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Departmarnt of School Educatian

@ Guvammant :31Tamilnadu
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Meaamements
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System

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Chapter l

ALGEBRA

1.1 Simple expressions with two variables
We have learnt about rectangle. Its area is l><b in which the letters &#39;l
and &#39;b&#39;
are variables.

Variables follow the rules of four fundamental operations of numbers.
Let us now translate a few Verbal phrasesinto expressionsusing Variables.
0.-.

-.

.*

.-.

The following table will help us to learn some of the words (phrases)that can
be used to indicate mathematical operations:

Write the algebraic expressionsfor the following:
1)

Twice the sum of m and n.

2)

b decreasedby twice a.

3)

Numbers x and y both squared and added.

4)

Product of p and q added to 7.

5)

Two times the product of a and b divided by 5.

6)

x more than two-third of y.

7)

Half a number x decreasedby 3.

8)

Sum of numbers m and 11decreasedby their product .

9)

4 times x less than sum of y and 6.

10)

Double the sum of one third of a and m.

11)

Quotient of y by 5 added to x.

Solution:
1)2(m
+n)
3)

x2 + y2

5)

7)
9)
yr

th §(i)ga:imeg
jb. at
W3 (ii)Sdmultipsliedby
the
sum
ofaiandVVy7
(iii)Twice
indecreased
n.3V
(iv)Four
dmesx
divided
by
&#39;
V
(v))Five
times
pLrnu1tip1ied
byg3
times
q. T M»"
Exercise
Choose

the correct

1 .1

answer:

The sum of 5 times x, 3 times y and 7

(A)5(x+3y+7)

(B)5x+3y+7

Chapter 1
2 less than the product of y and z
(A)2-~yz

(B)2+yz

(C) yz2

(D)

2y~z

(V) Half of 1) addedto the product of 6 and q

<A>
g +6q

(B)p+621 (C)§<p
+6q> (D)§<6p
+q)

2. Write the algebraic expressionsfor the following using variables, constantsand
arithmetic operations:
(i)
(ii)

Sum of x and twice y.
Subtraction of z from y.

(iii) Product of x and y increasedby 4
(iv)

The difference between 3 times x and 4 times y.

(v) The sum of 10, x and y.

(vi) Product of p and q decreasedby 5.
(vii)

Product of numbers m and n subtracted from 12.

(viii) Sum of numbers a and b subtractedfrom their product.
(ix) Number 6 addedto 3 times the product of numbers c and d.
(X) Four times the product of x and y divided by 3.

1.2 Simple Linear Equations
Ma1arsuncle presented her a statue. She wants to
know the weight of that statue.Sheuseda weighing balanceto
measureits weight. Sheknows her weight is 40kg . Shends
that the weight of the statueand potatoesbalanceher weight.

i.e.,
Weight
ofPlus
Weight
ofEqual
Malars
statue

potatoes

weight

3
Table
Now

we will

think

about

1.1

a balance

to nd

the value

of

s.

Take away 15 from both sides.

Now the balance shows the weight of the statue.
s + 15 =

s + 15

40 (from Table 1.1)

15
s

40 15 (Taking away 15 from both the sides)
=

25

So the statue weighs 25 kg.
The statement s +15 = 40 is an equation. i.e., a statement in which two
mathematical expressionsare equal is called an equation.
In a balance, if we take away some weight from one side, to balance it we must
take away the same weight from the other side also.
If we add some weight to one side of the balance, to balance it we must add the
same weight on the other side also.
Similarly, an equation is like a weighing balance having equal weights on
each side. In an equation there is always an equality sign. This equality sign shows
that Value of the expression on the left hand side (LHS) is equal to the Value of the
expressionon the right hand side (RHS).
* Consider the equation x + 7 = 15
Here

LHS is x + 7
RHS

is 15

We shall subtract 7 from both sides of the equation
x + 7 7 = 15 - 7 (Subtracting 7 reducesthe LHS to x)
x= 8

(Variablex is separated)

Chapter 1
* Consider the equation n 3 = 10
LHSBn~3
RHS

is

10

Adding 3 to both sides, we get
n~3+3=lO+3

n = 13

(variable 11is separated)

* Consider the equation 4m = 28
Divide both sidesby 4

4_m:£
4

4

m=7

* Consider
theequation
-32:
=6
Multiply both sidesby 2
.33.

2X2 __
6X2
y=l2

So, if we add (or subtract) any number on one side of an equation, we have
to add (or subtract) the same number the other side of the equation also to keep
the equation balanced. Similarly, if we multiply (or divide) both sides by the same
non-zero number, the equation is balanced. Hence to solve an equation, one has to
perform the arithmetical operations according to the given equations to separatethe
variable from the equation.

Solve

3p + 4 = 25

Salutirm:

3p + 4 -- 4 = 25 ~ 4 (Subtracting4frombothsidesoftheequation)
3p=21

3%
=3-1 (Dividing
both
sides
by3)
p=7

Solve

7m - 5 = 30

Solution:

7m - 5 + 5 = 30 + 5 (adding 5 on both sides)

7m=35

77&#39;"
=375

(Dividing
both
sides
by7)

m=5

While solving equations,the commonly usedoperation is adding or subtracting
the same number on both sides of the equation. Instead of adding or subtracting a
number on both sides of the equation, we can transposethe number.
Transposing a number (i.e., changing the side of the number) is the same as
adding or subtracting the number from both sides. While transposing a number we
should changeits sign. Let us seesome examples of transposing.

Solve

2a - 12 = 14

Solution:
M:

Solve

i
o

5x + 3 = 18

Solution: Transposing +3 from LHS to RHS
5x = 18

3 (on Transposing +3 becomes - 3)

$5529
155. (Dividing
both
sides
by5)

Chapter 1

Solve 2(x + 4) = 12
Solution: Divide both sidesby 2 to remove the brackets in the LHS.

2(x+4):Q
2

2

x+4=6

x= 6 4

(transposing +4 to RHS)

x=2

Solve --3(m - 2) = 18
Solution. Divide both sidesby (~--3) to remove the brackets in the LHS.

--3(m--2):1
~~3
m -

-3
2 = ~ 6

m=

6+ 2

(transposing - 2 to RHS)

m=-4

So1Ve(3x+1)7

= 12

Solution:

(3x+1)7
3x+17

5x + 2x = 17

3

(transposing + 3 to RHS and - 2x to LHS)

Sum of three consecutive integers is 45. Find the integers.
Solution: Let the rst integer be x.
=>secondinteger = x + 1
Thirdinteger

=x+1+1=x+2

Theirsum

=x+(x+1)+(x+2)=45
3x+3=45
3x=42
x =

14

Hence, the integers are x = 14

x+1=15
x+2=16
A number when added to 60 gives 75. What is the number?
Solution:

Let the number be x.

The equation is

60 + x = 75
x = 75

60

x = 15

20 less than a number
Solution:

is 80. What

Let the number

The equation is

x

be

is the number?

x.

20 = 80
x = 80 + 20
x = 100

Chapter 1

L of a numberis 63.Whatis thenumber?
10

Solutian:

Let the number

be x.

The
equation
is
=63
L

_

10(x)><1063><10
x=630

A number divided by 4 and increasedby 6 gives 10. Find the number.
Solution.

Let the number be

The equation is

+6=
25.

x.

10

4

=10-6
:4
2E

4x4

the number

=

4x4

is 16.

Thendra1sage is 3 less than that of Revathi. If Thendra1sage is 18, What is
Revathi s age?
Solution: Let ReVathisage be x
=»Thendra1sage = x - 3

Exercise
.

Choose

the correct

answer.

If p+3

=9,thenp

(A) 12
If 12x

(B) 6
= 8,thenx
is

(A) 4

is
(C) 3

(B) 20

(C) -4

If%=7,then
q is
(A)13
(B)215
If 7(x

9):

12

(C)
42

35, thenx is

(A) 5

(B) -4

Three times a number

(A) 63

(C) 14

is 60. Then the number

(B) 57

is

(C) 180

Solve:
(i)x--5:7
(iv) b-3:-5
(vii)

(ii)a+3=1O
(v) -x:5

3-~x=8

(viii)

14n=10

(X) 20y=7
Solve

:

(i) 2x = 100

(ii) 31 = 42

(iv) 51 = 17a

(V) 5x =45

(vii)
(X)

7x

= 42

3x =

(viii)

10m =

30

18

x=7
.1.

(ii) g-= 5

(iii)

2
_-1-7=8 (v):52=2
3x+1=

4a5

10

=41

(vi)

(ii) 11+2x=-19

(v) 3(x+2)=

(iii)

15

(vi)

Chapter 1
The sum of two numbers is 33. If one number is 18, what is the other number?

A number increasedby 12 gives 25. Find the number.
If 60 is subtracted from a number, the result is 48. Find
5 times a number

the number.

is 60. Find the number.

3 times a number decreasedby 6 gives 18. Find the number.
The sum of 2 consecutiveintegersis 75. Find the numbers.
Rams father gave him ?70.Now he has T130. How much money did Ram have
in the beginning?
8 years ago, I was 27 yearsold. How old am I now?

jjjy

these

(1) y+18=--70

(ii)

(iii) L-5:-6
3

-300+x=10O

(iv) 2x+9=19

(v) 3x+4=2x+11

Fun game
Ram asked his friends Arun, Saranya and Ravi to think of a number and told
them

to add 50 to it. Then

he asked them

to double

it. Next

he asked them

to add 48 to

the answer. Then he told them to divide it by 2 and subtract the number that they had
thought of. Ram said that the number could now be 74 for all of them. Check it out if
Arun had thought of 16, Saranyahad thought of 20 and Ravi had thought of 7.

ofaiinumber

Takeawaythenumber
you thoughtof 1.

Algebra is a branch of

Mathematics that involves

alphabet, numbers and

mathematical operations.
A variable or a literal is a quantity which can take various numerical values.

A quantity which has a xed numerical value is a constant.
An algebraic expression is a combination of variables and constants connected
by the arithmetic operations.

Expressionsare made up of terms.
Terms having the same variable or product of variables with same powers are
called Like terms. Terms having different variable or product of variables with
different powers are called Unlike terms.

The degree of an expressionof one variable is the highest value of the exponent
of the variable. The degree of an expression of more than one variable is the
highest value of the sum of the exponents of the variables in different terms
A statement in which two expressions are equal is called an equation.
An equation remains the same if the LHS and RHS are interchanged.

The value of the variable for which the equation is satised is called the
solution of the equation.

Chapter 2

LIFE

MATHEMATICS

2.1 Percent

V "I

In the banners put up in the shopswhat do you understand by 25%, 20% ?
Ramus mother refers to his report card to analyze his performance in
Mathematics

in standard

VI.

His marks in Maths as given in his report card are
17/25,36/50,75/100,80/100,22/25,45/50
RamuIi.

marks
Sheisunable
tond hisbestmarkandhisleastmarkbyjustlooking
atthe
So, sheconverts all the given marks for a maximum of 100(equivalentfractions
with denominator 100) as given below:

Life

Mathematics

Now, all his marks are out of 100. So, sheis able to compare his marks easily
and is happy that Rarnu has improved consistently in Mathematics in standard VI.
Now let us learn about these special fractions.
Try and help the duck to trace the path through the maze from Start to End.
Is there more than one path?
(&#39;7
xgrvw
)2
9&#39;
I
5
E

at

1

I

No, there is only one path that can be traced from Start to End.
Total number of the smallest squares = 100
Number of shadedsquares = 41
Number of unshadedsquares = 59
Number of squarestraced by the path =
Now, look at the table below and ll in the blanks:

with itsiiiidenomiinator

:1Percent.

The word Percent is derived from the Latin word Percentum,which means

perhundredor hundredth
or outof 100.
Percentagealso means percent.
Symbol used for percent is %

Anyratio
x:y,where
y=100
iscalled
Percent.

3

Chapter 2
To Express Percent in Different Forms:

Shadedportion representedin the form of :
Ratio

5 : 100
-

17 : 100

5

17

Fraction

T1
00

T1
00

Percent

5%

l7%

Exercise

2.1

Write the following as a percent:

(1)20.100 (11)100

(111)
11divided
by100

(1v)100

Write the following percentas a ratio:
(i) 43%

(ii) 75%

(iii) 5%

(iv) 17%%

(v) 33%%

(iv) 70%

(V) 82%

Write the following percent as a fraction:
(i) 25%

(ii) 12%%

(iii) 33%

Shop
-I

In

Shop
11

Find
theselling
price
inpercentage
when
25%
discount
isgiven,
intherstshop.
What is the reduction in percent given in the second shop?
Which shop offers better price?

Life

Mathematics

2.2 To Express atFraction and a Decimal as 21Percent

We
know
that
% =5%,
116%
=1.2%, =175%.
TOCOIIVCIT
L1 O {O21
p¬ICCl1t

%represented
pictorially
can
beconverted
toapercent
asshown
below:
5

T5

Multiply the numerator and denominator by 10 to make the denominator 100

5x10 _ 50 _50%

10 X 10 _ 100

This
can
also
bedone
bymultiplying
%by100%
(-1-5O><10o)%
=50%
Try

thesej;1
1
},
1
1

.5i(i)%is0fiitheyiscyircleisshaded.
25%
Ofthecircileisyshaded.
rydrawjygp1.¢i;g1esiwithi
(i)_50%,
(ii)25%
iportiopnyshaded
in

l differentiwiii

T

Less
than1andmore
than100canalsoberepresented
asapercent.
§%
V

120%

Chapter 2
(i) Fractions with denominators

that can be converted to 100

Express
ias
apercent
5
Solution:

5 multiplied by 20 gives 100

3 x20 _ 60 __60%
5x20

"100

Express
6L
asapercent
4

Solution:
L._Q
4

64

4 multiplied by 25 gives 100

25><25_ 625 _ 625%
4><25

&#39;
100

(ii) Fractions with denominators

that cannot be converted to 100

Expressi
asapercent
7
Solution: Multiply by 100%

(%x100)%
=$90
=57%%
=57.14%
Express
Las
apercent
3
Solution: Multiply by 100%

<%xwo>%=<%>%
= 33% % (or) 33.33%

There are 250 students in a school. 55 students like basketball, 75 students

like football, 63 students like throw ball, while the remaining like cricket. What is

Life

Mathematics

Solution.
Total
number
ofstudents
=250
(8)

Number

of students

who like

basketball:

55

55outof250likebasket
ballwhichcanberepresented
asL250

Percentage
ofstudents
who
likebasket
ball=

><
l00)%

= 22%

Number

of students

who like

throw

ball:

63

63outof250likethrowballandthatcanberepresented
asQ
250
63
Percentage
ofstudents
wholikethrow
ball=(250
X100)%

=126%=252%
5
22% like basket ball, 25.2% like throw ball.

(iii) To convert decimals to percents

Express 0.07 as a percent
Solution:

Multiply by 100%
(0.07 X l00)%

= 7%

Aliter:

0.07= --7100= 7%

Express 0.567 as a percent
Solution:

Multiply by 100%
(0.567 X 100) % = 56.7%

- .
567 =
567
Amer.
0.567_-1000
mxmo
_ 56.7

.

Chapter 2
Think!

I 1. %ofyour
blood
iswater.
What
%ofyour
blood
isnotWater.
2. -gofyour
body
weight
ismuscle.
What
%ofbody
ismuscle?
About
%of
your
body
weight
iswater.
Ismuscle
weight
plus
water
weight
more or less than 100 %? What does that tell about your muscles?
Exercise
Choose

the correct

2.2

answer:

6.25 =

(A) 62.5%
0.0003

(B) 6250%

(C) 625%

(D) 6.25%

(B) 0.3%

(C) 0.03%

(D) 0.0003%

(B)%%

(C)0.25%

(D)5%

(C) 33%

(D) none of these

(c)50

(D)20

=

(A) 3%

_5_.=
20

(A)25%

The percentof 20 minutes to 1 hour is
(A) 33%

(B) 33

The percentof 50 paiseto Re. 1 is

(A)500

(B)%

Convert the given fractions to percents

i)%8 ii)59()iii)5%

iv)§-

Convert the given decimalsto percents
i) 0.36

ii) 0.03

iii) 0.071

iv) 3.05

V) 0.75

In a class of 35 students, 7 students were absent on a particular day. What
percentageof the studentswere absent?
Ram bought 36 mangoes.5 mangoeswere rotten. What is the percentageof the
mangoesthat were rotten?
In a classof 50, 23 were girls and the rest were boys. What is the percentageof
girls and the percentageof boys?
Ravi got 66 marks out of 75 in Mathematics and 72 out of 80 in Science.In
which subjectdid he scoremore?
Shyamsmonthly income is ?12,000.He saves?1,200Find the percent of his
savingsand his expenditure.

Life

Mathematics

2.3 To Express a Percent as a Fraction (or) a Decimal
i) A percent is a fraction with its denominator 100. While expressing it as a
fraction, reduce the fraction to its lowest term.

Express 12% as a fraction.
Solution:
12
100

L
(reduce the fraction to its lowest terms)

25
Percents
thathave
easy

Express
233%%as
afraction.

fractions

Solution:

.

50%= we

wawute
Findmorerof
thiskind

.7.

1

3x100 3

2%

1

33§%=.

Express
%%
asafraction
Solution:

1

_

1

_

1

4%-4><100-400

(ii) A percent is a fraction with its denominator 100.To convert this fraction to
a decimal, take the numerator and move the decimal point to its left by 2 digits.

Express 15% as a decimal.
Solution:

15%__1_5___.
10-0.15
Express 25.7% as a decimal.
Solution:

25.7%

25.7

100

0.257

Chapter 2
Math game - To make a triplet (3 Matching cards)
This game can be played by 2 or 3 people.
Write an equivalent ratio and decimal for each of the given percent in different
cards as shown.

Make a deck of 48 cards (16 such sets of cards) - 3 cards to represent one
particular Value in the form of %, ratio and decimal.
Shufe the cards and deal the entire deck to all the players.
Playershave to pick out the three cards that representthe sameValueof percent,
ratio and decimal and place them face up on the table.
The remaining cards are held by the players and the game begins.
One player choosesa single unknown card from the player on his left. If this
card completes a triple (3 matching cards) the 3 cards are placed face up on the table.
If triplet cannot be made, the card is added to the players hand. Play proceedsto the
left.

Players take turns to choosethe cards until all triplets have been made.
The player with the most number of triplets is the winner.
TO FIND

THE

VALUES

OF PERCENTS

Colour 50% of the circle green and 25% of the circle red.

50%
=T5-%
=-£ofthecircle
istobecoloured
green.
Similarly,
25%_£..2_5100 100-

iof thecircle
istobecoloured
red.

Life

Mathematics

Now,
trycolouring
%ofthesquare,
green
and
i ofthe
square,red.
Do you think that the green coloured regions are equal in
both the gures?
N0, 50% of the circle is not equal to 50% of the square.
Similarly the red coloured regions, 25% of the circle is not equal to 25% of

the
square.
Now,
lets
ndthe
Value
of50%
of?l00
and
50%
of?10.
What is 50% of ?100?

50%-.5_9...r
-100

What is 50% of ?10?
__i0__L
5O%
100 2

2

SoLof100l><10o-50
2
2

Lor1o~i><1o5
2
"2

50% of ?l00 = ?50

50% of ?10 = ?5

Find the value of 20% of 1000 kg.
Solution:

20%of 1000

100of1000

A

20
100
X 1000

20% of 1000 kg

200 kg.

Find
theValue
of-Q-%of
200.
Solution:

L

L

1000f200
=

1

><2o0

"

Chapter 2

Find the Valueof 0.75%of 40 kg.
Solution:

0.75%
0.75%
of40

0.75

100

100X40

0.75

1..
10-03
0.75%of 40kg

0.3kg.

In a class of 70, 60% are boys. Find the number of boys and girls.

Solution:
Total
number
ofstudents
70
Number of boys

60% of 70

166%
x70
42

Number of boys

42

Number of girls

Total students Number of boys
70

42

28

Number of girls

28

In 2010, the population of a town is 1,50,000.If it is increased by 10% in the
next year, nd the population in 2011.
Solution:

Population in 2010

1,50,000

Increase
inpopulation~1-16%
X1,50,
000

Life
Exercise
Choose

the correct

The common

2.3

answer:

fraction

(A)L10

Mathematics

of 30 % is

(B)L10

(C)L100

(D)A10

(C)%

(D)100

(B) 25

(C) 0.0025

(D) 2.5

(B) ?20

(C) ?30

(D) ?300

(B) ?7.50

(C) ?5

(D) ?100

The
common
fraction
of£-%
is
(A)§

(B)57

The decimal equivalent of 25% is
(A) 0.25
10% of ?300 is

(A) ?10
5% of

?l50

is

(A) ?7

Convert the given percentsto fractions:

1)9%

ii)75% iii)i%

iv)2.5% V)66%%

Convert the given percentsto decimals:
i) 7%

ii) 64%

iii) 375%

iv) 0.03%

v) 0.5%

Find the value of:

i) 75% of 24

ii) 33§% of ?72

iv) 72% of 150

V) 7.5% of 50kg

iii) 45% of 80m

Ram spent 25% of his income on rent. Find the amount spent on rent, if his
income is ?25,000.

A teamplayed 25 matchesin a seasonand won 36% of them. Find the number of
matcheswon by the team.
The population of a village is 32,000. 40% of them are men. 25% of them are
women

and the rest are children.

Find the number

of men and children.

The value of an old car is ?45,000.If the price decreasesby 15%, nd its new
price.
The percentageof literacy in a village is 47%. Find the numberof illiterates in the
village, if the population is 7,500.

Chapter 2
Think!

1) Is it true?
20% of 25 is same as 25% of 20.

2) The tax in a restaurant is 1.5% of your total bill.
a) Write the tax % as a decimal.

b) A family of 6 members paid a bill of ? 750. What is the tax for
their

bill

amount?

c) What is the total amount that they should pay at the restaurant?
2.4 Prot

and Loss

Ram & Co. makes a prot of ?l,50,000 in 2008.
Ram & Co. makes a loss of ?25,000 in 2009.

Is it possible for Ram & Co. to make a prot in the rst year and a loss in the
subsequentyear?
Different stagesof a leather product - bag are shown below:

Factory

Wholesale Dealer

Retailer

Where are the bags produced?
Do the manufactures sell the products directly?
Whom does the products reach nally?
FRECE MST

Mango ?l0 each
Apple ?6 each
Banana ?3 each

Orange?5 each

0PRECE
MS"?

Raja, the fruit stall owner buys fruits from the wholesale market and sells
them in his shop.
On a particular day, he buys apples,mangoesand bananas.

Life

Mathematics

Each fruit has two prices, one at each shop, as shown in the price list.
The price at which Raja buys the fruit at the market is called the Cost Price
(CE). The price at which he sells the fruit in his stall is called the Selling Price (S.P.).
From the price list we can say that the selling price of the applesand the mangoes
in the shop are greater than their respective cost price in the whole sale market. (i.e.)
the shopkeepergets some amount in addition to the cost price. This additional amount
is called the profit.
Selling Price of mango =
Selling price

Cost price

Cost Price of mango + Prot

=

Prot

Prot
Selling Price
15

Prot

Cost Price

10

?5

i.e.,Prot

Selling
Price
~=
Cost
Price,

In caseof the apples,
Selling price of apple > Cost price of apple, there is a prot.
Prot

=

S.P.
8

Prot

C.P.
6

?2

As we know, bananasget rotten fast, the shopkeeper wanted to sell them without
wasting them. So, he sells the bananasat a lower price (less than the cost price). The
amount by which the cost is reduced from the cost price is called Loss.
In case of bananas,

Cost price of banana > selling price of banana, there is a loss.
S.P. of the banana
S.P.

C.P. of the banana
C.P.

Loss

Loss

C.P.

S.P.

Loss

3

Loss

?l

2

Reduced

amount

Chapter 2
So, we can say that
°

When the selling price of an article is greater than its cost price, then
there is a prot.
Prot = Selling Price - Cost Price
When the cost price of an article is greater than its selling price, then
there is a loss.

Loss = Cost Price

To nd

-

S.P = C.P + Prot

-

S.P = C.P - Loss.
Prot

/ Loss

Selling Price

"/9

Rakeshbuys articles for?l0,000
and sells them for ?l1,000 and makes

Try these

a prot of ?1,000, while Rameshbuys

articlesfor P19009000
and56115
them

for?1.01.0o0
and
makes
aprotof 1) Any,fraction
withitsdenominator
100iscalled+___;_____

?l,000.

Bothof themhavemadethe 2)

1
2

same
amount
ofprot.Canyousay M
35%
= _____
__(Sin
ffaCti011)
7
bothof themarebenetedequally? Q05=____;___e
%

f=_ % r

No.

To nd who has gained more,
we need to compare their prot based on their investment.
We know that comparison becomes easier when numbers are expressed in
percent. So, let us nd the prot %
Rakesh makes a prot of ?1,000,when he invests ?l0,000.
Prot of ?l,000 out of ?l0,000

For each 1 rupee, he makes a prot of
Therefore for ?l00, prot
Prot

=

10%

=

1000

10,000

1000
10000

X100

Life

Mathematics

Ramesh makes a prot of ?l000, when he invests ?l,00,000.

Protofl000
outofl,00,000
= légggo
Prot

=

1000 X100_
_ l%
100000

So, from the above we can say that Rakesh is beneted more than Ramesh.
So, Prot Percentage Loss % is also calculated in the same way.

LossPercentage
= LOSS
X100:

00Aiiiiiiiii$£g;ii£iiéégentage
orLoss
Percentage
is
onthecostpriceofthearticle.

A dealer bought a television set for ?l0,000 and sold it for ?l2,000. Find the
prot / loss made by him for 1 television set. If he had sold 5 television sets,nd the
total prot/ loss

Solution:
Selling
Price
ofthe
television
set?l2,000
Cost Price of the television set

S.P.
Prot

?l0,000

C.P,there is a prot
S.P.

C. P.

12000

Prot
Prot on 1 television set
Prot

on 5 television

sets

Prot on 5 television sets

10000

?2,000
?2,000
2000

X 5

?l0,000

Sanjay bought a bicycle for ?5,000. He sold it for ?600less after two years.
Find the selling price and the loss percent.
Solutiam

Loss

Selling Price

?600

Cost Price
5000

Selling Price of the bicycle

Loss

Loss

600

?4400
Loss

CR X 100

__6_9_9_
5000
X100

12%
12%
A man bought an old bicycle for ?1,250. He spent ?250 on its repairs. He then
sold it for ?1400. Find his gain or loss %

Solution:
Cost
Price
ofthe
bicycle
?1,250
Repair Charges

?250

Total Cost Price

1250 + 250 = ?1,500

Selling Price
C.P.

Loss

?1,400
S.P., there is a Loss

Cost Price
1500
100

?100

1400

Selling Price

Life

Mathematics

A fruit seller bought 8 boxes of grapes at ?150each. One box was damaged.
He sold the remaining boxes at ?l90 each. Find the prot / loss percent.

Saiutiom
Cost
Price
of1box
ofgrapes
?150
Cost Price of 8 boxes of grapes

150X 8
?1200

Number of boxes damaged
Number

of boxes

sold

1
8

1

7

Selling Price of 1 box of grapes
Selling Price of 7 boxes of grapes

?190
190 x 7
?1330

S.P.

Prot

C.P, there is a Prot.

Selling Price
1330

Cost Price

1200

130

Prot

?130

Percentage
oftheprot Péogt
X100
130

1200X100

10.83
10.83
%
Ram, the shopkeeperbought a pen for ?50and then sold it at a loss of ?5.Find

Chapter 2

Selling price of the pen

Sara baked

cakes for the school festival.

The cost of one cake was ?55. She

sold 25 cakes and made a prot of ?l1 on each cake. Find the selling price of the
cakes and the prot percent.

Solution:
Cost
price
of1cake
?55
Number

of cakes sold

Cost price of 25 cakes
Prot
Prot

on 1 cake

on 25 Cakes
S.P.

25

55 x 25 = ?1375
?ll
11 X 25 = ?275
C.P. + Prot
1375 + 275

1,650
?1,650

Profit
Percentageof the prot

C.P X100
275

1375X100
20

Prot

Exercise
1.

Choose

the correct

20 %

2.4

answer:

i) If the cost price of a bag is 3575 and the selling price is ?625, then
there is a prot of ?

Life

Mathematics

If the selling price of a bag is ?235 and the cost price is ?200,then
there is a

(A) prot of ?235

(B) loss of ?3

(C) prot of ?35
(D) loss of ?200
Gain or loss percentis always calculatedon
(A) cost price
(B) selling price
(C) gain
(D) loss
If a man makesa prot of ?25on a purchaseof ?250, then prot% is
(A) 25

(B) 10

(C) 250

(D) 225

107.50 ==
Find the selling price when cost price and prot / loss are given.
i) Cost Price = 3450
Prot = ?80
ii) Cost Price = ?760
Loss = ?l40
iii) Cost Price = ?980
Prot = ?47.50
iv) Cost Price = ?430
Loss = ?93.25
V) Cost Price = ?999.75
Loss = ?56.25

Vinoth purchaseda house for ?27, 50,000. He spent 32,50,000 on repairs and
painting. If he sells the housefor ?33,00,000what is his prot or loss % ?
A shop keeper bought 10 bananasfor ?l00. 2 bananaswere rotten. He sold the
remaining bananasat the rate of (ll per banana. Find his gain or loss %
A shop keeper purchased 100 ball pens for ?250. He sold each pen for
?4. Find the prot percent.
A VegetableVendorbought 40 kg of onions for ?360. He sold 36 kg at ?ll per
kg. The rest were sold at ?4.50 per kg as they were not Very good. Find his
prot / loss percent.

#4

5.Choose
one
product
and
ndoutthe
different
(stages
it

pcroissesjfromthe
time
itisproduced
inthetfactory
to
thetimeit reachesgthe
customer.

Chapter
2
Think!
i Doyouthinkdirectselling
bythemanufacturer
himself
ismorebenecial
for
the costumers?

Discuss.

Do it yourself
1. A trader mixes two kinds of oil, one costing ?10Oper Kg. and the other costing
?80 per Kg. in the ratio 3: 2 and sells the mixture at ?101.20 per Kg. Find his *
prot or loss percent.

Sathish sold a camerato Rajeshat a prot of 10%. Rajesh sold it to John at a
loss of 12 %. If John paid ?4,840,at what price did Sathish buy the camera?

Theprot earned
by a booksellerby sellinga bookata prot of 5%is ?l5 i
more

than when

he sells it at a loss of 5%. Find

the Cost Price

of the book.

2.5 Simple interest
x

TakebackRs.2l},i)00
BANK

DepositRs.l0,0()0
now!

!&#39;
V

y

Invest

Rs.10,000 now

;

Get back
Rs. 20,000 in
5 years

Deposit ?l0,000 now. Get ?20,000 at the end of 7 years.
Deposit ?l0,000 now. Get ?20,000at the end of 6 years.
Is it possible?What is the reason for these differences?
Lokesh received a prize amount of ?5,000 which he deposited in a bank in
June 2008. After one year he got back ?5,400.
Why does he get more money? How much more does he get?
If ?5,000 is left with him in his purse, will he gain ?400?
Lokesh deposited?5,000 for 1 year and received ?5,400 at the end of the rst
year.

When we borrow (or lend) money we pay (or receive) some additional amount
in addition to the original amount. This additional amount that we receive is termed

Life

Mathematics

As we have seen in the above case, money can be borrowed deposited in
banks to get Interest.
In the above case, Lokesh received an interest of ?400.

The amount borrowed / lent is called the Principal (P). In this case,the amount
deposited - ?5,000 is termed as Principal (P).
The Principal added to the Interest is called the Amount (A).
In the above case,

Amount = Principal +Interest
=

?5000 + ?400 = ?5,400.

Will this Interest remain the same always?
Denitely not. Now, look at the following cases
(i)

If the Principal depositedis increasedfrom ?5,000 to ?l0,000, then
will

(ii)

the interest

increase?

Similarly, if ?5,000is depositedfor more number of years, then will
the interest

increase?

Yes in both the above said cases,interest will denitely increase.
From the above, we can say that interest dependson principal and duration of
time. But it also dependson one more factor called the rate of interest.
Rate of interest is the amount calculated annually for ?100
(i.e.)if rate of interest is 10%per annum, then interest is ?l0 for ?l00 for 1 year.
So, Interest dependson:
Amount deposited or borrowed Principal (P)
Period of time - mostly expressedin years (n)
Rate of Interest (r)

This Interest is termed as Simple Interest becauseit is always calculated on the
initial amount (ie) Principal.
Calculation

of Interest

If r is the rate of interest, principal is ?l00, then Interest

for1year

_.

__7___
100><1><1O0

I
for2 years__ 100X2X:100
__Z__
for3years__10O><3><1O0
L
fornyears._..100><n><100

Chapter 2

A

Interest
I
The other

formulae

I

:

=

Amount

=

A

Principal

P

derived

from

Pnr
foo
are

1001
Pn
1001
Pr
P = 1001
1&#39;&#39;

fl

771
Note: :1is always calculated in years. When n is given in months \ days,
convert it into years.

Try these

Fill in theblanks

Kamal invested ?3,000 for 1 year at 7 % per annum. Find the simple interest
and the amount received by him at the end of one year.

Life

Mathematics

Interest
(I) = f()i6
3000 ><1>< 7
100

?210
P + I
3000

+ 210

?3,210

Radhika invested?5,000 for 2 years at 11% per annum. Find the simple interest
and the amount received by him at the end of 2 years.

Solution:
Principal
(P) ?5,000
Number of years (n)

2 years

Rate of interest (r)

I

11 %
_

M

100
5000>< 2x11
100

1100

I
Amount (A)

?1,100
P+I
5000

A

+ 1100

?6,100

Find the simple interest and the amount due on ?7,500 at 8 % per annum for
1 year 6 months.

now

ts

I

Solution:

12 months = 1 years

?7,500
1__6_
12

6months
=%year
=-12»
year
I

3

I 3months
= 1-2-year

Chapter 2
_

Pnr
100

7500><%><8
100

7500><3><8
2><10O
900

?900
P+I
7500 + 900

?8,400
?900,
Amount
=?8,400
?7,500

~g~
years
8%

P(1+1E)~r()-

<M

75001+ 100

7500<1+
3X8>
2 ><1oo

750o(%§-)
300 X 28
8400

Life

Mathematics

Find the simple interest and the amount due on ?6,750for 219 days at 10 %
per annum.
Solution.

?6,750

219days

Know this

K year
=%year it365days= lyear
10%
V219
days
= %61g
year
m
gyear
219

100

73

6750><3><10

V 73days 3"65&#39;Y"ar

405

1

5><1oo

1

_1_year
5

y

?405
P + I
6750 + 405

7,155
A
Interest

?7,155
?405, Amount = ?7,155

Rahul borrowed ?4,000on 7th of June2006 andreturned it on 19thAugust2006.
Find the amount he paid, if the interest is calculated at 5 % per annum.
Solution:

Number of days,

P

?4,000

7

5%

June

July

Know this

24 (30_ 6) Thirty days 1 hath A
September,
April, June

andNovember.
All the

August

Total number of days

yrest have thirty

one

except
February.

Tl

Chapter 2

Find the rate percent per annum when a principal of ?7,000earns a S.I. of
?1,680 in 16 months.

Solution.
?7,000
16 months
16

E)

4

= §}

?1,680
7

1001
Pn
100>< 1680

7000
><%
100 X 1680 X 3
7000><4

18
18
%

Life

Mathematics

5%
?
A

P

11,000

10,000

1,000
?1000

1001
Pr
100>< 1000
10000><5

2 years.

P(1%%>

10000(1+-qgg)
L

1*20

20+n
20
20+n
20+n
fl

2 years

A sum of money triples itself at 8 % per annum over a certain time. Find the
number of years.

Salution:

Chapter 2

11

Number of years
Aliter:

Let
Principal
be
?100
Amount
3X100
?300
AP
300 - 100

?200.
1001 __ 100 X 200

Pr "

100><8

n _ %=25
Number of years

25.

A certain sum of money amountsto ?10,080in 5 years at 8 % . Find the principal.

Solution:

Life

Mathematics

10080

10080
><%
Imo

Principal

A certain sum of money amounts to ?8,880in 6 years and ?7,920in 4 years
respectively. Find the principal and rate percent.
Solution:

Amount at the end of 6 years

Principal + interest for 6 years
P+g=8%0

Amount at the end of 4 years

Principal + Interest for 4 years
P+g=7m0
8880

7920

960

Interest at the end of 2 years

?960

E
Interest at the end of 1st year

2
480

Interest at the end of 4 years

480 X 4
1,920

P + 14
P + 1920

7920
7920

P

7920 -

P

6,000

Principal

1920

?6,000

100]
I

pn
100 X 1920

Chapter 2
Exercise
Choose

the correct

2.5

answer:

Simple Interest on ?1000 at 10 % per annumfor 2 yearsis
(A) ?1000

(B) ?200

If Amount = ?11,500,
(A) ?500
6 months

(C) ?100

(D) ?2000

Principal: ?11,000, Interest is

(B) ?22,500

(C) ?11,000

(D) ?11,000

(B)i yr

(C)3 yr

(D)1yr

(B)§yr

(C)~§ yr

=

(A)§ yr
292 days =

(A)§~yr
IfP = ?14000

(D)Ǥ~yr

I = ?1000,A is

(A) ?15000

(B) ?13000

(C) ?14000

(D) ?1000

Find the S.I. and the amounton ?5,000 at 10 % per annumfor 5 years.
. Find the S.I and the amounton ?1,200at 12% % per annumfor 3 years.
. Lokesh invested?10,000 in a bank that pays an interest of 10 % per annum. He
withdraws the amount after 2 yearsand 3 months.Find the interest,he receives.
Find the amount when ?2,500 is invested for 146 days at 13 % per annum.

Find the S.I andamounton ?12,000from May 21"1999to August 2"1999at
9 % per annum.
Sathya deposited 36,000 in a bank and received ?7500 at the end of 5 years.
Find

the rate of interest.

Find
theprincipal
that
earns
?250
asS.I.in2%years
at10%perannum.
In how many years will a sum of ?5,000 amount to ?5,800 at the rate of
8 % per annum.
A sum of money doubles itself in 10 years. Find the rate of interest.

Life

Mathematics

A sumof money
doubles
itselfat12L2 % per annum over a certain period of
time. Find the number of years.
A certain sum of money amountsto ?6,372 in 3 years at 6 % Find the principal.

Acertain
sum
ofmoney
amounts
to?6,500
in3years
and
?5,750
in1%years
respectively . Find the principal and the rate percent.
Find S.I. and amount on ? 3,600 at 15% p.a. for 3 years and 9 months.

Find
theprincipal
that
earns
?2,080
asS.I.in3%years
at16%
p.a.

Think!

A

Find
therate
percent
atwhich,
asum
ofmoney
becomes
24 times
_
1n2 years.

AIf Ramneeds
?6,00,000
after10years,
howmuch
should
he
. invest
nowinabank
if thebank
pays
20%interest
p.a.

.

A fraction

whose

denominator

is 100 or a ratio

whose

second

term

is 100 is

termed as a percent.
Percent means per hundred, denoted by %
To convert a fraction or a decimal to a percent, multiply by 100.
. The price at which an article is bought is called the cost price of an article.
. The price at which an article is sold is called the selling price of an article.

. If the selling price of an article is more than the cost price, there is a prot.

Chapter 2
7. if the cost price of an article is more than the selling price, there is a loss.
8. Total cost price = Cost Price + Repair Charges/ Transportation charges.

9. Prot or loss is always calculated for the same number of articles or same
units.

Prot = Selling Price Cost Price
Loss = Cost Price Selling Price
-0
Profit ><100
Prot
/0 = CR

Loss%
= x100
. Selling Price = Cost Price + Prot
Selling Price = Cost Price
.
.
.
Simple
interest
ISI _ m100
A

P+l

P+ Pnr
100

P<1+a>
=

A-P

Loss

Measurements

MEASUREMENTS

3.1 "Irapezium

A trapezium
is aquadrilateral
withonepairof
oppositesidesareparallel.

------->-----ma

A

The distance between the parallel sides is the

height
ofthe
trapezium.
Here
the
sides
AD
and
BC
Fig.
3.1
are not parallel, but AB ll DC.
If the non parallel sidesof a trapezium are equal ( AD = BC ), then it is known
as an isoscelestrapezium.
D
C
Here

AA
AC

=
BD

AA+AD 180°
; AB+AC=180°

B

Area of a trapezium
ABCD is a trapezium with parallel sides AB and DC measuring a and b.
Let the distance between the two parallel sides be h. The diagonal BD divides the
trapezium into two triangles ABD and BCD.
Area of the trapezium
=

area of AABD

+ area of ABCD

-1-><AB><h+;><Dc><h
><h[AB + DC]
><h[a + b] sq. units

Find the area of the trapezium whose height is 10 cm and the parallel sides are

Solution

Given: h = 10 cm, a = 12 cm, b:

8 cm

Area
ofatrapezium
= %><
h(a+b)
= %><1o><(12
+8): 5><(2o)
Areaof thetrapezium= 100sq.cm2
The length of the two parallel sides of a trapezium are 15 cm and 10 cm. If its
area is 100 sq. cm. Find the distance between the parallel sides.

Solution
Given:
a=15
cm,
b=10
cm,
Area
=100
sq.
cm.
Area of the trapezium = 100

%h(a+b)
= 100
%><h><(15+10)
= 100
h><25

=

h

200

_@_

25-8

the distance between the parallel sides = 8 cm.

The area of a trapezium is 102 sq. cm and its height is 12 cm. If one of its
parallel sides is 8 cm. Find the length of the other side.

Solution
Given:
Area
=102
cm2,
h=12
cm,
a=8cm.
Area of a trapezium = 102

%h(a
+b) = 102
%><12><(8
+19)= 102
6 (8 + 19)
8+b

A

lengthof theotherside

t

102
17

9 cm

=>

b=178=9

Measurements

13C)
(iii)
Fig. 3.4

EF divides the trapezium in to two parts as shown in the Fig. 4.40 (ii)
From D draw DG _LEF. Cut the three parts separately.
Arrange three parts as shown in the Fig. 3.4 (iii)
The gure obtained is a rectangle whose length is AB + CD = a + b

and
breadth
is%(height
oftrapezium
)=%h
Area of trapezium = area of rectangle as shown in Fig. 4.40 (iii)
= length x breadth

(a+b)(%h)
%h(a
+b)sq.units
Exercise
Choose

the correct

answer.

The areaof trapezium is

(A)h(a+b)

3.1

sq. units

(B)%h(a+b) (C)h(a 1;)

(D)%h(a b)

In an isosceles trapezium
(A) non parallel sides are equal

(B) parallel sides are equal

(C) height: base

(D) parallel sides = non parallel sides

The sum of parallel sides of a trapezium is 18 cm and height is 15 cm. Then
its area is

(A) 105cm2

(B) 115cm2

(C) 125cm2

(D) 135cm2

The height of a trapezium whose sum of parallel sides is 20 cm and the area
80 cm2 is
(A) 2 cm

(B) 4 cm

(C) 6 cm

(D) 8 cm

2. Find the areaof a trapezium whose altitudes and parallel sidesare given below:
i) altitude = 10 cm, parallel sides= 4 cm and 6 cm
ii) altitude = 11 cm, parallel sides= 7.5 cm and 4.5 cm
iii) altitude = 14 cm, parallel sides= 8 cm and 3.5 cm

Theareaof a trapezium
is 88cm2andits heightis 8 cm.If oneof its parallelside
is 10 cm. Find the length of the other side.
4. A gardenis in the form of a trapezium.The parallel sidesare 40 m and 30 m.
The perpendiculardistancebetweenthe parallel side is 25 m. Find the areaof the
garden.

5. Areaof a trapezium
is 960cm2.Theparallelsidesare40cmand60cm.Findthe
distancebetweenthe parallel sides.

3.2
Circle
Inour
daily
life,
we
come
across
anumber
ofobjects
like
wheels,
coins,
rings,
bangles,giant wheel, compact disc (C.D.)
What is the shapeof the above said objects?
round, round, round

Now,
let
us
try
to
draw
a
circle.
///\\
Movi
Take
athread
ofany
length
and
xoneend
tightly
atE/
Point
K A

Yes, it is round. In Mathematics it is called a Circle.

apoint
0asshown
inthe
gure.
Tie
apencil
(or
achalk)
to Fixed
Pointj
0

the
other
end
and
stretch
the
thread
completely
toapoint
A.

/

Holding
thethread
stretched
tightly,
move
thepencil.\~=~....,,,,,,,,,,...»

Stop it when the pencil again reachesthe point A. Now see
the path traced by the pencil.
Is the path traced by the pencil a circle or a straight line?
Circle

Yes,

Parts

of a Circle

The xed point is called the

of the circle.

The constant distance between the xed point and
the moving point is called the
of the circle.
i.e. The radius is a line segmentwith one end point
at the centre

and the other

end on the circle.

It is denoted

by r.
A line segmentjoining any two points on the circle

Fig.3.5

Measurements

is a chord passing through the centre of the circle. It is denotedby d .
The diameter is the longest chord. It is twice the radius.(i.e. d = 2r)
The diameter divides the circle into two equal parts. Each equal part is a

Think

DO
you

it:

Howmany
diameters
can

pl

acircle
have
?

Circumference

.,.,. .

f radsisradii.

theradii
ofacircle
are
equa

of a circle:

Can you nd the distance covered by an

athleteif hetakestwo roundson a circulartrack.

_

Since it is a circular track, we cannot use AA
the ruler

to nd

out the distance.

So, what can we do ?

Take a one rupee coin.Place it on a paper
and

draw

its outline.

Remove

the

coin.

Mark

Fig. 3.7

a

point A on the outline as shown in the Fig. 3.7
Take a thread and x one end at A. Now place the
thread in such a way that the thread coincides exactly with
the outline.

Cut the other

end of the thread

when

it reaches

the point A.

Lengthof thethreadis nothingbutthecircumference

A

of the coin.

Fig. 3.8
So,
the distance around a circle is called the circumference of the circle, which is

denoted by C. i.e., The perimeterof a circle is known as its circumference.

f :

._.._.._....._.._.._.._....

V Try

._.._.._.y._,.,._.
._.._.k
._.W, ._.._.._.._.._....._....._.._,_._
._.
._.._.._.._.._.._.._.._.._., ._.._..__J,

Takeabottle
caporabangleor
anyother
circular
objects
and
nd thecircuymferenceglf
possible
nd therelation
between
thecircumferenceanid
thediameter
of thecircular
objects.
3

Relation

between

diameter

and circumference

of the circle

Draw four circles with radii 3.5 cm, 7 cm, 5 cm, 10.5 cm in your note book.
Measure their circumferences using a thread and the diameter using a ruler as shown
in the Fig. 3.9 given below.
51

C&#39;\°?.}f.".""1ce

Fill in the missing Valuesin table 3.1 and nd the ratio of the circumference to
the diameter.

..2

it 7cm

uI

Q

2
Table

3.1

What
doyouinfer
from
theabove
table?.
Isthisratio
(9)approximately
d

the same?

Yes
!
=3.14
=:»
C=(3.14)d
So, can you say that the circumference of a circle is always more than 3 times
its diameter

?

Yes !

In allthecases,
theratio9d is aconstant
andis denoted
bytheGreek
letter7:
(read
aspi). Itsapproximate
valueis -2-2or3.14.
where

d is the diameter

of a circle.

Measurements

fromtheabove
formula,
C=7rd
=7r(2r)
=>
The Valueof 7: is calculated by many mathematicians.

Babylonians
: 7r=3 d
Greeks
V:7:=-27%
or3.14
Archemides
; 3%
<7:<3% Aryabhata:
7:=622080308
(or)
3.1416
Now,weuse71:4:
-2; or 3.14

Find

out the circumference

of a circle

whose

diameter

is 21 cm.

Solution
Circumference

of a circle

7rd

Q
7 x21
Find

out the circumference

of a circle

whose

radius

is 3.5 m.

Solution
Circumference
of
acircle
27rr
2x%2
x3.5
2 x 22 x 0.5
22m

Soiutian

When a Wheelmakes one complete revolutions,
Distance

covered

in one rotation

circumference

= Circumference

of the wheel

=

of wheel

7rd units

= l1x63cm
7

198 cm

For one revolution, the distance covered
for 20 revolutions, the distance covered

198 cm
20 x 198 cm
3960 cm

39 m60 cm

A scooter
the radius

wheel

makes

50 revolutions

to cover

a distance

[100 cm:

of 8800

1 m]

cm. Find

of the Wheel.

Solution
Distance

travelled

Number

of revolutions

Distance

Circumference

Number
8800

27rr

>< Circumference

travelled
of

revolutions

50

i.e., 27rr

176

2X272
Xr

176

l76><7

7&#39;
7&#39;

2x22
28 cm
radius

of the Wheel = 28 cm.

The radius of a cart wheel is 70 cm. How many revolution does it make in
travelling a distance of 132 m.
Solution

Given: r = 70 cm, Distance travelled = 132 m.
Circumference

of a cart wheel

= 27rr

2x%%x7o
440 cm

Measurements
Distance
Number

travelled

of revolutions

Number

of revolutions

Distance

travelled

>< Circumference

Circumference
132 m
440

cm

_

(1m=100
cm,132
m=13200
cm)

:30
Number

of revolutions

= 30.

The circumference of a circular eld is 44 m. A cow is tethered to a peg at the
centre of the eld. If the cow can graze the entire eld, nd the length of the rope used
to tie the cow.

Solution
Length
ofthe
rope
Radius
ofthe
circle
Circumference
i.e., 27rr

zxgxr

44 m (given)
44

44

7
.r "_ 2><22"7m
44X7 _

Fig.3.10

The length of the rope used to tie the cow is 7 m.

The radius of a circular ower garden is 56 In. What is the cost of fencing it at
?l0 a metre ?
Solution

Length to be fenced = Circumference of the circular ower garden
Circumference of the ower garden = 27rr

= 2><%><56=352m
Length of the fence = 352 m
Cost of fencing per metre = ?10

The cost of fencing a circular park at the rate of ?5per metre is ?1100.What
is the radius of the park.
Solution

Cost of fencing

Circumference X Rate
Cost of fencing

Circumference

Rate

i.e.,27rr

E

27zr

5

220

.
Q
..2><
7 ><r
I

Radius of the park
7 V Activity

Circular Geoboard

CTakeasquare
Boardanddrawacircle.C

Fixnailsonthecircumference
ofthecircle.
( See
g)
Usingrubberband,formvariousdiameters,
chords,radii and
compare.

Exercise
Choose

the correct

3.2

answer:

The line segmentthat joins the centre of a circle to any point on the circle is
called

(A) Diameter

(B) Radius

(C) Chord

(D) None

A line segmentjoining any two points on the circle is called
(A) Diameter

(B) Radius

(C) Chord

(D) None

A chord passingthrough the centreis called
(A) Diameter
The diameter

(B) Radius
of a circle

(C) Chord

is 1 m then its radius

(A) 100 cm

(B) 50 cm

The circumference

of a circle

(A) 22 cm

(B) 44 cm

is

(C) 20 cm
whose radius

(D) None

(D) 10 cm

is 14 cm is

(C) 66 cm

(D) 88 cm

Measurements

2. Fill up the unknown in the following table:

Find the circumferenceof a circle whosediametreis given below:
(i) 35 cm

(ii) 84 cm

(iii) 119 cm

(iv) 147 cm

Find the circurriferenceof a circle whoseradius is given below:
(i) 12.6 cm

(ii) 63 cm

(iii) 1.4 m

(iv) 4.2 In

Find the radius of a circle whose circumferenceis given below:
(i) 110 cm
The diameter

(ii) 132 cm
of a cart wheel

(iii) 4.4 In

(iv) 11 m

is 2.1 In. Find the distance

travelled

when it

completes100 revolutions.
The diameterof a circular park is 98 In. Find the cost of fencing it at ?4per
metre.

A wheel

makes 20 revolutions

to cover

a distance

of 66 m. Find the diameter

of

the wheel.

The radius of a cycle wheel is 35 cm. How many revolutions doesit make to
cover a distance

of 81.40 m?

10. The radius of a circular park is 63 In. Find the cost of fencing it at ?12 per metre.
Area

of a circle

Consider the following
A farmer levels a circular eld of radius 70 In. What will be the cost of levelling?
What will be the cost of polishing a circular table-top of radius 1.5 In ?
How will you nd the cost ?

.DOyouk

Tond thecostwhatdoyouneed
tond actually? The n

Areaorperimeter
7
Area, area, area

.

enclose by

the circumference
of a

circle is a circular region.

Yes. In such caseswe need to nd the area of the circular region.
So far, you have learnt to nd the area of triangles and quadrilaterals that made
up of straight lines. But, a cirlce is a plane gure made up of curved line different from
other plane gures.

2

So, we have to nd a new approachwhich will make the circle turn into a gure
with straight lines.
Take a chart paper and draw a circle. Cut the circle and take it separately.Shade
one half of the circular region. Now fold the entire circle into eight parts and out along
the folds (seeFig. 3.11).

Fig. 3.11

Arrange the pieces as shown below.

Fig..
What is the gure obtained?
These eight pieces roughly form a parallelogram.
Similarly, if we divide the circle into 64 equal parts and arrange these, it gives
nearly a rectangle. (seeFig. 3.13)

l

.

Z=2-><
circumference
Fig. 3.13

What is the breadth of this rectangle?
The breadth of the rectangle is the radius of the circle.
i.e.,

breadth

b = r

(1)

What is the length of this rectangle ?
As the whole circle is divided into 64 equal parts and on each side we have 32
equal parts. Therefore, the length of the rectangle is the length of 32 equal parts, which
is half

of the circumference

of a circle.

Measurements

-12-[circumference
ofthecircle]
-é-[2727]
=7z&#39;r
.&#39;.l 7rr

Area of the circle

(2)

Area of the rectangle (from the Fig. 4.50)
l>< b

(7z&#39;r)
Xr

(from (1) and (2))

7rr2sq. units.

Find

the area of a circle

whose

diameter

is 14 cm

Solution
Diameter

So,

d

radius r
Area

of circle

Area

of circle

A goat is tethered by a rope 3.5 m long. Find the maximum area that the goat
can graze.

Sulutirm
Radius
ofthe
circle
Length
ofthe
rope
radius
r = 3.5m=%m
maximum area grazed by the goat: 7rr2sq. units.

__2__2__ll

The circumference of a circular park is 176m. Find the area of the park.

Salution
Circumference
176
m(given)
27rr

176

zxgxr

176

7

176><7
44

1&#39;&#39;

&#39;.r

28m

7rr2

Area of the park

%><28><28
22 X 4 X 28

2464 sq. m.
A silver wire when bent in the form of a square enclosesan area of 121 sq. cm.
If the same wire

is bent in the form

of a circle.

Find

the area of the circle.

Soiution
Let
abe
the
side
ofthe
square
Area of the square = 121 sq. cm. (given)

a2 = 121=>a=11cm
Perimeter of the square = 4a units
=

4 X 11 Cm

=

44 cm

Length of the wire = Perimeter of the square

(11><11=121)

Measurements

When a man runs around circular plot of land 10 times, the distance covered by
him is 352 m. Find the area of the plot.

Solutirzn
Distance
covered
in10
times
=352
m
Distance
covered
inone
time=% m=35.2
m
The circumference of the circular plot = Distance covered in one time
circumference
27rr

=

= 35.2 m
35.2

2><2;><r = 35.2

7

35.2><7

r

44

2

0.8 X 7
5.6 m

7rr2

Area of the circular plot

%><5.6><
5.6

22><0.8><5.6
98.56
m2
Areaof circularplot

98.56m2

A wire in the shapeof a rectangle of length 37 cm and width 29 cm is reshaped
in the form

of a circle.

Find

the radius

and area of the circle.

Solution
Length
ofthe
wire
=perimeter
ofthe
rectangle
= 2 [ length + breadth ]
= 2[37cm+29cm]=2><
=

132 cm.

66cm

Exercise

3.3

Find the areaof the circles whosediametersare given below:
(i) 7 cm

(ii) 10.5 cm

(iii) 4.9 m

(iv) 6.3 m

(take 7: =

Find the areaof the circles whoseradii are given below:

(i)1.2
cm (ii)14cm

(iii)4.2m (iv)5.6m (take
72&#39;
=272)

The diameter of a circular plot of ground is 28 m. Find the cost of levelling the
ground at the rate of ?3 per sq. m.
A goat is tied to a peg on a grassland with a rope 7 m long. Find the maximum
areaof the land it can graze.
A circle and a squareeachhave a perimeter of 88 cm. Which has a larger area?
A wheel goesa distanceof 2200 m in 100revolutions. Find the areaof the wheel.
A wire is in the form of a circle of radius 28 cm. Find the area that will enclose, if

it is bent in the form of a squarehaving its perimeter equal to the circumference
of the cirlce.

Theareaof circularplot is 3850m2.Findtheradiusof theplot.Findthecostof
fencing the plot at T10 per metre.
The radius of a circular ground is 70 m. Find the distance covered by a child
walking along the boundary of the ground.

Theareaof acirculareld is 154m2.Findthetimetakenby anathlatetocomplete
2 rounds if sheis jogging at the rate of 5 krn/hr.
11. How many circles of radius 7 cm can be cut from a paper of length 50 cm and
width

32 cm.

3.3 Area of the path way
In our day to - day life we go for a walk in a park, or in a play ground or even
around a swimming pool.
Can you representthe path way of a park diagrammatically ?
Have you ever wondered if it is possible to nd the area of such paths ?
Can the path around the rectangular pool be related to the mount around the
photo in a photo frame ?
Can you think of some more examples?
In this

section

we will

learn

to nd

- Area of rectangular pathway
- Area of circular pathway

Measurements

Area of rectangular pathway
(a) Area of uniform pathway outside the rectangle
Consider a rectangular building. A uniform ower garden is to be laid outside
the building. How do we nd the areaof the ower .
garden?
The uniform ower garden including the
building is also a rectangle in shape. Let us call
it as outer rectangle. We call the building as inner
rectangle.
Let Zand b be the length and breadth of the
building.
Area of the inner rectangle = lb sq.
units.

Let w be the width of the ower garden.
What is the length and breadth of the outer rectangle ?
The length of the outer rectangle (L) = w + l + w = (l + 2w) units
The breadth of the outer rectangle (B) = w + b + w = (b + 2w) units
area of the outer rectangle = L X B
= (l + 2w)(b + 2w) sq. units
Now, what is the area of the ower garden ?
Actually, the area of the ower garden is the pathway bounded between two
rectangles.
Area of the ower garden =

(Area of building and ower garden)
(Area of building)

Generally,
Area
ofthe
pathway
:(Area
ofouter
rectangle)
-(Area
ofinner
rectangle)
i.e. Area of the pathway = (l + 2w) ([9+ 2w) lb.

Theareaof outerrectangle
is 360m2.Theareaof innerrectangle
is 280m2.The
two rectangleshave uniform pathway between them. What is the area of the pathway?
Solution

Chapter
3
(360
280)
m2
=80
m2
Areaof thepathway 80m2

The length of a building is 20 In and its breadth is 10 In. A path of width 1 rn is
made all around the building outside. Find the area of the path.
Solution

11" mangle

<giV*>

l= 20rn

width, w

b=l0m

=1m
L

Area

=lxb

Area

=20mx10m

=l+2w
=20+2=22m

B

=b+2w

=200m2

=l0+2=l2m
Area

=(l +2w) (b+2w)

Area

=22rn><

12m

= 264 m2
Area of the path = (Area of outer rectangle) (Area of inner rectangle)

= (264200)m2=64m2
Areaof thepath = 64 m2

Aschool
auditorium
is45rnlong
and
27In

H it

wide.Thisauditorium
is surrounded
byavarandha V6 Ait As6A AAAAA
of width 3 m on its outside. Find the area of the

Varandha.
Also,
ndthecost
oflaying
thetiles
in AAAA
the Varandhaat the rate of ?l00 per sq. m.
Solutiem

Inner (given) rectangle
l

=45m

b

=27m

Area=45m><27rn

=1215m2

Outer rectangle
Width,

w

=3m

L

=l+2w
=45+6=5lm

B=b+2w

AA

Measurements

(i) Area of the Verandha = (Area of outer rectangle) (Area of inner rectangle)

= (16831215)m2
= 468 m2

Areaof theVerandha= 468m2(or)468sq.m.
(ii)

Cost of laying tiles for 1 sq. m = ?l00
Cost of laying tiles for 468 sq. m = ?100 X 468
= ?46,800

.&#39;.Cost
of laying tiles in the Verandha = ?46,800

(in) Area of uniform pathway inside a rectangle
A swimming pool is built in the middle 1 *
of a rectangular ground leaving an uniform
width

all around

it to maintain

the lawn.

If the pathway outside the pool is to be
grassed,how can you nd its cost?
If the area of the pathway and cost of 1,
grassing per sq. unit is known, then the cost
of grassing the pathway can be found.
Here, the rectangular ground is the outer rectangle where l and b are length and
breadth.

Area of the ground (outer rectangle) = l 19sq. units
If w be the width of the pathway (lawn), what will be the length and breath of
the swimming pool ?
The length of the swimming pool =
=

The breadth of the swimming pool:
=

l w

w

l2w

19 w

w

b - 2w

Area of the swimming pool (inner rectangle) = (l - 2w)(b ~ 2w) Sq. units
Area of the lawn = Area of the ground Area of the swimming pool.

Generally,

Area
ofthe
pathway
(Area
ofouter
rectangle)
-~
(Area
ofinner
rectangl

gggil
¢ ,¢¢ I;
.4,..;,v;_a_=r.~!,¢

The length and breadth of a room are
8 m and 5 m respectively. A red colour border

¢¢¢¢¢,

of uniform width of 0.5 m has beenpainted all f *
around

on its inside.

Find

the area of the border.

.

¢I; 1.1, ,V. . ,

olu Eton

.

¢

.

¢

&#39;,_.V3,.,,.w_,.._,09,,.,
,

.»~:~/

e ;-§.¥1l§e5:e¢:§e
*

Fig. 3.19

l

=8m

b

= 5 m

width,w
L

Area=8m><5m

=0.5m
= l - 2w

=(81)m

= 40 m2

B

= b - 2w
= (5

Area

1) m

= 7m x 4 m

= 28 m2
Area of the path

(Area of outer rectangle) (Area of inner rectangle)

(40 28)m2
= 12 m2

Areaof theborderpaintedwith redcolour= 12m2

A carpet measures3 m X 2 m. A strip of 0.25 m wide is cut off from it on all
sides.Find the area of the remaining carpet and also nd the area of strip cut out.
Solution

Outer rectangle

Inner rectangle

carpet before cutting the strip

carpet after cutting the strip
width, w

= 0.25 m

L =l-2w=(30.5)m
: 2.5
m
B =b2w=(20.5)m
=1.5
m

Measurements

Area of the strip cut out = (Area of the carpet)
(Area of the remaining part)

= (6 3.75)m2
= 2.25 m2

Areaof thestripcutout= 2.25m2
Note:

Exercise

3.4

A play ground 60 m X 40 m is extendedon all sidesby 3 In. What is the extended
area.

A school play ground is rectangularin shapewith length 80 m and breadth60 m.
A cementedpathway running all around it on its outside of width 2 m is built.
Find the cost of cementingif the rate of cementing 1 sq. m is ?20.
A gardenis in the form of a rectangleof dimension 30 m X 20 m. A path of width
1.5 m is laid all around the garden on the outside at the rate of ?l0 per sq. In.
What is the total expense.
A picture is painted on a card board 50 cm long and 30 cm wide suchthat there is
a margin of 2.5 cm along eachof its sides.Find the total areaof the margin.
A rectangularhall has 10 m long and 7 m broad. A carpet is spreadin the centre
leaving a margin of 1 m near the walls. Find the areaof the carpet.Also nd the
area of the un covered

oor.

The outer length and breadth of a photo frame is 80 cm , 50 cm. If the width of
the frame is 3 cm all aroundthe photo. What is the areaof the picture that will be
Visible?

Circular
Concentric

pathway
circles

Circles drawn in a plane with a common centre and different
radii

are called

concentric

circles.

Circular pathway
A track of uniform width is laid around a circular park for
walking purpose.
Can you nd the area of this track ?

pig 3_21

Yes.

Area

of the

track

is the

area bounded

between

two

concentric circles. In Fig. 4.59, O is the common centre of the two
circles.Let

the radius

of the outer

circle

be R and inner

circle

be r.

The shaded portion is known as the circular ring or the
circular pathway. i.e. a circular pathway is the portion bounded
between

two concentric

circles.

width of the pathway, w

R

i.e.,w

r units

Rr=:»R

= w+runits
r

= R

w units.

The area of the circular path = (areaof the outer circle)

(areaof the inner circle)

= 7rR2 -- 7rr2

= 7r(R2- r2) sq. units

The adjoining gure showstwo concentric circles. The radius
of the larger circle is 14 cm and the smaller circle is 7 cm. Find
(i)

The area of the larger circle.

(ii)

The area of the smaller circle.

(iii)

The area of the shadedregion between two circles.

Solution
i)Larger
circle
R

ii)Smaller
circle

14

r

=

7

7rR2

7z&#39;r2

3-7-1
x14x14

-¬73
x7x7

22 X 28

22 X 7

= 616 cm2

154 cm?-

iii) The area of the shadedregion
= (Area of larger circle) (Area of smaller circle)

(616 154)em? = 462cm2

Measurements

Salution
Given:
R=5cm,
r:3cm
Area of the remaining sheet

7r(R2~ rz)

3.14(52 32)
3.14 (25 9)
3.14 x 16

50.24 cm?

Outer

R

circle

Inner

= 5 cm

r

Area = 7zR2sq. units

= 3 cm

Area: 7rr2sq. units

=3.14><5><5 A
A =3.14><25
= 78.5 cm2

circle

4 =3.14><3><3
=3.14><.9 5

2

2 2 = 28.26 cm2

Areaoftheremaining
sheet
= (Area
ofouter
circle)
+(Area
ofinner
circle)
= (78.5-28.26) cm?

=50.24
cm2
2

Area
oftheremaining
sheet:
50.24
cm

A circularower gardenhasan area500m2.A sprinklerat thecentreof the
garden can cover an areathat has a radius of 12 m. Will the sprinkler water the entire
garden (Take 7: = 3.14)

Solution
Given,
area
ofthe
garden
Area
covered
by
asprinkler

Saiutioiz

Area
of
the
circular
path
=7r(R
+r)(R
-r) <

Given:r=50m,

w=2m,

R=r+w=50+2=52m

ii

= 3.14><(52+ 50)(52 - 50)
=

3.14 X 102 X 2

=

3.14 X 204

= 640.56 m2
The cost of levelling the path of area 1 sq m = ?5

Thecostof levellingthepathof 640.56m2 = ?5 X 640.56
= ?3202.80

the cost of levelling the path = ?3202.80
Exercise

3.5

A circus tent hasa baseradiusof 50 m. The ring at the centrefor the performance
by an artists is 20 min radius. Find the arealeft for the audience.
(Take 7: = 3.14)

A circular eld of radius30 m hasa circular path of width 3 m inside its boundary.
Find the areaof the path (Take 7r= 3.14)
A ring shapemetal plate has an internal radius of 7 cm and an externalradius of
10.5 cm. If the cost of material is ?5 per sq. cm, nd the cost of 25 rings.
A circular well has radius 3 m. If a platform of uniform width of 1.5 m is laid
around it, nd the areaof the platform . (Take 7: = 3.14)
A uniform circular path of width 2.5 m is laid outside a circular park of radius

56m.Findthecostof levellingthepathattherateof ?5perm2(Take7: = 3.14)
The radii

of 2 concentric

circles

are 56 cm and 49 cm.

Find

the area of the

pathway.

Theareaof thecircularpathwayis 88m2. If theradiusof theoutercircleis 8
m, nd the radius of the inner circle.

The cost of levelling the areaof the circular pathway is ? 12,012 at the rate of

? 6 perm2. Findtheareaof thepathway.

Measurements

Forjumula,i I

-1-x heightx sumof

2

a
Trapezium

.1.

parallel
sides
Perimeter

of the circle

2 xh><(a+b)sq.
units

=
27rr

units

2 x 72 x radius

Area

of the circle

7: x radius

=

7rr2sq.units

><radius

Area of the pathway

Area

i) area of the rectangular

rectangle Area of

pathway

inner rectangle

Area

ii) area of the circular
Circular Pathway

of outer

pathway

Area

of outer

circle

of inner

circle

= 7:(R2 2)sq.units
= 7r (R+r)

(Rr)

sq. units

GEOMETRY

4.1 Triangle: Revision
A triangle is a closed plane gure made of three line
segments.

In Fig. 4.1 the line

segments AB,

BC and CA

form a closed gure. This is a triangle and is denoted by

AABC. This triangle may be named as AABC or ABCA or B

Fig. 4.1

ACAB.

The line segments forming a triangle are the three sides of the triangle. In

Fig.4.l E, W and C1 are the threesidesof the triangle.
The point where any two of the three line segmentsof a triangle intersect is
called the Vertex of the triangle. In Fig. 4.1 A,B and C are the three Vertices of the
AABC.

When two line segments intersect, they form an angle at that point. In the

triangle in Fig. 4.1 E

and W intersect at B and form an angle at that Vertex.

This angle at B is read as angle B or 4B or AABC.

Thus a triangle has three

angles AA, AB and 4C.
In Fig. 4.1 AABC has
Sides

:

AB,,Cj

Angles

:

LCAB, 4ABC,4BCA

Vertices

:

A, B, C

The side opposite to the Vertices A, B, C are BC, AC and AB respectively. The
angle opposite to the side BC, CA and AB is 4 A, AB and AC respectively.

Geometry
4.2 Types of Triangles
Based

on sides

A triangle is said to be
Equilateral, when all its sides are equal.
Isosceles,when two of its sides are equal.
Scalene,when its sides are unequal.
Based on angles
A triangle is said to be
Right angled, when one of its angle is a right angle and the other two angles are
acute.

Obtuse - angled, when one of its angle is obtuse and the other two angles are
acute.

Acute angled, when all the three of its angles are acute.

4.3 Angle sum property of a triangle:

,.

1

these

Drawanytriangle
ABConasheet
ofpaper §
andmark
theangles
1,2 and3onbothsides
ofthe
tsrihanygle
Whhoisie
Sides
are
7cm,*5cm?
and13cm?

paper as shown in Fig. 4.2 (i).
A

(ii)

(iii)

Fig. 4.2

Cut a triangle ABC in a paper. Fold the Vertex A to touch the side BC as shown
in the Fig. 4.2 (ii) Fold the vertices B and C to get a rectangle as shown in the Fig. 4.2
(iii) Now you seethat 41, 42 and 43 make a straight line.

Chapter 4
From this you observe that
41+42
4A+4B+4

+43
C

=

180°

=

180°

The sum of the three angles of a triangle is 180°
Activity 2
Draw a triangle. Cut on the three angles.
Re arrange them as shown in Fig. 4.2 (ii). You
observe that the three angles now constitute
one angle. This angle is a straight angle and so
has measure

180°

The sum of the three angles of a triangle is 180°
Think

it.

1. Can you have a triangle with the three angles less than 60°?
2. Can you have a triangle with two right angles?

4.4 Exterior angle of a triangle and its property

Activity
3
A

A

(D

Fig.4.4

Draw a triangle ABC and produce one of its sides, say BC as shown in
Fig. 4.4 (i) observe the angles ACD formed at the point C. This angle lies in the
exterior

of AABC

formed

at vertex

C.

LBCA is an adjacent angle to AACD. The remaining two angles of the
triangle namely AA nd 4 B are called the two interior opposite angles.
Now cut out (or make trace copies of) AA and LB and place them adjacentto
each other as shown in Fig. 4.4 (ii)
You observe that these two pieces together entirely cover 4 ACD.

Geometry
From this we conclude that the exterior angle of a triangle is equal to the sum 0
the two interior opposite angles.
The relation between an exterior angle and its two interior angles is referred to
as the exterior angle property of a triangle.

Drawamang1é°°Ai§c°°;1d
ax:

Fig.4.4(i)g4ACD
formed
at thepointC. Nowtakea protractor

I?Try
these
and
measure
ALACDQAA
and
AB.

8 F 1(1
VS

Find
thesum
AA+A B.and
compare
it withthemeasure
of4

?4ACD. Doyouobserve
thatAACD= 4iA.+B? . 1(1in

L

In the given gure nd the value of x.

Solution
4CAB
4&#39;
4ABC
+4BCA
=180° (Since
sum
ofthe
three
40° + X + X = 180°
40° + 2x

= 180°

2x

= 180°

2x

= 140°

anglesof a triangleis 180°)
40°

x =L2Q=70°
The value

of x = 70°.

Two
angles
ofatriangle
are40°and
60°.Find
the
ed.

third angle.
Solution

Q

ARPQ + APQR +4QRP
x + 40° + 60°

180°
180°

x + 100°

180°

x

180°

(Since sum of the three angles of a
triangle is 180°)
100°

80°

The third angle x

80°

In the given gure, nd the measure of AA.

Chapter 4
3x + 120° =

180°

3x =

180°

120°

3x=60°
x=_6..Q.._
3

= 20°
&#39;.
LA:

2x=2x20°=40°

In the given gure. Find the value of x.
Solution

In the gure exterior angle = AABD = 110°.
Sum of the two interior opposite angle = ABCA + 4 CAB
= x + 50°

x + 50° =
x =
=
The value

110°

(Since the sum of the two interior opposite

110° 50°

angle is equal to the exterior angle)

60°

of x is 60°.

In the given gure nd the values of x and y.
Solution

In the give gure,
Exterior angle
50° + x
x

4 DCA = 130°
130°

(Since sum of the two interior opposite

130° 50° angle is equal to the exterior angle)
80°

InAABC,
AA
+AB
+AC
=180°
(Since
sum
ofthree
angles
ofatriangle
is180°)
50° + x + y =

180°

Aliter:
Geometry

AACB
+4DCA
=

180° (Since sum of the adjacentangles on a line is 180°)

y + 130°
y

180°
180° 130°
50°

In A ABC,
AA

+ AB

+ 4 C

50° + x + y

180° (Since sum of the three anglesof a triangle is 180°)
180°

50° + x + 50°

180°

100° + x

180°

x

180°

100°

80°

Three angles of a triangle are 3x + 5°, x + 20°, x + 25°. Find the measureof each

angle.
Solution
Sum of the three angles of a triangle
3x+5°+x+20°

+x+25°
5x + 50°
5x
5x
)6

180°

Exercise
Choose

the correct

4.1

answer.

The sum of the three anglesof a triangle is
(A) 90°

(B) 180°

(C) 270°

(D) 360°

In a triangle, all the three anglesare equal, then the measureof eachangle is

(A) 30°

(B) 45°

(C) 60°

(D) 90°

Which of the following can be anglesof a triangle?

(A) 50°,30°,105° (B) 36°,44°,90°

(C) 70°,30°,80°

(D) 45°,45°80°

Two anglesof a triangle are 40° and 60°, then the third angle is

(A) 20°

(B) 40°

(C) 60°

(D) 80°

In AABC, BC is producedto D andAABC = 500,

LACD = 1050,thenA BACwill beequalto
(A) 75°

(B) 15°

(C) 40°

(D) 55°

Statewhich of the following are triangles.

(1)4A=25°
4B=350
(ii) 41>=90°
4Q=30°
(iii) AX = 40° LY: 70°

4c=120°
4 R=50°
4 z = 80°

Two anglesof a triangle is given, nd the third angle.

(1)75°,45° (ii) 80°,30° (iii) 40°,90° (iv) 45°,85°
Find the Valueof the unknown x in the following diagrams:

4

Geometry

5. Find the Valuesof the unknown x andy in the following diagrams:

(Vi)

6. Three anglesof a triangle arex + 5°, x + 10° and x + 15° nd x.

Chapter 4

Points

to Remember

1. The sum of the three angles of a triangle is 180°.

2. In a triangle an exterior angle is equal to the sum of the two interior opposite
angles.

Practical Geometry

PRACTICAL
5.1 Construction

GEOMETRY

of triangles

In the previous class, we have learnt the various types of triangles on the
basis of their sides and angles. Now let us recall the different types of triangles
and someproperties of triangle.
Classication
No.

of triangles
Name of Triangle

Equilateral triangle

Three sides are equal
B

Isosceles
triangle

C

/I3\

AnytwoSides
areequal

QL_.AR

Scalene triangle

Sides are unequal

X
(Q Z
D

Any one of the angles is

Acute
angled
triangle acute
(less
than
900)

Obtuse angled triangle
.

obtuse
than 90°)
All
the (more
three angles
are

E

Right
angled
triangle

L

F

Any one of the angles is

right
angle
(900)

ome properties of triangle
1. The sum of the lengths of any two sidesof a triangle is greater
than the third

2.

side.

The sum of all the three angles of a triangle is 180°.

To construct a triangle we need three measurementsin which at least the
length of one side must be given. Let us construct the following types of triangles
with the given measurements.
(i) Three sides (SSS).

(ii) Two sides and included angle between them (SAS).
(iii) Two angles and included side between them (ASA).
(i) To construct a triangle when three sides are given (SSS Criterion)

Construct a triangle ABC given that AB = 4cm, BC = 6 cm and AC = 5 cm.
Solution
Given
AB

RoughDiagram

measurements
= 4cm

BC = 6 cm
AC = 5 cm.

B

Stepsfor construction
Step 1 : Draw a line segmentBC = 6cm
Step 2 : With B ascentre,draw an arc of radius4 cm abovethe line BC.
Step 3 : With C ascentre,draw an arc of 5 cm to intersectthe previousare at A
Step 4 : JoinAB andAC.
Now ABC is the required triangle.

Practical Geometry

Byusing
protector
measure
alltheangles
ofatriangle.
Find;
thesum
ofallthethree
angles
ofatriangle.
V
Q:DOyou)
kiwi?L. L L L
L L

L

1. A student attempted to draw a triangle with given
measurements PQ = 2cm, QR = 6cm, PR = 3 cm. (as in

RoughFigure

the rough gure). First he drew QR = 6cm. Then he drew
an arc of 2cm with Q as centre and he drew another arc
of radius 3 cm with R as centre.They could not intersect
eachto get P.
(i) What is the reason?
(ii) What is the triangle property in connection with this?

pg

atriangle.
Check
whether
allofthempass

a

through
asame
point.
This
point
isincentre.

(ii) To construct a triangle when Two sides and an angle included between
them are given. (SAS Criterion)

Construct a triangle PQR given that PQ = 4 cm, QR = 6.5 cm and APQR = 60°.
Solution

_

G1V6H
measurements

QR = 6.5cm
=

60°

Rough Diagram

Q

6.5 cm

6.5 cm

Stepsfor construction
Step 1 : Draw the line segmentQR = 6.5 cm.
Step 2 : At Q, draw a line QX making an angle of 60° with QR.
Step 3 : With Q as centre, draw an arc of radius 4 cm to cut the line (QX)
at P.
:

Join PR.

PQR is the required triangle.

: :-

§C0nstructa trianglewith their given measurements.-E
TTry
these
XY
=
6cm,
YZ
=
6cm
and
AXYZ:
70°.
Measure
the
angles
thetriangle oppositeto the equal sides..Whatdo you

(iii) To construct a triangle when two of its angles and a side
included between them are given. (ASA criterion)
Construct a triangle XYZ given that XY = 6 cm, AZXY
AXYZ = 100°. Examine whether the third angle
measures

= 30° and

50°.

Solution
RoughFigure Z

Given
measurements
XY
AZXY

= 6 cm
= 30°

= 1000

X

6 cm

Practical Geometry

6cm
: Draw the line segmentXY = 6cm.
: At X, draw a ray XP making an angle of 30° with XY.
: At Y, draw another ray YQ making an angle of 100° with XY. The
rays XP and YQ intersect at Z.
: The third angle measures50° i.e AZ = 50°.

AA

Try these

A

AQ
: 70°,
AR=40°.
&#39;

t

Him.-is
Use
theAngle
Sum
Property
ofatriangle.
t

Constructthe triangles for the following given measurements.
Construct APQR, given that PQ = 6cm, QR = 7cm, PR = 5cm.
Construct an equilateral triangle with the side 7cm. Using protector measureeach
angle of the triangle. Are they equal?
Draw a triangle DEF suchthat DE = 4.5cm, EF = 5.5cm and DF = 4.5cm. Can
you indentify the type of the triangle?Write the nameof it.
Constructthe triangles for the following given measurements.
Construct AXYZ, given that YZ = 7cm, ZX = 5cm, AZ = 50°.
Construct APQR when PQ = 6cm, PR = 9cm and AP = 100°.

Construct AABC given that AB = 6 cm, BC = 8 cm and AB = 90° measure
length of AC.
Constructthe triangles for the following given measurements.
Construct AXYZ,

when AX = 50°, AY = 70° and XY = 5cm.

Construct AABC

when AA = 120°, AB = 30° and AB = 7cm.

Construct ALMN, given that AL = 40°, AM = 40° and LM = 6cm. Measureand
write the length of sidesoppositeto the AL and AM. Are they equal?What type
of Triangle is this?

DATA

HANDLING

6.1 Mean, Median and Mode of ungrouped data
Arithmetic

mean

We use the word averagein our day to day life.
Poovini spendson an averageof about 5 hours daily for her studies.
In the month of May, the averagetemperature at Chennai is 40 degree celsius.
What

do the above

statements

tell us?

Poovini usually studies for 5 hours. On some days, she may study for less
number of hours and on other days she may study longer.
The averagetemperature of 40 degree celsius means that, the temperature in
the month of May at chennai is 40 degree celsius. Some times it may be less than 40
degree celsius and at other times it may be more than 40 degree celsius.
Average lies between the highest and the lowest value of the given data.
Rohit gets the following marks in different subjects in an examination.
62, 84, 92, 98, 74

In order to get the averagemarks scored by him in the examination, we rst
add up all the marks obtained by him in different subjects.
62+84+92+98+74=410.

and then divide the sum by the total number of subjects.(i.e. 5)

Theaverage
marks
scored
byRohit= 5--1Q
= 82.
5
This number helps us to understand the general level of his academic
achievement

and is referred

to as mean.

The average or arithmetic mean or mean is dened as follows.

86

Data Handling

Gayathri studies for 4 hours, 5 hours and 3 hours respectively on 3 consecutive
days. How many hours did she study daily on an average?
Solution:

Average study time =

Total number of study hours
Number of days for which she studied.

= _4__j:__5__&#39;_+___..3_
hours

Q

3

3
= 4 hours per day.
Thus we can say that Gayathri studies for 4 hours daily on an average.

The monthly income of 6 families are ? 3500,?2700,?3000, ? 2800, ? 3900 and
? 2100. Find the mean income.
Solution:

Average monthly income

Total

income

Number

of 6 familes
of families

? 3500 + 2700 + 3000 + 2800 + 3900 + 2100
6
18000

? 6

? 3,000.

The mean price of 5 pens is ?75. What is the total cost of 5 pens?

Soluti(m.&#39;
Mean

Total cost of 5 pens

Total
cost
of5pens
Number of pens

Mean ><Number of pens
?75><5
? 375

Median

Now, Mr. Gowtham arrangedthe studentsaccording to their height in ascending

order.
106,
110,
110,
112,
115,
115,115,
120,
120,
123,
125
The middle

value in the data is 115 because

this value divides

the students

into

two equal groups of 5 students each. This value is called as median. Median refers
to the Value 115which lies in the middle of the data.Mr. Gowtham decides to keep
the middle student as a referee in the game.
Median is dened as the middle value of the data when the data is arranged
in ascending or descending order.

;&#39;

Find
themedian
ofthefollowing:
40503060,80,70

y mesei
Find
theb actual
&#39;
.
distance
,, etween
your school and 1,

Arrangethegivendatain ascending
order.

house,, Find 33the

30,40,50,60»
70»
80Here the number

of terms

is 6 which

is even.

So the third

and fourth

terms

are

middle terms. The averagevalue of the these terms is the median.

(i.e)Median
= 50360=% - 55.
(i) When the number of observations

is odd, the middle number is the

median.

(ii) When the number of observations is even, the median is the average of

the two middlenumbers.
Find the median of the following data.
3, 4, 5, 3, 6, 7, 2.
Solution:

Arrange
thedatain ascending
order.

2,3,3,4,5,6,7
Thenumber
ofobservation
is7which
isodd.
.&#39;.The
middle

Find

Value 4 is the median.

the median

of the data

12, 14, 25, 23, 18, 17, 24, 20.

In highways,
the yellow

line
represents
the
median.

Data Handling
12, 14, 17, 18, 20, 23, 24, 25.
The number

of observation

is 8 which

is even.

Median is the averageof the two middle terms 18 and 20.
- _18+20_38__
Med1an
2
2 -19

Find the median of the rst 5 prime numbers.

Solution:
The
rstveprime
numbers
are
2,3,5,7,11.
The number
The middle

of observation
value

is 5 which

is odd.

5 is the median.

Mode
Look
atthe
following
example,
Mr. Raghavan,the owner of a ready made dress shop saysthat the most popular
size of shirts

he sells is of size 40 cm.

Observe that here also, the owner is concerned about the number of shirts of

different sizes sold. He is looking at the shirt size that is sold, the most. The highest
occurring event is the sale of size 40 cm. This value is called the mode of the data.
Mode is the variable which occurs most frequentiy in the given data.

Mode of Large data
Putting the same observation together and counting them is not easy if the
number of observation is large. In such caseswe tabulate the data.

Following are the margin of victory in the foot ball matchesof a league.
1, 3, 2, 5, 1, 4, 6, 2, 5, 2, 2, 2, 4, 1, 2, 3, 2, 3, 2, 3,
1, 1, 2, 3, 2, 6, 4, 3, 2, 1, 1, 4, 2, 1, 5, 3, 4, 2, 1, 2. Find the mode of this data.
Solution.

Now we quickly say that 2 is the mode. Since 2 has occurred the more number
of times, then the most of the matcheshave been won with a Victory margin of 2 goals.

Find the mode of the following data.
3, 4, 5, 3, 6, 7

Solution:

the
transport
inyour

place. 11

L

3 occurs the most number of times.
Mode

of the data is 3.

Find the mode of the following data.
2, 2, 2, 3, 3, 4, 5,5, 5, 6,6, 8

Solution:
2and
5occur
3times.
Mode

of the data is 2 and 5.

Find
themode
ofthefollowing
data
90,40,68,94,50,60.

Findthemode
V
oftheower.

Here there are no frequently occurring Values. Hence this data has no mode.

The number of children in 20 families are 1, 2, 2, 1, 2, 1, 3, 1, 1, 3
1, 3, 1, 1, 1, 2, 1, 1, 2, 1. Find the mode.
Solution:

Table

6.2

12 families have 1 child only, so the mode of the data is 1.

Data Handling
Exercise:
Choose

the correct

The arithmetic

(A) 5

6.!

answer:

mean of 1, 3, 5, 7 and 9 is

(B) 7

(C) 3

(D) 9

The averagemarks of 5 children is 40 then their total mark is
(A) 20

(B) 200

(C) 8

(D) 4

(C) 30

(D) 10

(C) 7

(D) 14

(C) 7

(D) 2

The median of 30,50, 40, 10, 20 is

(A) 40

(B) 20

The median of 2, 4, 6, 8, 10, 12 is

(A) 6

(B) 8

The mode of 3, 4, 7, 4, 3, 2, 4 is
(A) 3

(B) 4

The marks

in mathematics

of 10 students

are

56, 48, 58, 60, 54, 76, 84, 92, 82, 98.

Find the range and arithmetic mean
The weights of 5 people are
72 kg, 48 kg, 51 kg, 69 kg, 67 kg.
Find the meanof their weights.
Two vesselscontain 30 litres and 50 litres of milk separately.What is the
capacity of the vesselsif both sharethe milk equally?
The maximum temperaturein a city on 7 days of a certain week was 34.8°C,
38.5°C, 33.4°C, 34.7°C, 35.8°C, 32.8°C, 34.3°C. Find the meantemperaturefor
the week.

The meanweight of 10 boys in a cricket team is 65.5 kg. What is the total weight
of 10 boys.
Find the median of the following data.
6,14,5,

13,11, 7, 8

The weight of 7 chocolatebars in grams are
131, 132, 125, 127, 130, 129, 133. Find the median.

The runs scoredby a batsmanin 5 innings are
60, 100, 78, 54, 49. Find the median.
Find the median

of the rst

seven natural

numbers.

Pocketmoney receivedby 7 studentsis given below.
? 42, ? 22, ? 40, ? 28, ? 23, ? 26, ? 43. Find the median.

Find the mode of the given data.
3, 4, 3, 5, 3, 6, 3, 8, 4.

13. Twelve eggscollected in a farm have the following weights.
32 gm,40 gm, 27 gm, 32 gm, 38 gm, 45 gm,
40 gm, 32 gm, 39 gm, 40 gm, 30 gm, 31 gm,
Find the mode of the above data.

14. Find the mode of the following data.
4, 6, 8, 10, 12, 14

15. Find the mode of the following data.
12, 14, 12, 16, 15, 13, 14, 18, 19, 12, 14, 15, 16, 15, 16, 16,

15, 17, 13, 16, 16, 15, 13, 15, 17, 15, 14, 15, 13, 15, 14.

Points to Remembr

F

1. Average lies between the highest and the lowest value of the given data.
2.

Mean

=

sum of all the observations
total

number

of observations

3. Median is dened as the middle value of the data, when the data is arranged
in ascending or descending order.
4. Mode is the variable which occurs most frequently in the given data.

1. (i) B

(ii)

2. (i) x + 2y

(ii) y Z

A

(iii)

D

(iii)

xy + 4

(iv)

C

(V) A

(iv) 3x 4y (if 3x > 4y) or 4y 3x (if 4y > 3x)

(vi)
194
-5

(V) 10 +x + y

(vii)

12

mn

(X)

4xy
(viii)ab (a+ b)

(ix) 3cd+ 6

T

(iii)

Answers

(iii)
(viii)
(iii)

(iii)

x=15
(ii)

y=67

(iii)

z=5

(viii)

x=6

(iv)
(ix)

y=3

(xii)m=%
11.37, 38

(xiii)

7.

13

8.

12.

60

13.

x=3 (xiv)x =
108
35

9.

12

(X)
3 (xv)
10.

8

Answers
.(i)

100%

. (i) 36%

(ii)

18%

(iii)

525%

(iv)

66.67% (V) 45.45%

(ii)

3%

(iii)

7.1%

(iv)

305%

(V) 75%

A

(iv)

C

(V) B

. 20%
. 13.89%

. Girls 46%; Boys 54 %
. He got more marks in Science.
. Savings 10%; Expenditure 90%

. (i) B

(ii)

B

.0) 3%

(in g

(iii) Z53 (iv)735 (V)§

. (i) 0.07

(ii)

(iii)

. (i) 18

(ii) ?24

0.64

(iii)

6. 9 matches

. ? 38250

9. 3975 illiterates

(ii)

C

(iv) A

(V) B

(iv)

(iii) 36m

. ? 6250

. (i) A

3.75

0.0003 (V) 0.005

(iv) 108

(V) 3.75kg

7. 12,800 men; 11,200 children

(iii)

C

2. Prot = ? 24, Loss 2 ? 21;
Prot = ? 35.45, Loss = ? 3362, Loss = 3 7.50
3. (i) ? 530

(ii) ? 620

(iv) ? 336.75
4. Prot

10%

? 1027.50

(V) ? 943.50
5. Loss 12% 6.

1. (i) B
(iv) C

2. ?2,500;?7,500
5. ?2,630

(iii)

Prot

(ii)

60%

7. Prot

A

15%

(iii)

A

(V) A

3. ?450;?&#39;
1,650
6. ?216;?12,216

7. 5%

4. ?2,250
8. ?1,000

Answers

9. 2 years
12. ? 5,400
Unit

10. 10%

11. 8 years

13. 3 5,000; 10%

14. S.I. = 2025; ? 5,625 15. ? 4,000

~3

1. (i) B

(ii)

2. (i) 50 cm2

(ii) 66 cm2 (iii)

3. 12cm

4.

1. (i) B

(ii)

2. (i) d:

A

(iii)

875 m2

C

5.

(iii)

A

(iv) D

(V) D

70 cm, C = 220 cm

(ii) r = 28 cm, c =2176 cm
(iii) r = 4.9 cm, d = 9.8 cm
3. (i) 110cm

(ii)

264cm

(iii)

374 cm(iv)

462cm

4. (i) 79.2 cm

(ii)

396cm

(iii)

8.8 m

(iv)

26.4m

5. (i) 17.5 cm

(ii)

21 cm

(iii)

0.7 m

(iv)

1.75 m

6. 660 m

7. ?&#39;
1232

1. (1) 38.5 C1112

8. 1.05 m

9. 37

10. ? 4,752

(11) 86.625 CII12

(111)18.865 1112

(iv)

124.74 1112

2. (1) 4.525 C1112

(11) 616 C1112

(111)55.44 1112

(iv) 98.56 1112

. ? 1848

4. 154m2

5. circle haslargerarea

. 38.5 m2

7. 1936 cm2

8. r = 35, ? 2200

10.

63.36

Second

2. 31152

5. 40 m2, 30 m2

11.

10

Answers

1. (i) B
2. (i) AA:

2. 536.94 m2

3. ? 24,050

4. 21.195 m2

6. 2310 cm2

7. 6 m

8. 2002 m2

(ii)

C

(iii)

C

(iv) D

(V) D

25°, AB =: 35°, AC 2 120°

3. (i) 60°

(ii)

70°

(iii)

50°

(iv)

50°

4. (i) 70°

(ii)

60°

(iii)

40°

(iv)

30°

(Vi)

60°, 60°, 60°

(V) 65°, 65°

5. (i) y = 60°, x = 70°

(ii) y = 80°, x = 50°

(iii) y = 70°, x = 110°

(iv) x = 60°, y = 90°

(V) y = 90°, x = 45°

(Vi) x = 60°, y = 50°

6. x = 50°.

Unit

- 6

1. (i) A

(ii)

B

(iii)

2. Range is 50; A.M. 270.8
3. 61.4 kg.

4. 40 litres

5. 34.9°C

6. 655.0kg

7. 8

8. 130gram 9.60

12. 3

13. 32 gm and40 gm

14. no mode

10.4
15.15

11128

I can, I did

Student&#39;s
Activity Record
Subject :
Date Lesson
No.

Topic of the

Activities

Lesson

97