PR, T

To prove: Area of ∆PQR = area of ∆PTS Q MN

Proof: R S

Statement Reasons

1. Area of parm. QRSN = 1. Both are standing on same base

area of parm. MRST RS and RS||QT.

2. Area of ∆PQR = 1 area 2. Both are standing on same base
of parm. QRSN 2
QR and QR||PS.

3. Area of ∆PQR = 1 area 3. From statment 1 and 2.
of parm. MRST 2

4. Area of ∆PST = 1 area 4. Both are standing on same base
of parm. MRST 2
ST and ST||RP.

5. Area of ∆PQR = area of 5. From statement. 3 and 4.
∆PTS

Proved.

GREEN Mathematics Book-10 163

EXAMPLE 7 In the adjoining figure, it is given that A B E
Solution : BC||AE and area of ∆ABC = area of D
∆BDE. Prove that BE||CD.
C
Given : BC||AE and area of ∆ABC =
area of ∆BDE.

To prove: BE||CD

Construction: Join C and E.

Proof: Reasons
Statement

1. Area of ∆ABC = area of 1. Given
∆BDE

2. Area of ∆ABC = area of 2. DABC and DBCE stand on the

∆BCE same base BC and between the

same parallels BC and AE.

3. Area of ∆BDE = area of 3. from st. (1) and st (2).
∆BCE

4. BE||CD 4. From st. (3)

Proved.

A D
HC
EXAMPLE 8 In the given figure, ABCD is a parallelogram.

H is any point on BC. A and H are joined. AH

and DC are produced to the point E. B and E

are joined. Prove that : ∆CHD = ∆BHE. B

Solution : E

Given : ABCD is a parallelogram. H is any point on BC. AH is joined
and produced to the point E where DC meets after producing. BE is
joined.

To prove: ∆CHD = ∆BHE.

Proof:

Statement Reasons

1. ∆AHD = 1 ABCD 1. DAHD and ABCD stand
2 on the same base AD
and between the same
parallels AD and BC.

2. ∆AHD + ∆ABH + ∆CHD = 2. Whole part axiom.
ABCD

3. ∆ABH + ∆CHD = 1 ABCD 3. From st. (1) and st. (2)
2

164 GREEN Mathematics Book-10

4. ∆ABE = 1 ABCD 4. ∆ABE and ABCD stand
2 on the same base AB
and between the same
parallels AB and DE.

5. ∆ABE = ∆ABH + ∆BHE 5. Whole part axiom

6. ∆ABH + ∆BHE = 1 ABCD 6. From st. (4) and st. (5).
7. ∆ABH + ∆BHE =2 ∆ABH + 7. Form st. (3) and st. (6)

∆CHD

8. ∆BHE = ∆CHD 8. Form St. (7)

Proved.

EXERCISE 9.1

A. Very Short Questions
1. Find the area of the following figures.

A b. D C c. A D

a.

6cm8 cm
4cm
4cm
5cm B
MA B 5cm CE
B 6cm C M
C
6cm N

2. a. ABCD is a parallelogram in which altitude DN on D B

BC is 8cm and altitude DM on AB is 5cm. If the area

of the parallelogram ABCD is 40sq. cm, find the

length of the sides of the parallelogram.

A
M

D 5cm C

b. Calculate the area of

the given trapezium. A B
E
9cm

c. In the adjoining figure, ABCD is a trapezium A 15 cm D
in which AD||BC, ∠ABC = 90°, AB = 15cm 17 cm
and DC = 17cm. If the area of the trapezium
ABCD is 204 sq. cm, calculate the length of B C
AD.

GREEN Mathematics Book-10 165

B. Short Questions DP C

3. a. In the given figure, ABCD is a parallelogram. AB = A B
14cm, PM = 8cm and PM⊥AB. Calculate (i) area of M
∆PBA (ii) area of ABCD (iii) area of (∆APD + ∆BPC).

D F C
E
b. In the adjoining figure, ABCD is a quadrilateral
in which AC = 2BF = 3DE = 18cm. Find the area A
of quad ABCD.

B

A E
D
c. In the given figure, area of ∆BCD is 30sq.cm. Find
the area of ∆ABE. C
M
B E

d. In the adjoining figure, E is a mid-point of DC. If the D C
area ABCD is 26sq. cm., find the area of quadrilateral
ABCE. B

A

e. In the given figure, PQRS is a rhombus in which PR P S T
= 12cm an QS = 8cm. If PS is produced to T, find the R
area of ∆QRT. C B
E
Q D
C
D

4. a. In the adjoining figure, EF⊥AD, EF = 6cm F
and CE = 8cm. Calculate the area of ∆BCD.

A

E

A

b. In the given figure, ABCD is a square with its di-

agonal AC= 6cm. If AD is produced to E, calculate

the area of ∆BCE. B

166 GREEN Mathematics Book-10

AD

c. In the adjoining figure, the area of the tra-

pezium ABCD is 96 sq.cm. and the area of

∆ADC is 24 sq. cm. Calculate the area of

∆DEC. B C
E

C. Long Questions

5. a. A parallelogram ABCD and a rectangle EBCF are sanding on the same base BC
and between the same parallels EF and BC. Prove that : (i) ∆AEB ≅ ∆DCF (ii)
area of parallelogram ABCD = area of rectangle EBCF.

b. A parallelogram ABCD and a triangle ABE are standing on the same base AB
and between the same parallels AB and DC. Prove that area of parallelogram
ABCD is twice of the area of triangle ABE.

c. In the adjoining figure, MN||PQ. Prove M R SN

that area of ∆RPQ is equal to the area of Q
D
∆SPQ.
E
P

C

d. In the figure, alongside, AB||CD. Prove that area B

of ∆AEC is equal to the area of ∆BED.

A

6. a. M and N are the mid-points of the sides AB and AC of a triangle ABC. In ∆ABC,
CM and BN intersect at P. Prove that : ∆BPM = ∆CPN.

DC
P

b. In the given figure, AB||DC and AD||BC. P is any
point on AC. Prove that : ∆CDP = ∆CBP.

AB

c. X and Y are any points on the sides BC and DC of a parallelogram ABCD respec-
tively. Prove that: area of ∆AXD = area of ∆AYD + area of ∆BYC.

d. P is any point inside a parallelogram ABCD. Prove that the sum of the area of
triangles APD and BPC is equal to half of the area of the parallelogram ABCD.

A D
C
e. In the given figure, AD||BC and AB||DC. D is

joined to any point E on BC. AB and DE are pro- B E
duced to meet at F. Prove that: area of ∆CEF = area

of ∆ABE.

F

GREEN Mathematics Book-10 167

E D
C
7. a. In the figure alongside, AB||ED and AD||BC.
B
Prove that : area of . ∆EAB = area of ∆ACD. A C

D

b. In the given figure, ABCD is a parallelogram and F
AC||EF. Prove that : ∆ADE = ∆CDF. B
E
A E

A D
IH
c. In the adjoining figure, ABCD is a parallelogram. G
EF||AB and GH||BC. Prove that : area of AGIE FC
= area of CHIF. C

B B
D
D
E
d. In the given figure, AB||DC and AE = ED. Prove

that : ∆BEC = 1 trapezium ABCD. E
2 A

A

8. a. In the given figure, ABCD is a quadrilateral and

AC||DE. Prove that : area of ∆ABE is equal to

the area of quadrilateral ABCD. B

C

A

b. In the figure alongside, D and E are the mid-

points of AB and AC so that DH||EF. Prove that: D E
1
DEFH = 2 ∆ABC.

BC
HF

c. In ∆ABC, D and E are mid-points of AB and AC respectively. F is any point on
BC. Prove that area of ∆ABC is equal to 4 times area of ∆DEF.

168 GREEN Mathematics Book-10

A

d. In the figure, AD||BE, BC||ED and AE||DC. Prove that

area of ∆ABE = Area of ∆ADC. B MN C

ED
A

e. In the given figure, D, E, F and H are the mid-points of E
BC, AD, CE and BF respectively. Prove that: ∆EFH = 1
8 H F
∆ABC. D C
B
A B

9. a. In the given figure, AD||BC, AB||DC||EF and DE||AF.D K C
Prove that : ABCD = DEFK. H

EF

AB

b. In the adjoining figure, AB||MN||DC. M N
Prove that : area of ∆AND = area of ∆BMC.
C
D ED

A F
C
c. In the given figure, ABCD and BEFG are two paral-
lelograms. Prove that ABCD = BEFG in area.

B

d. In the adjoining figure, ABCD is a trapezi-D G C
um where AB||DC. E and F are the mid F
points of BD and AC respectively. Prove that
: ∆BDF = ∆ACE. E

e. In the given figure, A is a mid-point of PR A B
P A
and B is any point on QR. QC||BA. Prove C

that : ∆BCR = 1 ∆PQR. B
2

Q R

GREEN Mathematics Book-10 169

10

Constructions

Estimated Teaching Periods : 6

Phocylides, Greek gnomic poet of Miletus, contemporary of Theognis of
Megara, was born about 560 BC.

Contents

10.1 Construction

Objectives

At the end of this unit, student will be able to :
construct triangles and different types of quadrilaterals.
construct a triangle and compare with rectangle, square and parallelogram.

Materials

Geometry box, geo board, chart paper, pencil, white paper, etc.

170 GREEN Mathematics Book-10

10.1 Construction

1. Construct ∆ABC in which a = 5 cm, b = 6.3 cm and c = 6cm. Construct ∆DBC equal in

area to ∆ABC and having Rough

a. an angle of 45° A

b. a side of length 7cm. 6cm 6cm 6.3cm

Solution : (a) an angle of 45° BC
5cm

∴ The required triangle DBC is constructed:

Stpes for above construction: A DY
X

Step – I Draw a base BC (a) = 5.cm. From
the point B, draw an arc with
radius AB = 6cm and from the
point C, draw another arc with
radius AC = 6.3 cm, so that these
two arcs cut at the point A. Join
B and C both to A to form the
∆ABC.

Step – II From the vertex A, draw a line B 45° C
XY parallel to BC making ∠XAB
= ∠ABC.

Step – III At B, draw an angle of CBD = 45°.
So, that BD meets XY at D.

Step – IV Join C and D to form a ∆DBC which is the required triangle whose area is equal to
the area of ∆ABC.

∴ DBC is the required triangle. Rough
(b) a side of length 7cm.
A
D'
X A DY
6.3cm
BC
7cm 5cm

7cm

BC
GREEN Mathematics Book-10 171

∴ The required triangle BDC (or BD'C) is constructed:

Stpes for above construction:

Step – I Draw a base BC (a) = 5cm. From the point B, draw an arc with radius AB = 6cm and
from the point C, draw another arc with radius AC = 6.3 cm, so that these two arcs
cut at the point A. Join B and C both to A to form the ∆ABC.

Step – II From the vertex A, draw a line XY parallel to BC making ∠XAB = ∠ABC.

Step – III From the point B, draw an arc with radius BD (or BD') = 7cm cutting XY at D (or D').

Step – IV Join B and C to D (or D') to form ∆BDC (or ∆BD'C). It is the required triangle whose
area is equal to the area of ∆ABC.

2. Construct a triangle ABC in which a = 4.9cm, b = 6.2cm and c = 5cm. Construct a
parallelogram equal in area to the ∆ABC and having an angle of 60°.

Solution : A E FY Rough
A
X

c = 5cm b = 6.2 cm

B C
a = 4.9 cm

B 60°
C

D

∴ The required parallelogram EDCF is constructed:
Stpes for above construction:
Step – I Construction of a ∆ABC. Draw a base BC = a = 4.9cm. From the point B, draw an arc

with radius AB = c = 5cm and from C draw another arc with radius AC = b = 6.2cm.
These two arcs intersect each other at A. Join B and C both to A to form a ∆ABC.

Step – II Through the vertex A, draw a line XY parallel to BC making ∠XAB = ∠ABC.

Step – III Take a middle point D on the base BC with the help of compass making the right
bisector of BC.

Step – IV Through D, draw a line DE making ∠CDE = 60°, where DE cuts XY at E.

Step – V From the point E, draw an arc of radius EF = DC which cuts XY at F. Join C and F
to form a parallelogram CDEF whose area is equal to the area of the ∆ABC. Thus,
CDEF is the required parallelogram.

172 GREEN Mathematics Book-10

3. Construct a triangle ABC in which ∠A = 45°, ∠C = 75° and AB = 5cm. Construct a
rectangle equal in area to the ∆ABC.

Solution :

QC X Rough
C
P
75°

45° B
A 5cm

∠A + ∠B + ∠C = 180°

or, 45° + ∠B + 75° = 180°

or, ∠B = 180° – 120° = 60°

R
A 5cm B

∴ The required rectangle PARQ is constructed:

Stpes for above construction:

Step – I Construction of a ∆ABC. Draw a base AB = 5cm. From the point A and B, draw an
angle of 45° and 60° respectively.

Step – II Through the vertex C, draw a line CX parallel to AB making ∠XCB = ∠CBA.

Step – III Take a middle point R on the base AB with the help of compass making the right
bisector of AB.

Step – IV Through R, draw a line RQ where RQ cuts CX at Q.

Step – V From the point Q, draw an arc of radius AR = QP which cuts CX at P. Join P and A
to form a rectangle PARQ whose area is equal to the area of the ∆ABC.

4. Construct a triangle ABC in which a = 4.9cm, b = 6.2cm and c = 5cm. Construct a
parallelogram equal in area to ∆ABC so that a side of the parallelogram is 8.5cm.

Solution :

Stpes for construction: Rough
A
Step – I Construction of a ∆ABC. Draw a base BC = a = 4.9cm. From
point B draw an arc with radius AB = c = 5cm and from C draw B a = 4.9cm C
another arc with radius AC = b = 6.2cm. These two arcs intersect c = 5 cm b = 6.2 cm
each other at A. Join B and C both to A to form a ∆ABC.

Step – II Through the vertex A, draw a line XY parallel to BC making
∠XAB = ∠ABC.

GREEN Mathematics Book-10 173

Step – III Find the middle point D X A E FY
on the base BC with the
help of compass making
the right bisector of BC.

Step – IV Through D and C draw
two arcs radii DE = CF =
8.5cm cutting XY at E and
F respectively.

Step – V Join D and E, C and F to B C
form a parallelogram
CDEF, which is the D
required parallelogram,
equal to the area of the
∆ABC. Thus, CDEF is the
required parallelogram.

5. Construct a parallelogram ABCD in which AB = 5.5cm, BC = 5 cm and ∠ABC = 60°.
Construct a parallelogram EBCF equal in area to the given parallelogram having a
side of length 6.8cm.

Solution :

Y E DF Rough
A
Y
A D

5.5cm 60°
5.5cmB 5cm C X

60° X
B C

5cm

Step – I Draw a line BX and cut an arc BC = 5cm at BX.

Draw an angle of XBY = 60° at B. From the point B, draw an arc with radius AB = 5.5
cm to cut BY at A. Again from the point A, draw an arc with radius AD = 5cm and
from C draw an arc with radius CD = 5.5cm. These two arcs cut at the point D. Join
AD and CD to form a parallelogram ABCD.

Step – II From the point B, draw an arc with radius BE = 6.8 cm to cut AD at E. Join B and E.

Step – III From the point E, draw an arc with radius EF = BC = 5cm to cut AD produced at
F. Join C and F to form a parallelogram EBCF which is the required parallelogram
whose area is equal to the parallelogram ABCD. Thus, EBCF is the required
parallelogram.

174 GREEN Mathematics Book-10

6. Construct a parallelogram ABCD in which AB = 5.5cm, BC = 5 cm and ∠ABC = 60°.
Construct a triangle BEF equal in area to the given parallelogram having a side of
length 7.8cm.

Solution : Rough

Y Y D
A A

ED
5.5cm
5.5cm 60°
B 5cm C X

60° C F
B

5cm

Step – I Draw a line BX and cut an arc BC = 5cm at BX.

Draw an angle of XBY = 60° at B. From the point B, draw an arc with radius AB = 5.5
cm to cut BY at A. Again from the point A, draw an arc with radius AD = 5cm and
from C draw an arc with radius CD = 5.5cm. These two arcs cut at the point D. Join
AD and CD to form a parallelogram ABCD.

Step – II From the point B, draw an arc with radius BE = 7.8 cm to cut AD at E. Join B and E.

Step – III From the point C, draw an arc with radius BC = CF = 5cm to cut F. Join E and F
to form a triangle BEF which is the required triangle whose area is equal to the
parallelogram ABCD. Thus, BEF is the required triangle.

7. Construct a quadrilateral ABCD in which AB = 3.9cm, BC = 4.3cm, CD = 2.7cm,
AD = 3.5 cm and the diagonal AC = 4cm. Construct a triangle equal in area to the
quadrilateral ABCD.

Solution : Rough

A 3.5cm A
3.5cm
D

3.9cm
2.7cm

3.9cm
2.7cm
4cm

4cm
DB C
4.3cm

B 4.3cm C E
GREEN Mathematics Book-10 175

Step – I Construction of a quadrilateral ABCD: Draw a base BC = 4.3cm. From the point B,
draw an arc with radius AB = 3.9cm and from C draw another arc with radius AC =
4cm. So, that these two arcs cut at the point A. Join B and C to A to form a ∆ABC.

From the point A, draw an arc with the radius AD = 3.5 cm and from C draw another
arc with the radius CD = 2.7cm, so that these two arcs cut at D. Join A and C both to
D to form another ∆ACD.

Thus the quadrilateral ABCD with given measurements is constructed.

Step – II From the vertex D, draw a line DE parallel to the diagonal AC of the quadrilateral
ABCD to meet BC produced at E.

Join A and E so that ∆ABE is the required triangle whose area is equal to the area of
the quadrilateral ABCD.



EXERCISE 10.1

A. Long Questions

1. a. Construct a triangle ABC in which a = 6.2cm, b = 5.4cm, c = 7.2cm. Construct a
triangle equal in area to the DABC having an angle of 45°.

b. Construct a triangle ABC in which BC = 5cm, AC = 4.8cm and AB = 5.8cm.
Construct a triangle equal in area to the DABC having an angle of 60°.

c. Construct a triangle ABC is which AB = 5cm, BC = 6.8 cm and CA = 7.8cm.
Construct the triangle equal in area to the DABC having an angle of 75°.

2. a. Construct a triangle ABC in which AB = 5.2cm, BC = 5.8cm, CA = 6.5cm.
Construct a triangle equal in area to the triangle ABC having a side of 7.2cm.

b. Construct a triangle ABC in which AB = 5.4cm, BC = 6.4cm, CA = 6cm. Draw a
triangle equal in area to the DABC having a side of 7cm.

c. Construct a triangle ABC is which a = 6.4cm, b = 5.8cm, c = 5cm. Draw a
triangle equal in area to the DABC with one of its side 6.4cm.

3. a. Construct a triangle ABC in which AB = 6cm, BC = 5cm, CA = 4cm. Construct
a parallelogram equal in area to the DABC having an angle of 60°.

b. Construct a triangle ABC in which a = 7.8cm, b = 7.2cm and C = 6.3cm.
Construct a parallelogram equal in area to the DABC having an angle of 45°.

c. Construct a triangle ABC in which ∠B = 45°, ∠C = 60° and a = 6.3. Reduce the
triangle to a parallelogram of equal area having one of its angle 75°.

4. a. Construct a triangle ABC in which AB = 6.2cm, BC = 5.8cm and AC = 5.6cm.
Construct a parallelogram equal in area to the triangle ABC having a side
equal to 7.4cm.

b. Construct a triangle ABC in which AB = 7cm, BC = 6.5cm and ∠B = 60°.
Construct a parallelogram equal in area to the triangle ABC having a side of
8.8cm.

176 GREEN Mathematics Book-10

c. Construct a triangle ABC in which AB = 5cm, BC = 6cm, ∠B = 60°. Construct a
parallelogram equal in area to the triangle ABC having a side of 7.5cm.

5. a. Construct a triangle ABC in which BC = 6.8cm, AC = 6.5cm, AB = 6cm.
Construct a rectangle equal in area to the DABC.

b. Construct a triangle XYZ in which XY = 6.5cm, ZX = 7.6cm and YZ = 6.9cm.
Construct a rectangle equal in area to the DXYZ.

c. Construct a triangle PQR in which PQ = 5.5cm, QR = 6.4cm and ∠Q = 60° and
construct a rectangle equal in area to the DPQR.

6. Construct a parallelogram whose adjacent sides are 4cm and 6cm and the
included angle of 45°. Construct another parallelogram equal in area to the given
parallelogram whose :

a. one angle is 60° b. one of its side is 7cm c. an angle is of 75°

7. Construct a parallelogram PQRS having PQ = 5cm, PR = 8cm and QS = 7cm.
Construct another parallelogram PQMN having

a. a side of 7cm. b. a side of 7.5cm. c. ∠NQP = 75°

8. a. Construct a quadrilateral PQRS in which PQ = 4.2cm, QR = 6.5cm, RS = 8cm,
PS = 4.9cm and SQ = 7cm. Reduce the quadrilateral PQRS to a triangle of equal
area.

b. Construct a quadrilateral ABCD in which AB = 6.5cm, BC = 5.5cm, CD = 4.5cm,
∠B = 120° and ∠C = 60°. Construct an equivalent triangle with AB as base and
vertex on BC.

c. Construct a quadrilateral PQRS in which PQ = 5.6cm, QR = 6.4cm, RS = 6.6cm,
PS = 5.2 cm and diagonal QS = 6cm. Locate a point on the line produced QR so
that area of ∆PQT = area of quadrilateral PQRS. Measure the length of QT.

9. a. Construct a quadrilateral ABCD in which AB = 5.5cm, BC = 6.5cm, AD = 5cm,
BD = 7cm and CD = 5cm. Construct a triangle equal in area to the quadrilateral
ABCD.

b. Construct a quadrilateral PQRS in which PQ = QR = 5.8cm, RS = SP = 5cm ,
and ∠P = 75°. Then construct a triangle of equal area to the given quadrilateral
PQRS.

c. Construct a quadrilateral PQRS in which PQ = 5.6cm, QR = 6.4cm, RS = 6.6cm,
SP = 7.2 cm and diagonal QS = 6cm. Construct a triangle equal in area to the
quadrilateral.

10. a. Construct a parallelogram ABCD in which AB = 5cm, AC = 8cm, BD = 6cm.
Construct a triangle equal in area to the quadrilateral.

b. Construct a parallelogram PQRS in which PQ = 5cm, PR = 7cm and
∠RPQ = 30°. Construct a triangle equal in area to the parallelogram.

GREEN Mathematics Book-10 177

11

Circle

Estimated Teaching Periods : 25

Euclidean geometry is a mathematical system attributed to the Alexandrian
Greek mathematician Euclid, which he described in his textbook on
geometry: the Elements. Euclid's method consists of assuming a small set
of intuitively appealing axioms, and deducing many other propositions
(theorems) from these.

Contents

11.1 Review
11.2 Important axiom
11.3 Theorems related to arcs and the angles subtended by them
11.4 Tangents to a circle

Objectives

At the end of this unit, students will be able to :
identify different parts of a circle.
state the relation of arcs, centre angle, segment angle and chord of a circle.
state properties of cyclic quadrilaterals.
state the properties of tangents.

Materials

Models of circle, chart of circle with the labelling of its different parts
Instrument box, chart paper, geo-board, etc.

178 GREEN Mathematics Book-10

11.1 Review

A circle is a closed figure bounded by a curve line traced out by a moving A1 A
point which is equidistant from a fixed point.

The fixed point is called centre of the circle. In the given figure, O is the O A3
centre of the circle. There are some usual terms of the circle which are A2
given below.

Circumference A

The outer circular line of a circle is called the circumference B
of the circle. The circumference of the circle is also known O
as the perimeter of the circle. In the given figure curve D
line ABCD is the circumference of the circle.
C

Radius Q P
S
A line segment which joins the centre of the circle and O
any point on the circumference of the circle is called R
radius of the circle. In the figure, OP is a radius of the
circle. The plural form of the radius is radii. The radii of
a circle are equal. In the given figure,

OP = OQ = OR = OS

Diameter S

A line segment which is drawn from a point of the P Q
circumference to the other point of the circumference O
through the centre of the circle is called the diameter of
the circle. In the figure, PQ and RS are the diameters of R
the circle. The diameter of the circle is denoted by d. The
length of the diameter is twice the radius of the circle.

∴ d = 2r, where r is the radius of the circle. A diameter
divides a circle into two equal parts.

GREEN Mathematics Book-10 179

Semi-circle

A diameter divides a circle into two equal halves and A X B
each half is called the semi-circle. In the figure, AXB and
AYB are semi-circles. O
Y

Arc

A part of the circumference of a circle is called an arc. It C

is denoted by symbol' '. In the figure, ACB is a part of A B
the circumference of the circle with centre O. So, ACB

is called an arc. There are minor arc and major arc. A O
minor arc is less than the half of the circumference and

major arc is greater than the half of the circumference.

So, ACB is a minor arc and ADB is a major arc. D

Chord A B
Q
A line segment which is formed by joining any two points O
on the circumference of the circle is called a chord of the P S
circle. In the figure, PQ, AB and RS are the examples of
the chords. A diameter is the longest chord of the circle. R

Segment AB
O
A part of the circle which is enclosed by an arc and a
chord is called the segment of a circle. The part of the
circle which is less than semi-circle is called minor
segment. In the figure, the shaded part is the minor
segment. The part of the circle which is greater than a
semicircle is called the major segment. In the figure, the
un-shaded part is the major segment. The minor and
major segments are known as the alternate segments of
each other of the circle.

180 GREEN Mathematics Book-10

Sector O
PQ
The region enclosed by any two radii and the arc of a
circle is called a sector of the circle. In the figure, the C
shaded region POQ is an example of a sector. There are B
minor sector and major sector. The shaded region is a A
minor sector and un-shaded region is a major sector.
O
Concentric circles
C1 C2 C3
Two or more circles having the same centre but different
radii are called the concentric circles. In the figure, C1,
C2 and C3 are three circles having the same centre O but
different radii OA, OB and OC respectively. So, these
three circles are concentric circles.

Intersecting circles A
PQ
If two circles with different centres intersect each other at
two points, then the circles are called intersecting circles. B
In the figure two circles with centres P and Q intersect
each other at two points A and B. These two circles are
known as the intersecting circles. The points are called
intersecting points. A chord AB is formed by joining the
points A and B. It lies in both circles. So, it is called the
common chord. A line is formed joining the centres P and
Q. The line PQ is called the line of the centre. The line of
centre is perpendicular bisector of the common chord.

Overlapping circles A

Two or more circles which overlap each other are known B
as the overlapping circles. In the figure, the circles. C1
and C2 are overlapping circles. The concentric circles are OO
also overlapping circles.
C1 C2

GREEN Mathematics Book-10 181

11.2 Important Axiom

i. Two circles are equal if their radii or diameters are equal.
ii. One and only one circle can be drawn with a given centre and a given radius.
iii. Radii or diameters of the equal circles are equal.
iv. A line cannot intersect a circle at more than two points.

11.3 Theorems related to arcs and the angles subtended by them

a. Central angle and corresponding arc S
O
In the given figure, O is the centre of a circle in which
OP and OQ are two radii and PRQ is an arc. P Q
R
The arc PRQ subtends an angle POQ at the centre 'O'.

So, ∠POR is called central angle and PRQ is its
corresponding arc. Since both the angle and the
arc are described by the same amount of rotation,
therefore the degree measure of PRQ and ∠POQ
are same.

∴ ∠POQ PRQ or ∠POQ ≡ PRQ .

b. Inscribed angle and corresponding arc B

In the given figure, O is the centre of a circle in o C
which AB and BC are two chords. The chords AB A
and BC meet at a point B on the circumference of
the circle. So, they form an angle ABC at B on the m
circumference. The angle on the circumference is
called the inscribed angle. ∠ABC is an inscribed
angle and AMC is its corresponding arc.

Now, for the relation between an inscribed angle
and its corresponding arc.

In ∆AOB, ∠OAB = ∠OBA [∴ OA = OB] B

∠OAB + ∠OBA + ∠AOB = 180°

or, ∠OBA + ∠OBA + ∠AOB = 180°

or, 2∠OBA = 180° – ∠AOB ................. (i) o C
A
In ∆BOC, ∠OBC = ∠OCB [∴ OB = OC]

∠OBC + ∠OCB + ∠BOC = 180°

or, ∠OBC + ∠OBC = 180° – ∠BOC

or, 2∠OBC = 180° – ∠BOC ................ (ii)

182 GREEN Mathematics Book-10

Adding (i) and (ii), we have

2 ∠OBA + 2∠OBC = 180° – ∠AOB + 180° – ∠BOC

or, 2 ∠OBA + ∠OBC) = 360° – (∠AOB + ∠BOC)

or, 2 ∠ABC = 360° – Reflex ∠AOC

or, 2 ∠ABC circumference – ABC ∴ Circumference 360° and Ref. ∠AOC ABC

or, 2 ∠ABC AC

∴ ∠ABC 1
AC
2

Thus, the degree measurement of the angle on

the circumference of a circle is equal to the degree

measurement of half of its corresponding arc. T

c. Concyclic points PS

The points which lie on the circumference of a circle are o R
called concyclic points. In the adjoining figure, P, Q, R, Q
S and T are concyclic points.

W Z

d. The figure formed by joining any four concyclic points

is called cyclic quadrilateral. In the figure, WXYZ is a o
cyclic quadrilateral.
XY

Theorem – 5

Arcs subtended by equal angles at the centre of the circle are equal.
Experimental Verification

Step: I With the help of a compass and a pencil, two o o
Step: II circles of different radii with centre at 'O' 40° 40° S
are drawn. In each figure, with the help of a 25° 25° S
protractor two equal angles POQ and ROS at P P R
the centre are drawn. Q
R
With the help of thread, arcs PQ and RS of Q Fig. II
each figure are measured and the results are
tabulated below: Fig. I

Fig. ∠POQ ∠ROS PQ RS Result
I 25° 25° PQ = RS
II 40° 40° PQ = RS

Conclusion: The above experiment shows that arcs subtended by equal angles at the
centre of the circle are equal.

GREEN Mathematics Book-10 183

Converse of theorem - 5

Angles subtended by two equal arcs of a circle at the centre are equal.

Experimental Verification

Step: I With the help of a compass and a pencil o N o
two circles of different radii with centre X N
Y
at 'O' are drawn. With the help of M M
Fig. I Y
compass two equal arcs XY and MN in
each figure are drawn. Points X, Y, M, X Fig. II

and N are joined with the centre O.

Step: II With the help of a protractor, central
angles XOY and MON of each figure are
measured and the results are tabulated
below :

Fig ∠XOY ∠MON Result

I ∠XOY = ∠MON

II ∠XOY = ∠MON

Conclusion: The above experiment shows that the angles subtended by equal arcs at the
centre of circle are equal.

Theorem – 6

Arcs cut off by equal chords of a circle are equal. Or, If two chords of a circle are equal,
the corresponding arcs are equal.

Experimental Verification

Step: I With the help of a compass and a pencil, M JM
two circles of different radii with centre
O are drawn. In each figure, two equal J 3.5cm 4cm
chords JK and MN are drawn.
o o

Step: II With the help of a thread, JK and MN are 3.5cm N 4cm
measured in each circle and the results are
tabulated below : K KN

To verify : JK = MN Fig. I Fig. II

Table of observation:

Fig. JK MN JK MN Result
JK = MN
I 3.5cm 3.5cm
II 4cm 4cm

Conclusion: The above experiment shows that if two chords of a circle are equal, the
corresponding arcs are equal.

184 GREEN Mathematics Book-10

Converse of theorem - 6

If two arcs of a circle are equal, the corresponding chords are equal.

Experimental Verification

Step: I With the help of a compass and a pencil, C AC
two circles of different radii with centre O A o

are drawn. In each figure, AB = CD are o BD
also drawn. Then the chords AB and CD are
Fig. II
drawn.

Step: II With the help of a divider and a scale, chords BD
AB and CD are measured and the results are
Fig. l
tabulated below :

To verify: AB = CD (Chord AB = Chord CD)
Table of observation:

Fig Chord AB Chord CD Result

I Chord AB = Chord CD

II Chord AB = Chord CD

Conclusion: The above experiment shows that if two arcs of a circle are equal, the
corresponding chords are equal.

Theorem – 7

The angle at the centre of a circle is double the angle at the circumference standing on the
same arc.

Experimental Verification

Step: I With the help of a compass and a pencil, two C A C
Step: II circles of different radii with centre O are O
drawn. In each figure, ∠AOB at the centre and A O
∠ACB at the circumference angles standing on Fig. I B
the same arc AB are drawn.
B
With the help of protractor, ∠AOB and ∠ACB
of each figure are measured and the results are Fig. II
tabulated below.

To verify : ∠AOB = 2∠ACB
Table of observation:

Fig. <AOB ∠ACB Result

I ∠AOB = 2∠ACB

II ∠AOB = 2∠ACB

Conclusion: The above experiment shows that angle at the centre of a circle is double of
the angle at the circumference standing on the same arc.

GREEN Mathematics Book-10 185

Theoretical proof P

Given : The central angle AOB and the circumference angle APB O
are standing on the same arc ADB of a circle APBD with CB
centre at 'O'. D

To prove : ∠AOB = 2∠APB A

Construction : P and O are joined and it is produced to C.

Proof :

Statements Reasons

(1) ∠OAP = ∠OPA (1) OA = OP (radii of the same circle)

(2) ∠AOC = ∠OAP +∠OPA (2) In ∆AOP, exterior angle = sum of its opposite
interior angles.

(3) ∴ ∠AOC = 2∠OPA (3) From statements (1) and (2)

(4) Similarly, ∠BOC = 2∠OPB (4) Reasons similar as statements (i), (2), and (3).

(5) ∠AOC + ∠BOC = 2 (∠OPA + ∠OPB) (5) Adding statements (3) and (4)

(6) ∠AOC + ∠BOC = ∠AOB and ∠OPA + ∠OPB (6) By whole parts axiom.
= ∠APB

(7) ∴ ∠AOB = 2∠APB (7) From statements (5) and (6)

Proved

Theorem – 8

The angle in a semi-circle is a right angle.

Experimental Verification

Step: I With the help of a compass and a C C
o oB
pencil, two circles of different radii BA
Fig. I
with centre O and diameter AB are
drawn. In each figure, ∠ACB is A

drawn.

Step: II With the help of a protractor, ∠ACB Fig. II
of each figure is measured and the

results are tabulated below :

To verify : ∠ACB = 90°

Table of observation:

Fig ∠ACB Result
I 90°
∠ACB = 90°
II 90°

Conclusion: The above experiment shows that the angle in a semi-circle is a right angle
(verified).

186 GREEN Mathematics Book-10

Theoretical proof C
oB
Given : ACB is an angle in the semi-circle with centre at O and

diameter AB. A

To prove : ∠ACB = 90°

Proof :

Statements Reasons
(1) ∠AOB = 2∠ACB (1) The central angle is double of the

(2) ∠AOB = 180° angle at the cercumference standing
on the same arc AB.
(3) 2∠ACB = 180° (2) ∠AOB is a straight angle
(3) From statements (1) and (2)
∴ ∠ACB = 180° = 90°
2 Proved



Theorem – 9

The angles at the circumference of a circle standing on the same arc are equal. OR Angles
in the same segment of a circle are equal.

Experimental Verification

Step: I With the help of a compass and a
pencil, two circles of different radii
with centre O are drawn. In each C D C
diagram, the inscribed angles ACB A B
and ADB in the same segment ACDB oA
standing on the same arc AB are B o
drawn. D
Fig. I
Step: II With the help of a protractor, Fig. II
∠ACB and ∠ADB of each figure
are measured and the results are
tabulated below.

To verify : ∠ACB = ∠ADB

Table of observation:

Fig ∠ACB ∠ADB Result

I
II ∠ACB = ∠ADB

Conclusion: The above experiment shows that the angles in the same segment of a circle
are equal.

GREEN Mathematics Book-10 187

Theoretical proof

Given : ∠ACD and ∠ABD are the inscribed angles in the same A C
segment ACBD of a circle with centre at O. B

To prove : ∠ACD = ∠ABD o
D
Construction : A and D are joined with O.

Proof:

Statements Reasons
(1) ∠AOD = 2∠ACD
(1) Being ∠AOD is a central angle and ∠ACD is
(2) ∠AOD = 2∠ABD an inscribed angle standing on the same-arc
(3) ∴ 2∠ACD = 2∠ABD AD.

i.e, ∠ACD = ∠ABD (2) Same as (1)

(3) From statement (1) and (2)

Proved

Theorem – 10

The opposite angles of a cyclic quadrilateral are supplementary. OR Angles in conjugate
segments of a circle are supplementary.
Experimental Verification

Step: I With the help of a compass and a AA
pencil, two circles of different radii
with centre O are drawn. In each o oD
circle, quadrilateral ABCD is drawn. D

Step: II With the help of a protractor, <A, B
<B, <C, and <D of each diagram
are measured and the results are C B C
tabulated below :
Fig. l Fig. u

To verify : ∠A + <C = 180° and <B + ∠D = 180°

Table of observation:

Fig ∠A ∠C ∠A + ∠C ∠B ∠D ∠B + ∠D Result
I
∠A + ∠C = 180°
II ∠B + ∠D = 180°

∠A + ∠C = 180°
∠B + ∠D = 180°

Conclusion: The above experiment shows that the opposite angles of a cyclic quadrilateral
are supplementary (verified).

188 GREEN Mathematics Book-10

Theoretical proof P

Given : PQRS is a cyclic quadrilateral and O is the centre of the circle. o
S
To prove : (i) ∠P + ∠R = 180°
R
(ii) ∠Q + ∠S = 180° Q

Construction : Q and S are joined to O.
Proof :

Statements Reasons
(1) Obtuse ∠QOS = 2∠QPS
(1) The central ∠QOS and the inscribed ∠QPS
(2) Reflex ∠QOS = 2∠QRS are standing on the same arc QRS

(3) Obtuse ∠QOS + reflex ∠QOS (2) The central ∠QOS and the inscribed ∠QRS
= 2∠QPS + 2 ∠QRS are standing on the same arc QPS
=2(∠QPS + ∠QRS)
(3) Adding statements (1) and (2)
(4) ∠QOS + reflex ∠QOS = 360°
(4) Angles at point in one complete rotation.
(5) ∴ 2(∠QPS + ∠QRS) = 360° (5) From statements (3) and (4).
or, ∠QPS + ∠QRS = 180°
i.e. ∠P + ∠R = 180° (6) By similar reasons as above.

(6) Similarly, ∠Q + ∠S = 180° P

Alternative method

Given : PQRS is a cyclic quadrilateral and O is the centre of the circle.

To prove : (i) ∠P + ∠R = 180° Q o
S
(ii) ∠Q + ∠S = 180°
Proof: R

Statements Reasons

(1) 1 (1) By the relation of the inscribed
∠QPS ≡ 2 QRS angle and its corresponding arc

(2) 1 (2) By reason similar to above
∠QRS ≡ 2 QPS (3) Adding statements (1) and (2)

11 (4) Circumference of the circle in
∠QPS + ∠QRS ≡ 2 (QRS + 2 QPS) degree measures is 360°
(3) 11

≡ 2 (QRS + QPS) ≡ 2 (Circumference)

1
(4) ∠QPS + ∠QRS = 2 × 360°

i.e, ∠P + ∠R = 180°

(5) ∴ ∠Q + ∠S = 180° (5) By reasons similar to above

Corollary : If one side of a cyclic quadrilateral is produced, the exterior angle is equal to
the opposite interior angle of the cycle quadrilateral.

GREEN Mathematics Book-10 189

Worked Out EXAMPLES

EXAMPLE 1 Find the values of x, y and z in the following figures.
Solution : a. b. c.

EXAMPLE 2 BB
Solution :
x y 50° C o

o A o A zC
A 50° y
C 130°

a. (i) ∠OCA = ∠OAC = 50° B
∵ In ∆AOC, OA = OC radii of the same circle

(ii) y + 50° + 50° = 180° Sum of the interior angles of ∆OAX

∴ y = 180° – 100° = 80° Inscribed angle y and central angle x
1 are standing on the same AC

(iii) ∠ABC = 2 ∠AOC
1

x = 2 × 80°

∴ x = 40°

b. (i) ∠ABC = 90° An angle fomed on semicircle

(ii) ∠ABC + ∠BAC + ∠ACB = 180° Sum of the interior angles of ∆ABC

or, 90° + y + 50° = 180°

∴ y = 180° – 140° = 40°

c. (i) Reflex ∠AOC = 2∠ABC Central angle AOC and inscribed angle
ABC are standing on the same arc AC
= 2 × 130°

= 260°

(ii) z + Reflex ∠AOC = 360° Angles at one complete rotation.

z + 260° = 360° AB32° C
∴ z = 360° – 260° = 100°
X
What is the size of ∠CBX?
Here, ∠ADB = 32° then, ∠ACB = ∠XCB = 32° 85°
So, In ∆BXC,
O
D

or, ∠CBX + ∠BXC + ∠XCB = 180° Sum of interior angles of ∆BXC

or, ∠CBX + 85° + 32° = 1 80°

∴ ∠CBX = 63°

190 GREEN Mathematics Book-10

EXAMPLE 3 What is the size of ∠BAC?

A O 55° B

Solution : The angle in the circumference of a semi-circle = 90°

∠ACB = 90° C

Now, ∠BAC = 180° – ∠ACB – ∠ABC Sum of interior angles of ∆ACB

= 180° – 90° – 55°

= 90° – 55°

= 35°

∴ The size of ∠BAC = 35°

EXAMPLE 4 In the given figure, O is the centre of the circle. AB// PQ and
∠PAB = 140°. Find the measure of ∠ABQ and obtuse ∠AOQ.

Po

AQ
B

Solution : ∠PAB = 140° and AB // PQ .

i. ∠PQB + ∠PAB = 180° ∠PQB and ∠PAB are the opposite angles of
or, ∠PQB + 140° = 180° the cyclic quadrilateral ABQP

∴ ∠PQB = 40°

ii. ∠ABQ +∠PQB = 180° ∠ABQ and ∠PQB are co-interior angles
or, ∠ABQ + 40° = 180° because AB//PQ

∴ ∠ABQ = 140° Central ∠AOQ and inscribed angle ABQ
iii. Reflex ∠AOQ = 2 ∠ABQ are standing on the same are APQB.
= 2 × 140°

= 280°

iv. Obtuse ∠AOQ + Reflex ∠AOQ = 360°

or, Obtuse ∠AOQ + 280° = 360°

∴ Obtuse ∠AOQ = 360° – 280° = 80°

GREEN Mathematics Book-10 191

EXAMPLE 5 In the given diagram, PQR is an isosceles triangle with PQ = PR
Solution : and a circle is drawn with PQ as a diameter. Prove that QS = RS.

Given : In ∆PQR, PQ = PR. O is centre P
and PQ is a diameter of a circle.

To prove : QS = RS o

Construction : Points P and S are joined. Q SR
Proof :

Statements Reasons

1. 1. Being ∠PSQ is in semi-circles with
∠QSP = 90° diameter PQ.

2. In ∆PSQ and ∆PSR 2. i. Each of 90°, from statement (1)
PS⊥QR.
i. ∠PSQ = ∠PSR (R)
ii. PQ = PR (H) ii. Given
iii. PS = PS (S) iii. Common side

3. ∴ ∆PSQ ≅ ∆PSR 3. By R.H.S. theorem

4. ∴ QS = SR 4. Corresponding sides of ≅ ∆s from (3)

Proved.

EXAMPLE 6 In the adjoining figure, if O is the centre of m

a circle to chords XY and MN produced to nq
meet at P, then prove that ∠XOM – ∠YON o

Solution : = 2 ∠XPM. : O is centre of a circle, where p y x
Given

XY and MN are produced to

meet at P.

To prove : ∠XOM – ∠YON = 2∠XPM

Construction : From N, NQ//YX is drawn which meets the circle at Q.
Proof:

Statements Reasons

1. ∠XOM ≡ XQM 1. Being the relation of the central angle and
corresponding arc of a circle

2. ∠YON ≡ YN 2. Same as (1)

3. YN ≡ QX 3. Arcs intercepted between NQ//YX

4. ∠YON ≡ XQ 4. From statements, (2) and (3)

5. ∠XOM – ∠YON ≡ XQM – XQ 5. Subtracting statements (4) from (1)

6. ∠XOM – ∠YON ≡ QM 6. XQM – XQ ≡ QM , from figure

7. QM ≡ 2∠QNM 7. By the relation of the inscribed angle and
corresponding arc

8. ∠QNM = ∠XPM 8. Corresponding angles by construction

9. ∴ ∠XOM – ∠YON = 2∠XPM 9. From statements (5), (7) and (8)

Proved.
192 GREEN Mathematics Book-10

EXAMPLE 7 In the given figure, O is the centre of the circle, PQ the diameter
a
and AO⊥PQ, prove that ∠PBR = ∠PAO.

Solution : Given : PQ is the diameter of a circle r
with centre at O and AO⊥PQ

To prove : ∠PBR = ∠PAO p oq
Construction : R and Q are joined

Proof: b

Statements Reasons

1. ∠PBR ≡ ∠PQR 1. Being inscribed angles
standing on the same arc PR

2. ∠PRQ = 90° 2. An angle in a semi-circle

3. ∠RPQ + ∠PQR = 90° 3. Sum of two acute angles of
right angled triangle

4. ∠APO + ∠PAO = 90° 4. Same as 3

5. ∠RPQ + ∠PQR = ∠APO + ∠PAO 5. From statement (3) and (4)

6. ∴ ∠PQR = ∠PAO 6. From statement (5)

7. ∠PBR = ∠PAO 7. From statements (1) and (6)

Proved.

EXAMPLE 8 In the given figure, TB is an angle bisector of ∠ATC. Prove that :
Solution : AC//MN.

Given : In the given diagram, TB bisects ts
∠ATC i.e. ∠ATB = ∠BTC
mn
To prove : AC//MN ac

Proof: b

Statements Reasons

1. ∠ATB = ∠BTC 1. Given TB bisects∠ATC

2. ∠ATB = ∠ASB 2. Angles on the same segment and
∠BTC = ∠ASB standing on the same arc AB

3. From statements (1) and (2)
i.e ∠MTN = ∠MSN 3.

GREEN Mathematics Book-10 193

Points M, N, S and T are From (3), line MN subtends equal
angles at T and S
concyclic 4.
4. Angles on the same segment and
standing on the same arc SN
i.e. MNST is a cyclic Angles on the same segment and
standing on the same arc SR
quadrilateral From statements (5) and (6)
From statements (7)
5. ∠SMN = ∠STN 5.
Proved.
∠STC = ∠SAC 6.
6.
7.
i.e, ∠STN =∠SAC 8.
7. ∠SMN =∠SAC
8. ∴ MN||AC

EXAMPLE 9 In the figure, chords AB and CD are perpendicular to each other.
Solution : Prove that : ∠AOD + ∠BOC = 180°.

C

Given : In a circle of center O, the chords
AB and CD are perpendicular to
each other. O

To prove : ∠AOD + ∠BOC = 180°. A B
M

Construction : A and C are joined. D

Proof:

Statements Reasons

1. ∠ACM + ∠CAM = 90° 1. Sum of acute angles of right
i.e. ∠ACD + ∠CAB = 90° angled DAMC

2. ∠AOD = 2 ∠ACD Being central ∠AOD and
2. inscribed angle ACD are standing

on the same AD

3. ∠BOC = 2∠CAB 3. Same as reason 2

4. ∠AOD + ∠BOC = 2∠ACD + 2∠CAB 4. Adding statements 2 and 3

5. ∠AOD + ∠BOC = 2(∠ACD + ∠CAB) 5. From statement 4

6. ∠AOD + BOC = 2 × 90° 6. From statement 1 and 5
7. ∴ ∠AOD + ∠BOC = 180° 7. From statement 6

Proved.

194 GREEN Mathematics Book-10