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Chapter-30
Mensuration I
(Area and Perimeter)
Mensuration
Mensuration is the science of measurement of the
length of lines, areas of surfaces and volumes of
solids. Its knowledge is of immense use to the
surveyor, architect and engineer. In this chapter,
we will deal with the perimeter and area of the
plane figures.
Perimeter
The perimeter of a figure is the sum of length of
all its sides. This is measured in the units of
length. For example, cm, m, etc.
Area
The area of any figure is the amount of surface
enclosed within its bounding lines. This is
measured by the number of square cm or square
metres (or other units of square measures) it
contains.
Measures of Area
Area, being the product of two linear measurements
(see in the following pages), is expressed in square
units of linear measure. The following will give
you the measures of area in the Metric System:
(i) Square Centimetre: A square centimetre is
the amount of surface enclosed within a square of
which each side is one centimetre in length.
The term square metre, square hectometre,
square kilometre etc are used in the same sense.
(ii) Square Metre: The area of a region formed
by a square of side 1 metre is called a square metre
and is written as 1 m2. The basic unit of length in SI system is a metre
(1 m) and the basic unit of area in SI system is a
square metre (1 m2).
Since,
1 m = 100 cm
1 m2 = 1 m × 1 m = 100 cm × 100 cm
= 10000 cm2
Thus, 1 m2 = 10000 cm2.
(iii) Square Decimetre: The area of a region
formed by a square of side 1 decimetre (1 dm) is
called a square decimetre and is written as 1 dm2.
Since,
1 m = 10 dm
1 m2
= 1 m × 1 m = 10 dm × 10 dm = 100 dm2
Also, 1 dm = 10 cm
1 dm2 = 1 dm × 1 dm = 10 cm × 10 cm = 100 cm2
(iv) Square Millimetre: The area of a region
formed by a square of side 1 millimetre (1 mm) is
called a square millimetre and is written as 1 mm2.
Since 1 cm = 10 mm
1 cm2 = 1 cm×1 cm = 10 mm × 10 mm
= 100 mm2
Also, 1 m2 = 10000 cm2
1 m2 = 10000 mm × 100 mm = 1000000 mm2
(v) Square Decametre or an Are: The area of a
region formed by a square of side one decametre (1
dam) is called a square decametre or an arc is
denoted by 1 dam2 or 1a .
Since 1 dam = 10 m
1 dam2
= 1 dam × 1 dam = 10 m × 10 m = 100 m2
Thus, 1 dam2 = 100 m2 or 1 are = 100 m2
(vi) Square Hectometre or Hectare: The area of
a region formed by a square of side 1 hectometre (1
hm) is called a square hectometre or a hectare and
is denoted by 1 hm2 or 1 ha.
Since, 1 hm = 100 m
1 ha = 1 hm × 1 hm = 100 m × 100 m
= 10000 m2
Also, 1 hm = 10 dam
1 ha = 10 dam × 10 dam
= 100 dam2 = 100 a.
Thus, 1 ha = 10000 m2 or 1 ha = 100 a.
(vii) Square Kilometre: The area of a region
formed by a square of side 1 kilometre (1 km) is
called a square kilometre and is written as 1 km2
.Since, 1 km = 1000 m
1 km2 = 1000 m × 1000 m = 1000000 m2
Since 1 ha = 10000 m2. Therefore,
1 km2 =10000
1000000 ha = 100 ha
Also, 1 km2 = 10000 are
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502 Concep t o f A r i t hm e t i c
K KUNDAN
Ex. 1: F i n d t h e a r e a , i n h e ct a r e, o f a f i el d
w h o s e l e n g t h i s 2 4 0 m a n d b r e a d t h
1 1 0 m .Soln: We have, Length of the field = 240 m
Breadth of the field = 110 m.
Area of the field = (240 × 110) m2
= 26400 m2 =10000
26400 hectare
[ 10000 m2 = 1 hectare]
= 2.64 hectare
Ex. 2: W h a t w i l l h a p p e n t o t h e a r ea o f a
r e c t a n g l e i f
(i ) i t s l en g t h i s d ou b l ed a n d br ea d t h i s t r e b l ed .
(i i ) i t s l en g t h a n d b r ea d t h a r e
d o u b l e d ?
S o l n : Let l cm and b cm be the length and
breadth of the rectangle. Further, let A be
the area of the rectangle. Then,
A = l × b ...... (i)
Solved Examples
We have introduced various standard units of
area. Each can be converted into others as given
below.
Conversion of Units
Units of Length Units of Area
1 cm = 10 mm 1 cm2 = (10 × 10) mm2
= 100 mm2
1 m = 10 dm 1 m2 = (10 × 10) dm2
= 100 dm2
1 dm = 10 cm 1 dm2 = (10 × 10) cm2
= 100 cm2
1 m = 100 cm 1 m2 = (100 × 100) cm2
= 10000 cm2
1 dam = 10 m 1 dam2 = (10 × 10) m2
= 100 m2 = 1 are
1 hm = 100 m 1 hm2 = (100 × 100) m2
= 10000 m2
= 1 hectare
1 km = 10 hm 1 mm2 = (10 × 10) hm2
= 100 hm2
1 km = 1000 m 1 km2 = (1000 × 1000) m2
= 1000000 m2
Rectangle and Square
(i) Rectangle
A rectangle is a four-sided figure having all its
angles right angles. The page of this book,the faces
of a brick, the floor of a room are rectangles, for
their opposite sides are equal and their angles are
right angles.
The sides of a rectangle are usually called its
length and breadth. In the rectangle, the length
and breadth are uneqaual.
( i ) Perimeter of a rectangle = 2 (Length +Breadth)
( ii) Area of a rectangle = (Length × Breadth)
From this formula we get,
(a) Length =Breadth
Area
(b) Breadth = Length
Area
( i i i ) Diagonal of a rectangle
= 22 (Breadth)(Length)
ABC is a right-angled triangle.
AC = 22 BCAB
(ii) Square
A square is a four-sided figure having all its angles
right angles and all its sides are equal. In a square,
the length is equal to the breadth.
( i ) Perimeter of a square = 4 × length
= 4 × one of its sides
( ii) Area of a square = side × side = (side)2
From this formula we get,Side = Area
( i i i ) Diagonal of a square = )Area(2
If ABCD be a square, then
AC = 22 BCAB = 22 ABAB = AB2
[ AB = BC]
AC = 2AB
From the above, we have
AB =2
AC
Area = (AB)2 =2
1(AC)2
Hence, the area of a square is half the square
of its diagonal.
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503Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
or, 28922 y x ..... (i)
And, Perimeter = 46 cm
or, 2 (x + y ) = 46 cm
or, x + y = 23 .... (ii)Now, x + y = 23
or, (x + y )2 = 232
or, x 2 + y 2 + 2xy = 529
or, 289 + 2xy = 529
or, 2xy = 529 - 289
or, 2xy = 240
or, xy = 120
Hence, Area = xy = 120 cm2.
Ex. 6: T h e l e n g t h o f a r e ct a n g l e ex c e ed s i t s
w i d t h b y 8 c m a n d t h e a r e a o f t h e
r e c t a n g l e i s 2 4 0 s q c m . F i n d t h e
d i m en s i o n s o f t h e r e ct a n g l e .
S o l n : Let the breadth of the given rectangle be x
cm. Then, length = (x + 8) cm.
Now, Area = 240 cm2
or, length × breadth = 240or, (x + 8) x = 240
or, x 2 + 8x – 240 = 0
or, x 2 + 20x – 12x – 240 = 0
or, x (x + 20) – 12 (x + 20) = 0
or, (x + 20) (x – 12) = 0
or, x = 12 or x = –20
But x cannot be negative. So, x = 12.
Hence, breadth = 12 cm and length
= (12 + 8 =) 20 cm.
Ex. 7: T h e c os t o f p l o u g h i n g a r ec t a n g u l a r
f i e l d a t 8 5 p a i s e p er s q u a r e c en t i m et r e
i s R s 6 2 4 .7 5 . F i n d t h e p er i m et e r o f t h e
f i e l d i f i t s si d e s a r e i n t h e r a t i o 5 : 3 .
S o l n : Let the length and the breadth of the
rectangular field be 5x cm and 3x cm in
length. Then,Area = 5x × 3x = 15x 2 cm2 .... (i)
It is given that the cost of ploughing the
rectangular field at the rate of Rs100
85 is
Rs 624.75.
Area of the rectangular field
= msq perRate
cost Total
=85
10075.624
100
85
75.624
= 735 cm2 .... (ii)
From (i) and (ii), we get15x 2 = 735 or x 2 = 49 or x = 7
Hence, the sides of the rectangular field
are 5x = (5 × 7) = 35 cm and 3x = (3 × 7 =)
21 cm respectively.
Perimeter = {2 (35 + 21) =} 112 cm
(i) We have,
New breadth = 2l ; New length = 3b
A1 = New Area
= 2l × 3b = 6 (l × b ) = 6A
[Using (i)]Hence, the area of the new rectangle is 6
times the area of the old rectangle.
(ii) We have,
New length = 2l , New breadth = 2b
A1 = New Area
= 2l × 2b = 4 (l × b ) = 4A
[Using (i)]
Hence, the area of the new rectangle is 4
times the area of the old rectangle.
Ex. 3: W h a t w i l l h a p p en t o t h e a r e a o f a
s q u a r e i f i t s s i d e i s ( i ) d o u b l ed ( i i )
h a l v e d ?
So l n : (i) Let the side of a square be x cm. Then,
Area = A = x 2 cm2 ....(i)
When the side is doubled, then,
side of the new square = 2x cm. A
1 = Area of the new square
= (2x 2) cm2 = 4x 2 cm2 = 4A
[Using (i)]
Thus, if the side is doubled, then area
becomes 4 times.
(ii) When the side is halved, then,
Side of the new square =2
x cm.
A2 = Area of the new square
= A4
1
4
1cm
2
222
x
x
[Using (i)]
Thus, if the side is halved, then area
becomes one-fourth.Ex. 4: T h e l e n g t h o f a r e c t a n g l e i s t w i c e i t s
b r ea d t h . F i n d t h e d i m e n s i on s o f t h e
r e ct a n g l e i f i t s a r ea i s 2 8 8 s q cm .
Soln: Let the breadth of the given rectangle be x
cm.
Then, length = 2x cm [Given].
Area = 2x × x = 2x 2
or, 2x 2 = 288
[ Area = 288 sq cm (given)]
or, x 2 = 144
or, x = 12 cm
Hence, length = 24 cm, and breadth 12
cm.
Ex. 5: I f t h e d i a g o n a l o f a r ec t a n g l e i s 1 7 cm
l o n g a n d t h e p er i m et e r o f t h e r e ct a n g l e
i s 46 cm . F i nd t he a r ea o f t h e r ec t ang l e.
Soln: Let the length and breadth of the given
rectangle be x cm and y cm respectively.
Th en ,
Diagonal = 17 cm
or, 1722 y x
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504 Concep t o f A r i t hm e t i c
K KUNDAN
Ex. 8: T h e l en g t h a n d b r e a d t h o f a
r e ct a n g u l a r f i e l d a r e i n t h e r a t i o 3 : 2 .
I f t h e a r ea o f t h e f i e l d i s 3 4 5 6 c m 2 ,
f i n d t h e c o st o f f en c i n g t h e f i el d a t R s
3 . 5 0 p e r cm .Soln: Let the length and breadth of the
rectangular field be 3x and 2x cm
respectively. Then,
Area of the rectangular field
= (3x × 2x ) cm2 = 6x 2 cm2
Also, area of the rectangular field
= 3456 cm2 (given)
6x 2 = 3456
or, x 2 =6
3456
or, x 2 = 576
or, x = 576 = 24
Length = (3 × 24) cm = 72 cm
Breadth = (2 × 24) cm = 48 cm Perimeter of the field
= 2 × (length + breadth)
= [2 × (72 + 48)] cm = 240 cm
Rate of fencing = Rs 3.50 per cm
Cost of fencing = Rs (240 × 3.50)
= Rs 840
Ex. 9: F i n d t h e a r ea o f a r e ct a n g u l a r p l o t o n e
s i d e o f wh i c h i s 4 8 cm and i t s d i a g o n a l
5 0 c m .
Soln: Let the other side be x cm. Since ABC is
a right triangle, therefore,
AC2 = AB2 + BC2
or, 502 = 482 + x 2
or, x 2 = (50)2 – (48)2
or, x 2 = (50 + 48) (50 – 48)
or, x 2 = 98 × 2or, x 2 = 142
or, x = 14
Thus, the other side of the rectangle is
14 cm
Area of the rectangle = (48 × 14) cm2
= 672 cm2
Ex. 10: The a r ea o f squa r e ABCD i s 16 cm 2 . F i nd
t h e a r e a o f t h e s q u a r e jo i n i n g t h e m i d -
p o i n t s o f t h e s i d e s.
Soln: We have,
Area of square ABCD = 16 cm2
Each side of square ABCD
= 16 cm = 4 cm
In APS, we have
AP = AB2
1 = 2 cm and
AS = AD2
1
= 2 cm
Also, PS2 = AP2 + AS2
[Using Pythagoras theorem]
PS = 243244 22
Thus, eac h side of square PQRS is of
length 24 cm.
Area of the square PQRS
= 224 cm2 = 32 cm2
Ex. 11: T h e l en g t h o f a r e ct a n g u l a r f i el d i s
i n c r e a se d by 5 0% a n d i t s br e a d t h i s
d e c r e a s ed b y 5 0 % t o f o r m a n e w
r e ct a n g u l a r f i e l d . F i n d t h e p er c en t a g e
c h a n g e i n t h e a r ea o f t h e f i e l d .Soln: Let x and y denote the length and breadth
of the given field. Then, its area is given
by A = xy
Increase in length
= 50% of the original length of x
=2100
50 x x
New length = x x
x 2
3
2
Decrease in breadth
= 50% of the original breadth
= 50% of y =2100
50 y y
New length =22
y y y
A1 = Area of the new field
= xy y
x 4
3
22
3
Change in Area
= xy xy xy 4
1
4
3AA1
Percentage change in the area
= 1004
1
100A
AA1
xy
xy = 25%
Hence, there is 25% decrease in the area.
Ex. 12: A r e ct a n g u l a r c ou r t y a r d i s 3 m 7 8 cm
l o n g a n d 5 m 2 5 cm b r o a d . I t i s d e s i r ed
t o p a v e i t w i t h s q u a r e t i l e s of t h e s ame
s i z e. W h a t i s t h e l a r g es t s i z e o f t h e
t i l e t h a t c a n b e u s ed ? A l s o , f i n d t h e
n um ber o f s u c h t i l e s .
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505Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
Soln: Clearly, the size of the tile should be a
factor of both the length and breadth of
the courtyard. Therefore, the size of the
largest size tile should be the HCF of 378
cm and 525 cm.
7
213
633
1893
3782
7
213
1053
5255
Now,
378 = 2 × 33 × 7 and 525 = 5 × 32 × 7
Hence, HCF of 378 and 525 is 3 × 7 = 21
Thus, the size of the largest size square
tile = 21 cm
Area of a tile
= (side)2 cm2 = (21)2 cm2 = 441 cm2
Area of courtyard = (378 × 525) cm2
= 198450 cm
2
Now,
Number of tiles =tileaof Area
courtyardof Area
=441
198450 = 450
Ex. 13: The f l o or o f a r ec t a n g u l a r h a l l i s t o b e
c ov er e d w i t h a c a r p et 1 5 0 cm w i d e. I f
t h e l en g t h a n d b r ea d t h o f t h e h a l l a r e
2 0 m a n d 1 8 m r e sp ec t i v el y , f i n d t h e
c os t o f t h e ca r p et a t t h e r a t e of R s 1 0
p e r me t r e .
Soln: We have,
Area of the hall = (20 × 18) m2 = 360 m2
Width of the carpet = 150 cm = 1.5 m
Length of the carpet
= 5.1
360
carpettheof Width
halltheof Area = 240 m
Rate of the carpet = Rs 10 per metre.
Cost of the carpet = Rs (240 × 10)
= Rs 2400.
Ex. 14: F i n d t h e h e i g h t o f t h e wa l l w h o se l en g t h
i s 4 m a n d w h i c h c a n b e co v er e d b y
2 4 0 0 t i l es o f si z e 2 5 cm b y 2 0 cm .
Soln: Area of a tile = (25 × 20) cm2 = 500 cm2.
Area of 2400 tiles = (2400 × 500) cm2
= 1200000 cm2
=10000
1200000 m2
]m1cm10000[ 22
= 120 m2.
Let the height of the wall be h metres.
Th en ,
area of the wall = 4h m2.
Since 2400 tiles completely cover the wall.
Therefore,
Area of the wall = Area of 2400 tiles
or, 4h = 120
or,4
120
4
4
h
[Dividing both sides by 4]
or, h = 30
Hence, the height of the wall is 30 metres.
Ex. 15: F i n d t h e p er i m et e r o f a r e ct a n g u l a r f i el d
w h o s e l e n gt h i s f o u r t i m e s i t s w i d t h
a n d wh i c h h a s a n a r e a eq u a l t o 3 0 9 7 6
cm 2 .
Soln: Let the width of the field be b cm. Then,
Length of the field = 4b cm.
Area of the field = (b × 4b ) cm2 = 4b 2 cm2
But, area of the field is given as 30976
cm2.
4b 2 = 30976
or, 77444
309762 b
or, b 2 = (88)2or, b = 88
Length of the field = 4b cm
= (4 × 88) cm = 352 cm
Width of the field = b cm = 88 cm
Hence,
Perimeter of the field
= 2 (length + breadth)
= 2 (352 + 88) cm = 880 cm
Ex. 16: A 5 m w i d e l a n e wa s p a v ed w i t h b r i c k s
o f s i z e 2 0 cm b y 1 5 cm . I f t h e r a t e o f
b r i c k s wa s R s 7 5 0 p e r t h o u s a n d a n d i f
b r i c k s w o r t h R s 4 9 5 0 0 w e r e u s ed f o r
p a v emen t s , f i n d t h e l e n gt h o f t h e l a n e.
Soln: We have,
Rate of bricks = Rs 750 per thousand
Total cost of bricks = Rs 49500
Number of bricks =750
49500 thousand
= 66 thousand
= 66000
Area of one brick = (20 × 15) cm2
= 300 cm2
Area covered by 66000 bricks
= 66000 × 300 cm2
= 19800000 cm2
=10000
19800000 m2 = 1980 m2
22 m1cm 10000
Hence, area of the lane = 1980 m2
.It is given that the width of the lane is 5 m.
Length of the lane =
width
Aream
=5
1980 m = 396 m
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506 Concep t o f A r i t hm e t i c
K KUNDAN
Ex. 17: The l en g t h a n d b r e a d h o f a p l a y g r o u n d
a r e 7 5 m 2 0 c m a n d 3 4 m 8 0 c m ,
r e s p ec t i v e l y . F i n d t h e c os t o f l e v el l i n g
i t a t R s 1 . 5 0 p e r s q u a r e m e t r e . H o w
l o n g w i l l a b oy t a k e t o g o t h r e e t i m es r o u n d t h e f i el d , i f h e w a l k s a t t h e r a t e
o f 1 . 5 m / s ec .
Soln: We have,
Length of the playground
= 75 m 20 cm = 75.20 m
Breadth of the playground
= 34 m 80 m = 34.80
Area of the playground
= 75.20 × 34.80 m2 = 2616.96 m2
Cost of levelling = Rs 2616.96 × 1.50
= Rs 3925.44
Perimeter of the playground
= 2(length + breadth)
= 2(75.20 + 34.80) m
= (2 × 110) m = 220 m.
Total distance to be covered by the boy = 3 (Perimeter of the playground)
= 3 × 220 m = 660 m
Speed of the boy = 1.5 m/sec.
Time taken by the boy =5.1
660 sec
Speed
Distance Time
= 440 sec =60
440 minutes
=3
22 minutes
= 7 minutes 20 secondsEx. 18: The c a r p e t f o r a r o om 6 . 6 m b y 5 . 6 m
cos t s Rs 3 9 6 0 a n d i t w a s made f r om a
r o l l 7 0 c m w i d e . Fi n d t h e c os t o f t h e
c a r p e t p e r me t r e .
Soln: We have,
Area of the carpet = 6.6 × 5.6 m2
= 36.96 m2.
Width of the roll = 70 cm = 0.7 m
Length of the roll =Width
Area =
7.0
96.36
= 52.8 m
Cost of the carpet = Rs 3960.
Cost of the carpet per metre
= Rs 8.52
3960 = Rs 75.
Hence, the carpet costs Rs 75 per metre.
Ex. 19: T h e a r ea o f a r ec t a n g u l a r f i e l d i s
c a l c u l a t e d t o b e 2 0 0 m 2 wh en i t s s i d es
a r e m ea s u r ed w i t h a f a u l t y m et r e r o d .
I f t h a t me t r e r o d i s a ct u a l l y 0 . 9 0 me t r e
l o n g , f i n d t h e co r r e c t a r e a o f t h e f i el d .
Soln: Let the actual length and breadth of the
rectangular field be l and b metres
respectively.
The faulty metre-rod measures 0.90 metre
as 1 metre.
It will measure 1 metre as90.0
1 metre
It will measure l metres as l 90.0
1
metres.
Thus, according to the faulty metre-rod
the length of the field is90.0
l metres.
Similarly, breadth of the field measured
by the faulty rod =90.0
b metres.
Area of the field =81.090.090.0
lb b l
But, the faulty metre-rod measures the area
of the field as 200 m2.
20081.0
lb
or, lb = 200 × 0.81 m2
or, lb = 162 m2
Hence, the correct area of the field is 162
m2.
Areas of Paths
Ex. 20: A r e ct a n g u l a r g r a s sy l a w n m ea s u r i n g
3 0 m b y 2 8 m i s t o be su r r o u n d ed
ex t er n a l l y b y a p a t h w h i c h i s 2 m w i d e .
F i n d t h e co s t o f l ev el l i n g t h e p a t h a t
t h e r a t e o f R s 5 p er s q u a r e me t r e .
Soln: Let ABCD be the grassy lawn, and let
PQRS be the external boundaries of the
path.
We have,
length AB = 30 m, breadth BC = 28 m Area of lawn ABCD = 30 × 28 m2
= 840 m2
Length PQ = (30 m + 2 m + 2 m)
= 34 m
Breadth QR = 28 m + 2 m + 2 m
= 32 m
Area PQRS = 34 × 32 m2 = 1088 m2
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507Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
Now, Area of the path
= Area PQRS - Area of the lawn
= (1088 - 840 =) 248 m2
Cost of levelling the path
= Rs (248 × 5) = Rs 1240Ex. 21: A g r assy p l o t i s 80 m × 60 m . Two c r oss
p a t h s ea c h 4 m w i d e a r e c o n st r u c t ed
a t r i g h t a n g l e s t h r o u g h t h e c en t r e of
t h e f i e l d , s u ch t h a t e a c h p a t h i s
p a r a l l e l t o o n e o f t h e s i d e s o f t h e
r e ct a n g l e . Fi n d t h e t o t a l a r e a u s ed a s
p a t h . A l so , f i n d t h e co st o f g r a v el l i n g
t h em a t R s 5 p er s q u a r e me t r e .
Soln: Let ABCD and EFGH be the cross paths.
We have, AB = 80 m and BC = 4 m
Area of path ABCD = (80 × 4) m2
= 320 m2
Again, EF = 60 m and FG = 4 m
Area of path EFGH = (60 × 4) m2
= 240 m2
Clearly, area PQRS is common to both the
paths.
We have,
Area PQRS = (4 × 4 =) 16 m2
Total area used as path = Area of path
ABCD + Area of path EFGH - Area PQRS
= (320 + 240 - 16) = 544 m2
We have, rate of gravelling the path
= Rs 5 per square metre
Total cost of gravelling the path
= Rs (5 × 544) = Rs 2720
Ex. 22: Ca l cu l a t e t he a r ea o f t he shaded r eg i on
s h own i n t h e f i g u r e g i v en b e l ow :
S o l n : We have,
Area of the rectangle ABCD
= (60 × 48) m2 = 2880 m2
Area of a square at one of the corners
= (8 × 8) m2 = 64 m2
Area of the four squares
= 4 × 64 m2 = 256 m2
Hence, required area = Area of therectangle ABCD – Area of the four squares
= (2880 - 64 =) 2816 m2
Ex. 23: Ca l c u l a t e t h e a r e a o f t h e f i g u r e gi v e n
b e l o w .
S o l n : Complete the rectangles ABCQ and DEFR
by drawing dotted lines NQ and PR.
Now,
Area of rectangle ABCQ = (12 × 3) m2
= 36 m2
Area of rectangle DEFR = (12 × 3) m2
= 36 m2
Area of rectangle PNQR = (9 × 4) m2
= 36 m2
[ PN = PH + HM + MN = (3 + 3 + 3) m
and, QN = AQ – AN = (12 – 8) m]
Area of rectangle LMHK = (6 × 3) m2
= 18 m
2
Hence, required area
= 36 m2 + 36 m2 + 36 m2 + 18 m2
= 126 m2
Ex. 24: A t a b l e cover , 4 m × 2 m , i s sp r ead on
a m e et i n g t a b l e . I f 2 5 cm o f t h e t a b l e
c ov er i s h a n g i n g a l l a r o u n d t h e t a b l e ,
f i n d t h e c os t o f p o l i s h i n g t h e t a b l e t o p
a t R s 2 . 2 5 p e r s q u a r e me t r e .
Soln: To find the cost of polishing the table top
we have to find its area for which we
require its length and breadth.
We have,
Length of the cloth = 4 m
Breadth of the cloth = 2 m
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508 Concep t o f A r i t hm e t i c
K KUNDAN
Since 25 cm width of cloth is outside the
table on each side. Therefore,
Length of the table = (4 – 2 × 0.25) m
= 3.5 m
Breadth of the table = (2 – 2 × 0.25) m = 1.5 m
Area of the top of the table
= (3.5 × 1.5) m2
It is given that the cost of polishing the
table top is at the rate of Rs 2.25 per square
metre. Therefore, cost of polishing the top
= Area × Rate per sq metre
= Rs (3.5 × 1.5 × 2.25)
=
4
9
2
3
2
7
= Rs 11.81
Ex. 25: T h e r e i s a s q u a r e f i e l d w h o s e s i d e i s
4 4 m . A s q u a r e f l ow e r b e d i s p r e p a r ed
i n i t s c en t r e l e a v i n g a g r a v el p a t h a l l
r o u n d t h e f l ow er b ed . T h e t o t a l c os t o f l a y i n g t h e f l ow er b ed a n d g r a v e l l i n g t h e
p a t h a t R s 2 . 7 5 a n d R s 1 . 5 0 p er s qu a r e
me t r e r e spec t i v e l y , i s Rs 4904 . F i nd t h e
w i d t h o f t h e gr a v e l p a t h .
Soln: Let the width of the gravel path be x
metres. Then, each side of the square
flower bed is (44 - 2x ) metres.
Now, area of the square field
= (44 × 44 =) 1936 m2
Area of the flower bed = (44 – 2x )2 m2
Area of the gravel path = Area of the
field - Area of the flower bed
= 1936 – (44 – 2x )2
= 1936 – (1936 – 176x + 4x 2)
= (176x – 4x 2) m2
Cost of laying the flower bed = (Area of
the flower bed) (Rate per sq m)
=100
275)244( 2 x
=2
)244(4
11
x = 2)22(11 x
Cost of gravelling the path
= (Area of the path) × (Rate per sq m)
= )44(6100
150)4176( 22 x x x x
It is given that the total cost of laying the
flower bed and gravelling the path is Rs
4904
11 (22 – x )2 + 6 (44x – x 2 ) = 4904
or, 11 (484 – 44x + x 2) + (264x – 6x 2)
= 4904
or, 5x 2 – 220x + 5324 = 4908
or, 5x 2 – 220x + 420 = 0
or, x 2 – 44x + 84 = 0
or, x 2 – 42x – 2x + 84 = 0
or, x (x – 42) – 2 (x – 42) = 0
or, (x – 2) (x – 42) = 0
or, x = 2 or x = 42
But x 42, as the side of the square is 44
m. Therefore, x = 2.
Hence, the width of the gravel path is 2
metres.
Ex. 26: The l en g t h a n d b r e a d t h o f a r ec t a n g u l a r
f i e l d a r e i n t h e r a t i o o f 7 : 4 . A p a t h 4
m w i d e r u n n i n g a l l a r o u n d o u t s i d e i t
h a s a n a r ea o f 4 1 6 m 2 . Fi nd t he l eng t h
a n d b r e a d t h o f t h e f i el d .
Soln: Let the length and breadth of the field be
7x and 4x metres respectively. Then,
Area of the field = (7x × 4x ) m2
= 28x 2 m2
Length of the field (including path)
= (7x + 8) m
Breadth of the field (including path)= (4x + 8) m
So, Area of the field and path together
= (7x + 8) (4x + 8) m2
Area of the path
= [(7x + 8) (4x + 8) – 28x 2] m2
= (88x + 64) m2
It is given that the area of the path is 416
m2
88x + 64 = 416
or, 88x = 416 - 64
or, 88x = 352
or, x = 4m
Hence, length of the field = 7x = 28 m
breadth of the field = 4x = 16 m
Ex. 27: A c h e ss b o a r d c o n t a i n s 6 4 e qu a l
s q u a r e s a n d t h e a r e a o f ea c h s q u a r e i s 6 . 2 5 cm 2 . A b o r d e r r o u n d t h e b oa r d
i s 2 c m w i d e. Fi n d t h e l en g t h o f t h e
s i d e o f t h e c h e s s b o a r d .
Soln: Let the length of the side of the chess
board be x cm. Then,
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509Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
area of 64 squares = (x – 4)2
(x – 4)2 = 64 × 6.25
or, x 2 – 8x + 16 = 400
or, x 2 – 8x – 384 = 0
or, x 2 – 24x + 16x – 384 = 0
or, (x – 24) (x + 16) = 0
or, x = 24 cm
Triangle
A figure bounded by the three sides is called a
triangle.
(i ) Area of a triangle =2
1 × Base × Height
ABC is the given triangle. Let BDEC be the
rectangle on the same base BC and on the
same height AF. Since AF is perpendicular
to BC, each of the figures ADBF, AECF is a
rectangle.
Now triangle ABF =2
1 rectangle ADBF
Now triangle ACF =2
1 rectangle AECF
By adding, we get
ABF + ACF
=2
1rectangle ADBF +
2
1rectangle AECF
ie ABC =2
1 rectangle BCED
=2
1 × BC × CE
=2
1 × BC × AF [As CE = AF]
=2
1 × Base × Height
Hence the area of a triangle is equal to half
the product of the base and the height.
From the above, we have
Base = Height
Area2
and Height = Base
Area2
( ii) Area of a Triangle when its sides are given
(Hero’s Formula) = ))()(( c s b s a s s
Where, is the area of the triangle; and a ,
b , c are its sides and s =2
1(a + b + c ) =
semi-perimeter of the triangle.
Thus, from half of the sum of the three
sides subtract each side separately. Multiply
the half sum and the three remainders
together. The square root of the product will
be the area of triangle.
Equilateral TriangleA triangle whose all the three sides are equal is
called equilateral triangle.
In an equilateral triangle ABC, a = b= c
s =2
a a a = a
2
3
Hence,
( i ) Area =
a
a a
a a
a a
2
3
2
3
2
3
2
3
=2222
3 a a a a =
2
4
3a
=43 (Side)2
( ii) Height =a
a 2
4
32
Base
Area2
= (Side)2
3
2
3a
Right-angled Triangle
A triangle having one of its angles equal to 90° is
called right-angled triangle.
The figure ABC is right-angled tri angle, angle
B being a right angle ie of 90°. Here, BC is the
base of the triangle, AB is the height of the triangle.
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510 Concep t o f A r i t hm e t i c
K KUNDAN
AC, the side opposite to the right angle, is called
the hypotenuse .
In case of a triangle ABC right-angled at B, AC2
= AB2 + BC2. This is known as the Pythagoras
Theorem. It may be stated in words thus:In a right-angled triangle the square described
on the hypotenuse is equal to the sum of the
squares on the other two sides.
Let b be the base, p be the perpendicular and h
be the hypotenuse of a right-angled triangle. Then,
( i ) Perimeter = (b + p + h )
( ii) Area =2
1× Base × Height =
p b
2
1
( i i i ) Hypotenuse = 22 p b
Isosceles Triangle
An isosceles triangle is one which has two of its
sides equal. Its third side is usually called the
base.
Let ABC be an isosceles triangle such that AB =
AC = b units and BC = a units.
Area of ABC
=22 (Base)
4
1side)(EqualBase
2
1
Isosceles Right-angled Triangle
For an isosceles right-angled triangle, each of
whose equal side is a , we have
( i ) Hypotenuse = a a a 222
( ii) Perimeter = a a a 2222
( i i i ) Area =2
1× Base × Height
=2
2
1
2
1a a a
Solved Examples
Ex. 28: T h e a r e a o f a r i g h t -a n g l e d t r i a n g l e i s
5 0 m 2
. I f o n e o f t h e l e g s i s 2 0 m , f i n d t h e l e n g t h o f t h e o t h e r l eg .
Soln: In a right-angled triangle, if one side is
the base, then the other side is its altitude
or height.
Let the given leg be the base. Then, the
other leg is the altitude.
Here, Area of the triangle = 50 m2
One leg of the triangle = 20 m
The other leg of the triangle
= Height of the triangle
=
20
502
Base
Area2m = 5 m
Ex. 29: F i n d t h e a r e a o f a n i s o sc el e s r i g h t -
ang l ed t r i a n g l e , i f one o f t he equa l s i des
i s 2 0 cm l o n g .Soln: We know that in an isosceles right angled
triangle, any one of the two sides which
are at right angle can be taken as the base
and the other perpendicular side is the
altitude.
There fore ,
base = 20 cm and altitude = 20 cm
So, area of the given triangle
=22 cm200cm2020
2
1
Ex. 30: The a r ea o f a t r i a n g l e i s eq u a l t o t h a t
o f a s q u a r e who s e ea c h s i d e mea s u r e s
6 0 m et r es . F i n d t h e si d e o f t h e t r i a n g l e
w h o se co r r e s p on d i n g a l t i t u d e i s 9 0
m e t r e s .Soln: We have,
Area of the square = (60 × 60) m2
= 3600 m2
Area of the square = 3600 m2
Altitude of the triangle = 90 m
Side of the triangle
= AltitudeingCorrespond
Area2
=
90
36002m = 80 m
Ex. 31: T h e b a se o f a t r i a n g u l a r f i e l d i s t h r ee
t i m e s i t s a l t i t u d e . I f t h e c os t o f
c u l t i v a t i n g t h e f i e l d a t R s 2 4 . 6 0 p e r
hec t a r e i s Rs 332 . 10 , f i n d i t s base and h e i g h t .
Soln: Let the altitude of the triangular field be x
metres.
Then, base = 3x metres (given).
Area =2
1(Base × Height)
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511Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
=2
1(3x × x )
= msq 2
3 2
x ...... (i)
It is given that the cost of cultivating the
field at the rate of Rs 24.60 per hectare is
Rs 332.10.
Area =60.24
10.332
Rate
cost Total
= 13.5 hectares
= (13.5 × 10000) sq m.
[ 1 hectare = 10000 sq m]
= 135000 sq m ....... (ii)
From (i) and (ii) , we have
1350002
3 2 x
or,321350002 x
or, x 2 = 90000
or, x = 300
Hence, height = 300 m and
base = 3x = 900 m.
Ex. 32: F i n d t h e a r ea o f a r i g h t -a n g l e d t r i a n g l e
w i t h h y p o t e n u se 2 5 cm and b a s e 7 cm .
Soln: Let ABC be the right-angled triangle with
base
BC = 7 cm and
hypotenuse AC = 25 cm.
Using Pythagoras Theorem, we have
AC2 = AB2 + BC2
or, (25)2 = AB2 + 72
or, AB2 = 252 – 72 = 625 – 49 = 576
or, AB = 24.
Hence, area of ABC =2
1 (Base × height)
=
2
1 (7 × 24) = 84 cm2.
Ex. 33: The l e n g t h o f t h e s i d e s f o rm i n g r i g h t -
a n g l e o f a r i g h t -a n g l e d t r i a n g l e a r e 5 x
cm an d (3x – 1 ) cm . I f t h e a r ea o f t h e
t r i a n g l e i s 6 0 cm 2 , f i n d i t s h y p o t e n u se.
Soln: Let ABC be a right-angled triangle with
right-angle at B. Let AB = 5x and BC
= 3x – 1.
Th en ,
Area of ABC =2
1(Base × Height)
or, 60 =2
1(AB × BC)
or, 60 =2
1× 5x (3x – 1)
or, 120 = 5x (3x – 1)
or, 24 = x (3x – 1)
or, 3x 2 – x – 24 = 0
or, 3x 2 – 9x + 8x – 24 = 0
or, 3x (x – 3) + 8 (x – 3) = 0
or, (x – 3) (3x + 8) = 0
or, x – 3 = 0 or, 3x + 8 = 0
or, x = 3
or, x =3
8
or, x = 3 [ x 3
8 ]
AB = 5x = (5 × 3 =) 15 cm and
BC = (3x – 1) = (3 × 3 – 1 =) 8 cm.
Now, AC2 = AB2 + BC2
[By Pythagoras Theorem]
or, AC2 = (15)2 + (8)2
or, AC2 = 289
or, AC = 17 cm.
Hence, hypotenuse = 17 cm.
Ex. 34: The p er i m et e r o f a r i g h t -a n g l e d t r i a n g l e
i s 60 cm . I t s hy po t enues i s 25 cm . F i nd
t h e a r e a o f t h e t r i a n g l e.
Soln: Let ABC be the given right-angled triangle
such that base = BC = x cm and
hypotenuse AC = 25 cm.
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512 Concep t o f A r i t hm e t i c
K KUNDAN
Now, perimeter = 60 cm
or, AB + BC + AC = 60
or, AB + x + 25 = 60 cm
or, AB = 35 – x .
By Pythagoras Theorem, we haveAB2 + BC2 = AC2
or, 222 25)35( x x
or, 0600702 2 x x
or, 0300352 x x
or, 030015202 x x x
or, 0)15)(20( x x
or, x = 20 or x = 15
If x = 20, then AB = 35 – x = 15 and
BC = x = 20
Area = AB)(BC2
1
=2150cm15)(20
2
1
If x = 15, then AB = 35 – x = 20 and
BC = x = 15
Area = AB)(BC2
1
= .cm150)2015(2
1 2
Hence, area = 150 cm2
Ex. 35: T h e a r e a o f a r i g h t - a n g l ed t r i a n g l e i s
6 0 0 s q cm . I f t h e b a s e o f t h e t r i a n g l e
ex c eed s t h e a l t i t u d e by 1 0 cm , f i n d t h e
d i m e ns i o n s of t h e t r i a n g l e .
Soln: Let the altitude of the given triangle be x cm long.
Then, base = (x + 10) cm.
Now, Area = 600 cm2
or,2
1(Base × Height) = 600
or,2
1(x + 10)x = 600
or, x 2 + 10x = 1200
or, x 2 + 10x – 1200 = 0
or, x 2 + 40x – 30x – 1200 = 0
or, x (x + 40) – 30 (x + 40) = 0
or, (x + 40) (x – 30) = 0
or, x = 30 or x = – 40But x cannot be negative. So, x = 30.
So, base = x + 10 = (30 + 10 =) 40 cm.
Since the triangle is right-angled.
Therefore,
(Hypotenuse)2 = (Base)2 + (Perpendicular)2
or, (Hypotenuse)2 = 402 + 302
or, (Hypotenuse)2 = 2500
or, Hypotenuse = 50 cm
Hence, the dimensions of the given
triangle are Base = 40 cm, Altitude = 30
cm and Hypotenuse = 50 cm.
Ex. 36: F i n d t h e p e r i m et e r o f a n eq u i l a t e r a l
t r i a n g l e w h o s e a r ea i s 3 4 cm 2 .
S o l n : Let each side of the triangle be a cm. Then,
its area is .4
3 2a
344
3 2 a
or, 163
4342 a
or, a = 4.
Hence, perimeter of the given triangle
= 3a cm = (3 × 4) cm = 12 cm.
Ex. 37: I f ea c h s i d e o f a n e qu i l a t e r a l t r i a n g l e i s i n c r e a se d b y 2 cm , t h e n i t s a r e a
in c reases by 3 3 cm 2 . F i nd t h e l eng t h
o f e a ch s i d e a n d i t s a r e a .
Soln: Let ABC be an equilateral triangle of side
a cm.
Th en ,
A1 = Area of ABC =
2
4
3a cm2 ......(i)
Let DEF be the new equilateral triangle of
side (a + b ) cm. Then,
A1 = Area of DEF =
2)(4
3b a cm2 .... (ii)
It is given that
33AA 12
or, 334
3)2(
4
3 22 a a
[Using (i) and (ii)]
or, )12(4
3)2(
4
3 22 a a
or, 12)2( 22 a a
or, 1244 22 a a a
or, 4a = 8
or, a = 2
So, length of each side of ABC = 2 cm.
And, area of ABC =2)2(
4
3 = 3 cm2.
Ex. 38: F i n d t h e a r e a o f a n i s os c el e s t r i a n g l e
h a v i n g t h e b a s e 6 cm a n d t h e l e n gt h
o f e a ch e q u a l s i d e 5 cm .
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513Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
Soln: We know that,
Area of an isosceles triangle
=2
1 × Base ×
22 (Base)
4
1side)(Equal
Here, base = 6 cm, equal side = 5 cm.
Area of the given triangle
=222 cm(6)
4
1(5)6
2
1
= 2cm9253
= 2cm163 = 12 cm2
Ex. 39: The base o f an i soscel e s t r i a n g l e i s 12
cm and i t s p er i m et e r i s 3 2 cm . F i n d i t s
a r e a .
Soln: We have, base = 12 cm and
perimeter =32 cm.
Let the length of each of the two equal
sides be b cm. Then,Perimeter = 32 cm
or, 2b + 12 = 32
or, 2b = 32 – 12
or, 2b = 20
or, b = 10
Thus, we have
Base = 12 cm and equal side = 10 cm.
Area of the given triangle
=2
1 × Base ×
22 (Base)4
1side)(Equal
=222 cm(12)
4
1(10)12
2
1
=
2
cm361006 = 2cm646
= 6 × 8 cm2 = 48 cm2
Alternative Method:
Let the length of the two equal sides be x
cm. Then,
Perimeter = 32 cm
or, 2x + 12 = 32
or, 2x = 32 – 12
or, 2x = 20
or, x = 10
Thus, the sides of the given triangle are a
= 10 cm, b = 10 cm and c = 12 cm, and 2s
= 32 cm.
s = 16 cm,
s – a = (16 – 10) cm = 6 cm,s – b = (16 – 10) = 6 cm and
s – c = (16 – 12) cm = 4 cm.
Area of the triangle
= ))()(( c s b s a s s
= 2cm46616
= 2cm6664
= 222 cm68
= (6 × 8) cm2
= 48 cm2
Ex. 40: F i nd t he per i me t e r o f an i soscel e s r i gh t -
a n g l ed t r i a n g l e h a v i n g a n a r ea o f 2 0 0
cm 2 .
Soln: Let ABC be an isosceles right-angled
triangle with right angle at B such that
AB = BC = a cm. Then,
Area of ABC =2
1(Base × Height)
=2
1(a × a ) =
2
2a
or, 200 =2
2a [ Area = 200 cm2 (given)]
or, a 2 = 400
or, a = 20
Now, AC2 = AB2 + BC2
[By Pythagoras Theorem]
or, AC2 = a 2 + a 2
or, AC2 = 2a 2
or, cm.a2AC Hence,
perimeter = AB + BC + AC
= 2a + a 2
= 40 + 20 2 [ a = 20]
= 40 + 20 × 1.41
= 68.2 cm.
Ex. 41: The a r ea of an i soscel e s t r i a n g l e i s 60
cm 2 a n d t h e l en g t h o f ea c h o n e o f i t s
eq u a l s i d e s i s 1 3 cm . F i n d i t s b a s e .
Soln: Let ABC be the given isosceles triangle in
which AB = AC = 13 cm. Draw AD
perpendicular from A on BC. Let BC = 2x
cm.
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514 Concep t o f A r i t hm e t i c
K KUNDAN
Then, BD = DC = x cm.
In ABC, we have
222 BDADAB [ By Pythagoras Theorem]
or, 222 AD13 x
or, 222 16913AD x x .
Now, area = 60 cm2
or, 60AD)(BC2
1
or, 6016922
1 2
x x
or, 60169 2 x x
or, 3600)169( 22 x x
or, 03600169 24 x x
or, 0)25)(144( 22 x x
or, x 2 = 144 or x 2 =25
x = 12 or x =5
Hence, Base = 2x = 24 cm or 10 cm
Ex. 42: T h e p er i m et e r o f a t r i a n g u l a r f i el d i s
1 4 4 m and t h e r a t i o o f t h e s i d es i s 3 :
4 : 5 . F i n d t h e a r ea o f t h e f i e l d .
Soln: Let a , b , c be the lengths of the sides of
the triangular field. Then,
a : b : c = 3 : 4 : 5
or a = 3x , b = 4x and c = 5x .
Now, perimeter = 144 m
or (3x + 4x + 5x ) = 144 m
or 12x = 144
x = 12
144 = 12
Thus, the sides of the triangle are
a = (3 × 12) m = 36 m,
b = (4 × 12) m = 48 m and
c = (5 × 12) m = 60 m
Now, s = )(2
1c b a
or
s =2
1 × (36 + 48 + 60) = 72
s – a = (72 – 36) = 36,
s – b = (72 – 48) = 24 and
s – c = (72 – 60) = 12
Area of field = ))()(( c s b s a s s
= 2m12243672
= 2222 m2626626
= 246 m26
= 63 × 22 = 864 m2
Ex. 43: F i nd t he per cen t age i n c r ease i n t he a r ea
o f a t r i a n g l e i f i t s e a ch s i d e i s d o u b l ed .
Soln: Let a , b , c be the sides of the old triangle
and s be its semi-perimeter. Then,
s =2
1 (a + b + c ).
The sides of the new triangle are 2a , 2b
and 2c . Let s be its semi-perimeter. Then,
s =2
1 (2a + 2b + 2c ) = a + b + c = 2s .
Let and be the areas of the old and
new triangles respectively. Then,
= ))()(( c s b s a s s and
= )2)(2)(2( c s b s a s s
= )22)(22)(22(2 c s b s a s s
[ s = 2s ]
= 4))()(( c s b s a s s
Increase in the area of the triangle
= – = 4 – = 3.
Hence, percentage increase in area
=
100
3 = 300%.
Ex. 44: T h e l e n gt h s of t h e si d e s of a t r i a n g l e
a r e 5 c m , 1 2 c m a n d 1 3 c m . Fi n d t h e
l e n g t h o f p e r p en d i c u l a r f r o m t h e
oppos i t e ve r t e x t o t he s i de wh ose l eng t h
i s 1 3 cm .
Soln: Here a = 5, b = 12 and c = 13.
s =2
1 (a + b + c )
=2
1 (5 + 12 + 13) = 15
Let A be the area of the given triangle.
Th en ,
A = ))()(( c s b s a s s
or, A = )1315)(1215)(515(15
or, A = 2cm231015 ....(i)
Let p be the length of the perpendicular
from vertex A on the side BC. Then,
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515Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
Area =2
1 × Base × Height
or, A = 2
1
× (13) × p ..... (ii)
From (i) and (ii), we get
2
1 × 13 × p = 30
13
60 p cm.
Hence, the length of the perpendicular
from the opposite vertex to the side whose
length is 13 cm is
13
60 cm.
Ex. 45: A f i e l d i n t h e f o r m o f a p a r a l l el o g r am
ha s o n e o f i t s d i a g o n a l s 4 2 m l o n g a n d
t h e p e r p en d i c u l a r d i s t a n c e o f t h i s
d i a g o n a l f r o m e i t h e r o f t h e ou t l y i n g
v e r t i c e s i s 1 0 m 8 dm (s ee F i g ) . F i n d
t h e a r ea o f t h e f i el d .
Soln: We have, AC = 42 m and DL = BM = 10 m
8 dm = 10 m 80 cm = 10.8 m.
Area of the field = 2 × Area of ACD
=2m10.842
2
12
= 453.6 m2
Ex. 46: ABCD i s a squa r e. F i s t h e m i d -po i n t o f
A B a n d B E i s o n e t h i r d o f B C . I f t h e
area of the F B E i s 1 0 8 s q cm , f i n d
t h e l e n g t h o f AC.
Soln: Let the side of the square be x cm. Since
F is the mid-point of AB.
So, BF = AF =2
x cm.
Also, BE is one third of BC.
BC =3
x cm
Now, Area of FBE = 108 cm2 (given)
or, 108BE)(FB2
1
or, 108322
1
x x
or, x 2 = 1296
or x 2 = 12 × 12 × 9
or, x = 12 × 3 = 36 cm.
In ABC, we have AC2 = AB2 + BC2
222 3636AC
or, 2362AC
or, 236AC
Hence, 236AC cm.
Parallelogram
A parallelogram is a four-sided figure whose
oppostie sides are parallel. Thus ABCD is a
parallelogram in which AB||DC and AD||BC.
Area of parallelogram ABCD = Base × Height
Let CDEF be a rectangle on the same base DC
and of the same height FC. Then since
parallelograms on the same base and of the same
height are equal in area.
Parallelogram ABCD = Rectangle CDEF
= CD × FC
= Base × Height
Hence area of a parallelogram is equal to the
product of its base and height.
Area = Base × Height
From the above we have,
( i ) Base of a parallelogram = Height
Area
or, Side of a parallelogram
= altitudeingCorrespond
Area
( ii) Height of a parallelogram =Base
Area
or, Altitude of parallelogram
= sideingCorrespond
Area
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516 Concep t o f A r i t hm e t i c
K KUNDAN
Solved Examples
Ex. 47: T h e ba s e o f a p a r a l l e l o g r am i s t h r i c e
i t s h ei g h t . If t h e a r e a i s 8 7 6 cm 2
, f i n d t h e b a s e a n d h ei g h t o f t h e
p a r a l l e l o g r a m .
Soln: Let the height of the parallelogram be x
cm. Then, base = 3x cm.
Area of the parallelogram
= (x × 3x ) cm2 = 3x 2 cm2
But, area of the parallelogram is given as
867 cm2.
3x 2 = 867
or, x 2 = 289
17289 x
Thus, height = 17 cm and base
= (3 × 17) cm = 51 cm.
Ex. 48: I n t h e f i g u r e gi v e n b el o w , A BCD i s a
p a r a l l e l o g r am , CM AB and BL AD .(i ) I f AB = 1 6 cm , A D = 1 2 cm a n d
CM = 10 cm , f i n d BL .
(i i ) I f AD = 1 0 cm , CM = 8 cm a n d
BL = 12 cm , f i n d AB.
Soln:(i) We have, base AB = 16 cm and
altitude CM = 10 cm.
Area of parallelogram ABCD
= Base × Altitude
= (16 × 10) cm2 = 160 cm2 .....(i)
Now, taking AD as the base, we have
Area of parallelogram ABCD
= Base × Altitude
= (12 × BL) cm2 .....(ii)
From (i) and (ii), we have
12 × BL = 160
cm13.3312
160BL
(ii) We have, AD = 10 cm, BL = 12 cm
Area of parallelogram ABCD
= Base × Height
= 10 cm × 12 cm = 120 cm2 ....(iii)
Now, taking AB as the base, we have
Area of parallelogram ABCD = AB × CM
= (AB × 8) cm2 ....(iv)
From (iii) and (iv), we get
AB × 8 = 120
or, AB =8
120 cm
or, AB = 15 cm.
Ex. 49: A f i e l d i n t h e f o r m o f a p a r a l l el o g r am
h a s b a se 1 5 d m a n d a l t i t u d e 8 d m .
F i n d t h e co st o f w a t e r i n g t h e f i e l d a t
t h e r a t e o f 5 0 p a i s e p e r s q u a r e me t r e .
Soln: We have,Base = 15 dm = (15 × 10) m
[1 dm = 10 m]
= 150 m
Altitude = 8 dm = (8 × 10) m = 80 m
Area of the field = (150 × 80) m 2
= 12000 m2
Rate of watering the field
= 50 paise per square metre
= Re2
1 per square metre
Cost of watering the field
= Rs
2
112000 = Rs 6000.
Ex. 50: I n t h e f i g u r e gi v e n b el o w , A BCD i s a parallelogram. DL AB , AB = 8 cm a nd
A D = 5 c m . I f t h e a r e a o f t h e
p a r a l l e l o gr am i s 2 4 cm 2 , f i n d A L .
Soln: We have, base = AB = 8 cm
Area = 24 cm2
Area = Base × Altitude
or, 24 = 8 × DL
or, DL =8
24cm = 3 cm.
Now, in ALD, we have
AD = 5 cm, DL = 3 cm.
By Pythagoras theorem, we have
AD2 = AL 2 + DL 2
or, 52 = AL 2 + 32
or, AL 2 = 52 – 32 = 25 – 9 = 16
or, AL 2 = 42
or, AL = 4
Rhombus
A rhombus is a parallelogram all of whose sides
are equal. In a rhombus the diagonals bisect each
other at right angles. Thus in the rhombus ABCD,
AB = BC = CD = DA and
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517Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
AO = OC, BO = OD
BOC = 90°
Area of the rhombus ABCD
= 2BCD = 4BOC
= 4 ×2
1 × BO × OC
=2
AC
2
BD2
=2
1 × BD × AC
=2
1 (Product of diagonals)
Hence, the area of rhombus is equal to the half
of the product of its diagonals.
Solved Examples
Ex. 51: F i n d t h e a r ea o f a r h o m b u s w h o s e
d i a g o n a l s a r e o f l en g t h s 2 0 cm and 1 8
cm .
Soln: Area of the given rhombus
=2
1 × Product of diagonals
=
1820
2
1cm2 = 180 cm2.
Ex. 52: The a r e a o f a r h om bu s i s 7 2 cm 2 . If i t s
p er i m et e r i s 3 2 cm , f i n d i t s a l t i t u d e.
Soln: We have, perimeter of the rhombus
= 32 cm
4 (side) = 32 cm
[ Perimeter = 4 (side)]
side =4
32cm = 8 cm
Now, area of the rhombus = 72 cm2
or, (Side × Altitude) = 72
or, 8 × Altitude = 72
Altitude =8
72cm = 9 cm
Ex. 53: F i n d t h e a l t i t u d e of a r h o m b u s w h o s e
a r e a i s 3 6 m 2 a n d p e r i m et e r i s 3 6 m .
S o l n : We have, perimeter of the rhombus = 36
m and, area of the rhombus = 36 m2
Now, side of the rhombus
=436
4Perimeter m = 9 m
Altitude of the rhombus
=9
36
Side
Area m = 4 m.
Ex. 54: F i nd t he a r ea o f a r hom bus whose each
s i d e i s of l e n g t h 5 m a n d o n e of t h e
d i a g o n a l s i s o f l en g t h 8 m .
Soln: Let ABCD be a rhombus whose each side
is of length 5 m. Let the diagonals ACand BD intersect at O. Let AC be 8 m.
Since the diagonals of a rhombus bisect
each other at right angles. Therefore,
AOB is a right-triangle
or, AB2 = OA2 + OB2
[Using: Pythagoras theorem]
or, 52 = 42 + OB2
or, OB2 = 25 – 16
or, OB2 = 9
or, OB2 = 32
or, OB = 3 m
BD = 2 × OB = (2 × 3)m = 6m.
Hence, area of rhombus ABCD
=2
1 × Product of diagonals
=
68
2
1m2 = 24 m2
Ex. 55: I f t h e a r ea of a r h ombus be 48 cm 2 an d
o n e o f i t s d i a g o n a l i s 1 2 c m , f i n d i t s
a l t i t u d e .
Soln: Let ABCD be a rhombus of area 48 cm2
and diagonal BD = 12 cm.
Now,
Area = 48 cm2
or, 48BDAC2
1
or, 4812AC2
1
or, 6 × AC = 48
AC =6
48cm = 8 cm
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518 Concep t o f A r i t hm e t i c
K KUNDAN
Since the diagonals of a rhombus bisect
each other at right angles.
cm4AC2
1OA
cm3BD2
1OB
Also, AB2 = OA2 + OB2
[Using Pythagoras Theorem]
or, AB2 = 42 + 32
or, AB2 = 16 + 9 = 25 = 52
or, AB = 5 cm
Since a rhombus is a parallelogram also.
Therefore,
Area of rhobmus = AltitudeAB2
1
or, Altitude52
148
or, Altitude = cm5
248 =
5
96cm = 19.2 cm
Ex. 56: I f t h e a r ea o f a r h ombu s b e 2 4 cm 2 an d
o n e o f i t s d i a g o n a l s be 4 cm , f i n d t h e
p e r i m e t e r o f t h e r h om b u s .
Soln: Let ABCD be a rhombus such that its one
diagonal AC = 4 cm. Suppose the diagonals
AC and BD intersect at O.
Now,
Area of rhombus ABCD = 24 cm2
or, 24BDAC2
1
or, 24BD42
1
or, 2 × BD = 24
or, BD = 12
Thus, we have, AC = 4 cm and BD = 12 cm
cm2AC2
1OA and
cm6BD2
1
OB
Since the diagonals of a rhombus bisect
each other at right angle. Therefore, OAB
is a right triangle, right angled at O.
By Pythagoras theorem, we have
AB2 = OA2 + OB2
or, AB2 = 22 + 62 = 40
or, cm102cm40AB
Hence, perimeter of rhombus
cm108cm1024ABCD .
Ex. 57: I f t h e s i d e o f a s qu a r e i s 4 m and i t i s
c on v e r t e d i n t o a r h ombu s who s e ma j o r
d i a g o n a l i s 6 m , f i n d t h e ot h er d i a g o n a l
a n d t h e a r ea o f t h e r h om b u s .
Soln: Let AB = 4 m be the side of a square ABPQ
which is converted into a rhombus ABCD
such that diagonal AC = 6 m.
Since the diagonals of a rhombus bisect
each other at right angle, therefore
3mAC2
1OA and AOB = 90°.
In OAB, we have
AB2 = OA2 + OB2
or, 42 = 32 + OB2
or, OB2 = 16 – 9
or, OB = m7
BC = 2OB = m72 .
Hence, area of rhombus ABCD
=BDAC
2
1
=22 m76m726
2
1
.
Trapezium
A trapezium is a four-sided figure having a pair of
opposite sides parallel. Thus ABCD is a trapezium
in which AB||DC.
Area of trapezium
=2
1 × Height × Sum of the parallel sides
=2
1 × (Distance between parallel sides) ×
(Sum of parallel sides)
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519Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
Draw AE and BF perpendicular from A and B to
DC.
Trapezium ABCD
= ADE + Rectangle ABFE + BFC
=2
1 × DE × AE + AE × EF +
2
1 × BF × FC
=2
1 × AE (DE + 2EF + FC) [ BF = AE]
=2
1 × AE (DE + EF + FC + AB) [ EF = AB]
=2
1 × AE × (DC + AB)
=2
1 × Height × Sum of the parallel sides
Hence, the area of a trapezium is equal to the
product of half the sum of parallel sides and height.
Solved Examples
Ex. 58: F i n d t h e a r e a of a t r a p e zi u m w h o s e
p a r a l l e l s i d e s a r e o f l en g t h s 1 0 cm and
1 2 cm a n d t h e d i s t a n c e b et w e en t h em
i s 4 cm .
Soln: We have,
Area of the trapezium =2
1× (Sum of the
parallel sides) × (Distance between the
parallel sides)
=
41210
2
1
cm2
=
422
2
1cm2 = 44 cm2
Ex. 59: F i nd t he a r ea o f t he f i gu r e g i ven bel ow :
Soln: We have,
Area of the given figure
= Area of trapezium ABCD + Area
of trapezium CDEF
=
2cm101521
2
1
2
cm1224152
1
=22 cm1239
2
1cm1036
2
1
= 180 cm2 + 234 cm2 = 414 cm2
Ex. 60: T h e a r e a of a t r a p e zi u m i s 1 8 0 c m 2
a n d i t s h e i g h t i s 9 c m . I f o n e o f t h e
p a r a l l el s i d e s i s l o n g er t h a n t h e o t h e r
b y 6 cm , f i n d t h e t w o p a r a l l e l s i d es .
Soln: Let one of the parallel sides be of length x
cm. Then, the length of the other parallel
side is (x + 6) cm.
Area of the trapezium
=
96
2
1x x cm2
=
962
2
1x cm2
= (9x + 27) cm2
But the area of the trapezium is given as
180 cm2.
9x + 27 = 180
or, 9x = 180 – 27 = 153
or, x =9
153 = 17
Thus, the two parallel sides are of lengths
17 cm and (17 + 6) cm = 23 cm.
Ex. 61: I f t h e p e r i m e t e r o f a t r a p e zi um b e 5 2
cm , i t s n o n -p a r a l l el s i d e s a r e eq u a l t o
1 0 c m ea c h a n d i t s a l t i t u d e i s 8 c m ,f i n d t h e a r ea o f t h e t r a p e z i um .
Soln: We have,
Perimeter of the trapezium = 52 cm
or, Sum of the parallel sides + Sum of the
non-parallel sides = 52 cm
or, Sum of the parallel sides + 2 × 10 = 52
or, Sum of the parallel sides
= (52 – 20)cm = 32 cm
Altitude of the trapezium = 8 cm.
Area of the trapezium
=2
1× (Sum of the parallel sides) × Altitude
=
832
2
1cm2 = 128 cm2.
Ex. 62: T h e p a r a l l e l s i d es o f a t r a p ez i u m a r e
2 0 cm and 1 0 cm . I t s n o n -p a r a l l e l s i d e s
a r e bo t h equa l , each bei ng 13 cm . F i nd
t h e a r e a o f t h e t r a p ez i um .
Soln: Let ABCD be a trapezium such that AB =
20 cm, CD = 10 cm and AD = BC = 13 cm.
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520 Concep t o f A r i t hm e t i c
K KUNDAN
Draw CL || AD and CM AB.
Now, CL || AD and CD || AB.
ALCD is a parallelogram.
AL = CD = 10 cm and
CL = AD = 13 cm.
In CLB, we have
CL = CB = 13 cm
CLB is an isosceles triangle.
LM = MB = BL 2
1 = 10
2
1 cm = 5 cm
cm10cm10)(20
AL ABBL
Applying Pythagoras theorem in CLM,
we have
CL 2 = CM2 + LM2
or, 132 = CM2 + 52
or, CM2 = 169 – 25 = 144
or, CM = 144 = 12
Area of CLB = CMBL 2
1
=
1210
2
1cm2 = 60 cm2.
Area of parallelogram ALCD = AL × CM= (10 × 12)cm2 = 120 cm2.
Hence, area of trapezium ABCD = Area of
parallelogram ALCD + Area of CLB
= (120 + 60)cm2 = 180 cm2.
Ex. 63: T h e p a r a l l e l s i d es o f a t r a p ez i u m a r e
2 5 cm an d 1 1 cm , wh i l e i t s n o n -p a r a l l e l
s i d es a r e 1 5 cm a n d 1 3 c m . Fi n d t h e
a r e a o f t h e t r a p e zi um .
Soln: Let ABCD be a trapezium such that AB ||
DC, AB = 25 cm, DC = 11 cm, AD = 15 cm
and BC = 13 cm.
Draw CE || DA and CF AB.
Clearly, AECD is a parallelogram.
Now, EB = AB – AE
= AB – DC [ AE = DC]
= (25 – 11) cm = 14 cm
Also, EC = AD = 15 cm. Thus, in ECB, we have
EB = 14 cm, EC = 15 cm and
BC = 13 cm
Let s be the semi-perimeter of the ECB.
Th en ,
1315142
1s cm = 21 cm.
Area of ECB
= ))()(( c s b s a s s
= )1321)(1521)(1421(21 cm2
= 86721 cm2
= 3232737 cm2
= 432 237 cm2
= 7 × 3 × 22 cm2 = 84 cm2 ....(i)
Also, Area of ECB =2
1(Base × Height)
= CF)(EB2
1
= )CF14(2
1 cm2
= (7 × CF) cm2 ....(ii)
From (i) and (ii), we get 7 × CF = 84
or, CF =7
84cm = 12 cm.
Area of parallelogram AECB
= Base × Height
= AE × EF
= (11 × 12) cm2 = 132 cm2
Now,
Area of trapezium ABCD = (Area of
parallelogram AECB) + (Area of ECB)
= (132 + 84) cm2 = 216 cm2
Note : We can find the area of trapezium
directly as follows:
We have, lengths of parallel sides
= 11 cm and 25 cm
Height of the trapezium = 12 cm
Area of trapezium
=21 × Height × (Sum of the parallel sides)
=
)1125(12
2
1 sq cm
=
3612
2
1 = [36 × 6] = 216 sq cm
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521Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
Quadrilateral
A quadrilateral is a plane figure bounded by four
sides. Thus ABCD is a quadrilateral.
Area of quadrilateral ABCD
=2
1 (Length of a diagonal) × (Sum of the lengths
of perpendiculars from the remaining two
vertices on the diagonal).
Join DB. Draw AE and CF perpendiculars to DB
Quadrilateral ABCD = ABD + BDC
=2
1 × BD × AE +
2
1 × BD × CF
=2
1 × BD × (AE + CF)
Hence, the area of quadrilateral is equal to the
product of one diagonals and half the sum of
perpendiculars drawn on it from the other two
vertices.
Note: (i) The area of the quadrilateral can also be
found if the lengths of all its sides and
one diagonal is known, then the area of
each of the two triangles into which the
diagonal divides the quadrilateral can be
found.
( ii) This formula is al so appl icable to a
rectangle, a square, a parallelogram, a
rhombus or a trapezium for which wehave special formulae.
Solved Examples
Ex. 64: T h e d i a go n a l o f a q u a d r i l a t e r a l i s 2 0
m i n l e n gt h a n d t h e p e r p en d i c u l a r s t o
i t f r om t h e oppos i t e ver t i c e s a r e 8 . 5 m
a n d 1 1 m . F i n d t h e a r e a o f t h e
q u a d r i l a t e r a l .
Soln: In quadrilateral ABCD, we have AC = 20
m. Let BL AC and DM AC such that
BL = 8.5 m and DM = 11 m.
Area of quadrilateral ABCD
=2
1× AC × (BL + DM)
=
)115.8(20
2
1 m2
= (10 × 19.5) m2 = 195 m2.
Ex. 65: I n q u a d r i l a t er a l A BCD s h ow n i n f i g u r e g i v en b el o w , A B | | D C a n d A D A B .
A l so , AB = 8 m , DC = BC = 5 m . F i nd t he
a r e a o f t h e qu a d r i l a t e r a l .
Soln: Clearly, ABCD is a trapezium, and AD
= CE is its height.
We have,BE = AB – AE = AB – DC = (8 – 5) m
= 3 m.
In BCE, we have
BC2 = BE2 + CE2
or, 52 = 32 + CE2
or, CE2 = 25 – 9
or, CE2 = 16
or, CE = 16 m = 4 m.
Area of quadrilateral ABCD
=2
1 × (AB + DC) × CE
=
4)58(
2
1 m2 = 26 m2.
Ex. 66: F i n d t h e a r e a o f t h e q u a d r i l a t e r a l
ABCD, i n w h i ch AB = 7 cm , BC = 6 cm ,
CD = 12 cm , DA = 15 cm an d AC = 9 cm .
Soln: The diagonal AC divides the quadrilateral
ABCD into two triangles ABC and ACD.
Area of quadrilateral ABCD
= Area of ABC + Area of ACD
For ABC, we have
s =
2
97611 cm
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522 Concep t o f A r i t hm e t i c
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Area of ABC = ))(()( c s b s a s s
= )911()711()611(11
= 24511
= 440 sq cm = 20.98 cm2
For ACD, we have
s =
2
1512918 cm.
Area of ACD
= )1518()1218()918(18
= 36918
= 18 × 3 cm2 = 54 cm2
Hence, area of quadrilateral ABCD
= (20.98 + 54) cm2 = 74.98 cm2.
Ex. 67: F i n d t h e a r ea o f a q u a d r i l a t e r a l A BCD
wh o s e s i d e s a r e 9 m , 4 0 m , 2 8 m an d 15 m r espec t i v e l y an d t he ang l e betw een
t h e f i r s t t w o s i d e s i s a r i g h t a n g l e .
Soln: Let ABCD be the given quadrilateral such
that ABC = 90° and AB = 9 m, BC = 40
m, CD = 28 m, AD = 15 m.
In ABC, we have AC2 = AB2 + BC2
[Using Pythagoras Theorem]
or, AC2 = 92 + 402 = 1681
or, AC = 41 m
Now, Area of ABC =2
1 (Base × Height)
=2
1 (AB × BC)
=2
1(9 × 40) m2 = 180 m2
In ACD, we have
AC = 41 m, CD = 28 m and DA = 15 m.Let a = AC = 41 m, b = CD = 28 m and c
= DA = 15 m. Then,
s =2
1 (a + b + c)
=2
1 (41 +28 + 15) = 42
Area of ACD = )()()( c s b s a s s
= )1542)(2842()4142(42
= 2714314 = 14 × 9 = 126 m2
Hence, Area of quadrilateral ABCD
= (Area of ABC) + (Area of ACD)
= (180 + 126) m2 = 306 m2
Regular Polygons
A polygon is figure bounded by more than four
straight lines. A polygon is said to be regular when
all its sides and angles are equal.
Polygon can be either convex or concave as
mentioned below.
A polygon in which none of its interior (internal)
angles is more than 180°, is known as a convex
polygon.
On the other hand, if at least one angle of apolygon is more than 180º then it is said to be
concave polygon.
We use the following terminology depending
upon the number of sides of a polygon.
Number of sides Polygon
5 Pentagon
6 Hexagon
7 Septagon
8 Octagon
9 Nonagon
10 Decagon
11 Undecagon
12 Dodecagon
(i) Area of a regular polygon of n sides having given
the length of a side and the radius of the inscribed
circle
=2
1 × number of sides (n ) × length of a side ×
radius of the inscribed circle
Let ABCDE be a regular polygon in which AB = a
and OG (the radius of the inscribed circle) = r
Join OA, OB, OC, OD, OE. Thus the polygon is
divided into as many triangles as its number of
sides.
Area of polygon
= Area of AOB × Number of sides of the
polygon
=2
1 × AB × OG × n = r a
n
2
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(a) Hexagon
It can be easily seen that triangle AOB is
equilateral.
Area of equilateral triangle AOB =2
4
3a
Area of hexagon ABCDEF =2
4
36 a
= 2
33 2a
(b) Octagon
Here it will be seen that the radius of the inscribed
circle
OG = OH + HG = 2
a
+ HG
HG = KB =2
a
OG =
2
12
22a
a a
Area of octagon ABCDEFLM = r a n
2
= 122
2
12
2
8 2
a a
a
(ii) The area of a regular polygon of n sides having
given the length of one side and the radius of thecircumscribed circle
=2
1× number of sides × side
22
2
SideRadius
Let ABCDE be a regular polygon of n sides with
AB = a and OB (the radius of the circumscribed
circle) = R
Draw OG perpendicular to AB. Then OG is the
radius of the inscribed circle.
OG =
2222
2RGBOB
a
Area of a regular polygon
22
222
a R a
n ar
n
In the case of a hexagon, R = a and n = 6.
area of a regular hexagon
22
2
2
33
22
6a
a a a
(iii) Some Important Results
(A) Interior Angle of a Regular Polygon: Each
interior angle of a regular polygon of n sides is
equal to
º
1802
n
n or 180° – (Exterior Angle).
(B) Exterior Angle of a Regular Polygon: Each
exterior angle of a regular polygon of n sides is
equal to
º360
n .
(C) In a convex polygon of n sides, we have:
(i ) Sum of all interior angles
= (2n – 4) right angles
= (n – 2) × 180°
( ii) Sum of all exterior angles = 4 right angles
( i i i ) Number of diagonals of a polygon of n sides
=
n
n n
2
)1(
Some Particular cases:
Regular Polygon Internal Angle Tr iang le 60°
Quadrilateral 90°
Pentagon 108°
Hexagon 120°
Octagon 135°
Nonagon 140°
Decagon 144°
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524 Concep t o f A r i t hm e t i c
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(D) Circum-circle of a Regular Polygon: A
regular polygon can be inscribed in a circle which
is known as the circum-circle or circumscribing
circle. The centre of this circle is also the centre
of the polygon and the radius is known as thecircum-radius which is generally denoted by R.
If a is the length of each side of a regular
polygon and R is the circum-radius, then we have
the following results:
(a) R =
n a 180ºcosec2
(b ) Area of the polygon =
n na
º180cot
4
1 2
or, Area of the polygon
=
n n n
º180cos
º180sinR2
(c) Area of the circum-circle of an n-sided
regular polygon =
n a
º180cosec
4
22
Particular cases:
l Area of a regular hexagon
=22
2
3330ºcot4
6 a a
l Area of circum-circle of a regular
hexagon = a 2
l Perimeter = 6a . Each angle = 120º.
(E) In-circle of a Regular Polygon: A regular
polygon can also circumscribe a circle. A circle
having centre at the centre of a regular polygon
and touching all sides of it is called the in-circle.
If a is the length of a side of a regular polygon
and r is the radius of the in-circle, then we have
the following results:
(a) r =
n
a 180ºcot
2
(b ) Area of polygon =
n nr º180tan2
(c) Area of the in-circle of an n-sided regular
polygon =
n a
º180cot
4
22
Particular cases:
l Radius of in-circle of a regular hexagon
= a a
2
3º30cot
2
l Area of the in-circle of a regular hexagon
=2
2
4
3
2
3a a
Solved Examples
Ex. 68: F i n d t h e a r e a o f a r e gu l a r h e x a g on
wh o s e s i d e i s 1 0 cm l o n g .
S o l n : Area of a regular hexagon = 2side2
33
Here, side = 10 cm.
Hence, Area of the given regular hexagon
= 22 cm3150102
33
Ex. 69: T h e a r e a o f a r e g u l a r h e xa g o n i s 6 0 0
cm 2 . D et e rm i n e i t s p er i m e t e r .
Soln: We know that the area of a regular
hexagon is equal to 2side2
33.
Area = 23600 cm
or, 3600)side(2
33 2
or,33
31200)side( 2
or, (side)2 = 400
or, side = 20 cm.
So, perimeter = 6 (side) = (6 × 20) cm
= 120 cm.
Ex. 70: F i nd t o t he nea r est m e t r e t he si de o f a
r e g u l a r o c t a g o n a l e n c l o su r e w h o s e
a r e a i s 1 h ec t a r e .
Soln: Area of a regular octagon = 2212 a
2212 a = 1 hectare.
a 2 =)21(2
10000
sq cm
or, a 2 = 2701 sq m approx.
a = 46 metres approx.
Ex. 71: A s qu a r e a n d a r e g u l a r h e x a g o n h a v e
eq u a l p e r i m e t e r s . Compa r e t h e i r a r ea s .
Soln: Let P be the perimeter of both a square
and a regular hexagon.
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525Mensu r a t i o n I (A rea an d Per i me t e r )
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Th en ,
side of the square =4
P and
side of the regular hexagon =6
P.
A1 = Area of the square = (side)2 =
16
P2
A2 = Area of the regular hexagon
= 22
2P
24
3
6
P
2
33sdie
2
33
32
3
243
16
2
2
2
1 P
P
A
A .
Hence, areas of the square and the
hexagon are in the ratio 32:3 .
Ex. 72: T h e s i d e of a r eg u l a r p en t a g on i s 1 0
cm . F i n d i t s a r e a .[Tak e Cot 36°= 1.37 63 ]
Soln: The area of an n -sided regular polygon is
2sidesº180
cot4
n
n .
Here, n = 5 and side = 10 cm
Area of the pentagon =4
5(cot 36°) (10)2
= 125 cot 36° cm2
= 125 × 1.3763
= 172.04 cm2
Ex. 73: F i n d t h e d i f f e r e n c e b e t w e en t h e a r e a
o f a r eg u l a r h e x a g o n ea c h o f w h o s e
s i d e i s 7 2 cm a nd t h e a r e a of t h e ci r c l e
i n s cr i b ed i n i t .
(Tak e7
2 2 ).
Soln: We know that the area of an n -sided
regular polygon is
n na
º180cot
4
1 2 and
area of the incircle is
n a
º180cot
4
1 22;
where a is the side of the polygon.
Here, a = 72 and n = 6.
Required area
=
n a
n na
º180cot
4
1º180cot
4
1 222
= º30cot727
22
4
1º30cot726
4
1 222
= 2cm42.1221937776
= (13468.42 - 12219.42) m2
= 1249 cm2.
Ex. 74: A r e g u l a r h e x a g on i s i n s c r i b e d i n a
c i r c l e of r a d i u s 5 cm . F i n d t h e a r e a o f
t h e c i r c l e wh i c h i s o u t s i d e t h e h ex a g o n .
[Use = 3 .14 an d 3 = 1 . 73 ]
Soln: Area of the circumcircle
= (radius)2 = × 52 = 25 cm2
= 25 × 3.14 cm2 = 78.5 cm2
We know that the area of a regular polygon
of n sides is given by
n n n
180cos
180sinR2
;
where R is the radius of the circumcircle.
Here, n = 6, R = 5 cm.
So, Area of the regular hexagon
= 6 × 52 × sin 30° × cos 30° cm2
= 150 ×2
3
2
1 cm2
= 64.875 cm2
Hence,
required area = Area of the circumcircle
– Area of the hexagon
= 78.5 - 64.875 cm2
= 13.625 cm2
Circle
A circle is a geometrical figure consisting of all
those points in a plane which are at a given
distance from a fixed point in the same plane.
The fixed point is called the centre of the circle
and the constant distance is known as its radius.
(plural radii)
In the given figure, O is the centre and r is the
radius of the circle. A circle with centre O and
radius r is generally denoted by C (O, r ).
The word ci rcle is of ten used for the
circumference.
(i) Some Important Terms
(a) Circular Region: The part of the circle that
consists of the circle and its interior is called the
circular region.
A circular region is also called a circular disc
as shown in the figure given below.
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526 Concep t o f A r i t hm e t i c
K KUNDAN
(b) Chord of Circle: A line segment joining
any two points on a circle is called a chord of the
circle. It should be noted that a chord is not a part
of the circle.
In the figure given below, PQ is a chord of thecircle.
(c) Diameter: A chord passing through the centre
of a circle is known as its diameter.
Note that a circle has many diameters and a
diameter of a given circle is one of the largest chords
of the circle. Also, all diameters are of the same
length.
(d) Semi-Circle: Clearly, if d is diameter of a
circle of radius = r , then d = 2r.
A diameter of a circle divides the circumference
of a circle into two equal parts each of which is
called a semi-circle.
(e) Quadrant: Two perpendicular diameters of
a circle divide its circumference into four equal
parts each of which is known as a quadrant.
(f) Concentric Circles: Circles having the same
centre but with different radii are said to be
concentric circles. The figure given below shows two concentric
circles.
(g) Congruent Circle: Two circles are said to
be congruent if and only if either of them can be
superpassed on the other so as to cover it exactly.
It follows from the above definition that two circles
are congruent if and only if their radii are equal.
(ii) Circumference of a Circle
The perimeter of a circle is called its circumference. The ratio of the circumference of a circle and
its diameter is always constant.
The ratio
Diameter
nceCircumfere = 3.14 (approximately)
This ratio is denoted by (Pi).
Thus, we have
= 3.14 (approximately) =7
22 (approximately)
Now, Diameter
nceCircumfere
=
or,r 2
C = C = 2r
Thus, circumference C of a circle of radius r is
given by C = 2r
If d denotes the diameter of the circle.
Then, d = 2r
C = d
Note: The number is not a rational number, but
its value upto two decimal places coincides
with7
22. So, we take the value of as
7
22.
In the remaining part of this chapter, unless
stated otherwise, the value of will be taken
as 7
22
.
(iii) Area of a Circle
In this section, we shall first obtain the formula
for the area of a circle and then the same will be
used to solve some simple problems.
To obtain the formula for the area of a circle, let
us consider the following.
Draw a circle of any radius (say, 2 cm) on a thin
card-board. Cut it out, and by folding divide it into
four equal sectors. Cut these four sectors out, and
bisect each of them by folding.
Now you have got eight sectors, arrange these
as shown in the figure below.
Next bisect each sector, as before, and so get 16
equal sectors. Rearrange these as shown in the
figure below.
Now notice that as the number of sectors is
increased, each arc is decreased; so that
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527Mensu r a t i o n I (A rea an d Per i me t e r )
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( i ) the outlines AB, DC tend to become straight
lines, and
( ii) the angles at D and B tend to become right
angles.
Thus, when the number of se ctor s isindefinitely increased the figure ABCD ultimately
becomes a rectangle whose length is the semi-
circumference of the circle , and whose breadth is its
radius .
Hence area of the circle
= )radius()ncecircumfere(2
1
=22
2
1r r r
Hence, area A of a circle of radius r cm is
given by
A = r 2
Also, r =
A
Note:Area of a semi-circle
=2
1 (Area of the circle) =
2
2
1r
Area of a quadrant of a circle
=4
1 (Area of the circle) =
2
4
1r .
(iv) Area Enclosed By Two Concentric
Circles
If R and r are radii of two concentric circles, then
Area enclosed by the two circles
)R)(R()R(R 2222 r r r r
(v) Sector—Its arc and Area
The angle at the centre of a circle contains four
right angles or 360°. Hence, if through the centre
of a circle we draw 360 radii making equal angles
with one another, 360 angles of 1 degree each
would be formed at the centre. Since equal angles
at the centre are subtended by equal arcs, the
whole circumference would be divided into 360equal arcs and the area of the circle would be
divided into 360 equal sectors.
Hence,
(a) Arc of a sector of 1° =360
1× circumference
Arc of a sector of D° =360
D× circumference
Hence, Arc of the sector
= circletheof nceCircumfere
360
angleSector
(b ) Area of a sector of 1° =360
1× area of circle
Area of a sector of D° =360
D× area of circle
Hence, Area of the sector
=360
angleSector × Area of the circle
(c) To show that area of sector
=2
1× radius × length of arc.
Proof. Area of a sector of D°
=360
D× area of circle
=360
D×
2
1 × radius × circumference
=360
D× circumference ×
2
1× radius
= arc ×2
1 × radius
=2
1radius × length of arc
Note: A quadrant is a part of the circle contained
between two perpendicular radii. Hence a quadrant
is a sector of 90°.
(vi) Area of Segment
Any chord of a circle, which is not a diameter,
such as AB, divides the circle into two segments,
one greater and one lesser than a semi-circle.
Greater segment is called major segment and
lesser segment is called minor segment.
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528 Concep t o f A r i t hm e t i c
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It will be seen from the figure that area of
segment ACB = sector OACB – OAB.
The area of the segment ADB will be found by
subtracting the area of the segment ACB from the
area of the circle.Note: Area of a minor segment of angle in a
circle of radius r is given by
A =
θ
θ r sin
2
1
360
2 (Always Remember)
(vii) Some Particular Cases
(a) Area of a semi-circle : The sector of a semi-
circle is 180°.
Area of a semi-circle
=22
2
1
360
180r r
(b) Area of a quadrant : The sector angle of a
quadrant of a circle is 90°.
Area of a sector of circle
=22
4
1
360
90r r
(c) Angle described by minute-hand in 60
minutes = 360°.
Angle described by minute-hand in one
minute =
60
360 = 6°.
Thus, minute-hand rotat es th rough an
angle of 6° in one minute.
(d ) Angle described by hour-hand in 12 hours
= 360°
Angle described by hour-hand in one
minute =
2
1
60
30
Thus, hour -han d rot at es thr ough
2
1
6012
360 in one minute.
Solved Examples
Ex. 75: The r a t i o o f t h e r a d i i o f t w o ci r c l e s i s
2 : 5 . W h a t i s t h e r a t i o o f t h e i r
c i r c um f e r e n c e s?
Soln: We have, ratio of radii = 2 : 5. So, let the
radii of two circles be 2 r and 5r
respectively.
Let C1 and C
2 be the circumference of two
circles of radii 2r and 5r respectively.
Th en ,
r r 422C1 and r r 1052C2
5
2
10
4
C
C
2
1
r
r
or, C1 : C
2 = 2 : 5.
Ex. 76: A p i e c e o f w i r e i n t h e f o r m o f a
r e ct a n g l e 8 . 9 cm l o n g a n d 5 . 4 cm b r o a d
i s r e s h a p ed a n d b en t i n t o t h e f o r m o f
a c i r c l e . Fi n d t h e r a d i u s o f t h e ci r c l e.
Soln: We have,Length of the wire
= Perimeter of the rectangle
= 2 (l + b )
= 2 × (8.9 + 5.4) cm
= 28.6 cm
Let the wire be bent into the form of a
circle of radius r cm. Then,
Circumference = 28.6 cm
or, 6.282 r
or, 6.287
222 r
or, r =222
76.28
cm
or, r =10222
7286
cm = 4.55 cm.
Ex. 77: A co p p er w i r e, when b en t i n t h e f o rm o f
a squa r e , encl o ses an a r ea o f 48 4 cm 2 .
I f t h e s ame w i r e i s b en t i n t h e f o rm o f
a c i r c l e, f i n d t h e a r e a e n cl o s ed b y i t .
(Use7
2 2 ).
Soln: Area of the square = 484 cm2.
Side of the square
= 484 cm = 22 cm
[ Area = (side)2 side = Area )
So, perimeter of the square = 4 (side)= (4 × 22) cm = 88 cm.
Let r be the radius of the circle. Then,
Circumference of the circle = Perimeter of
the square
or, 882 r
or, 887
222 r
or, r = 14 cm
Area of the circle =
22 14
7
22r cm2
= 616 cm2
Ex. 78: T h e d i am et e r o f t h e w h e el o f a c a r i s
7 7 c m . H ow m a n y r e v ol u t i o n s w i l l i t
m a k e t o t r a v el 1 2 1 k m .Soln: We have,
diameter of the wheel of the car = 77 cm
Circumference of the wheel of the car
= d =
77
7
22cm = 242 cm
Note that in one revolution of the wheel,
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or, r 7
222 = 220
r = 35 m
Since the track is 7 metres wideeverywhere. Therefore,
R = Outer radius = r + 7 = (35 + 7) m
= 42 m.
Outer circumference = 2R
=
42
7
222 m = 264 m.
Rate of fencing = Rs 2 per metre
Total cost of fencing
= (Circumference × Rate)
= Rs (264 × 2) = Rs 528.
Ex. 84: A c i r c u l a r g r a s s y p l o t o f l a n d , 4 2 m i n
d i a m e t e r , h a s a p a t h 3 . 5 m w i d e
r u n n i n g r o u n d i t o n t h e o u t s i d e. F i n d
t h e c o st o f g r a v el l i n g t h e p a t h a t R s 4
p e r s q u a r e me t r e .
Soln: Radius of the plot =
2
42m = 21m.
Radius of the plot including the path
= (21 + 3.5) m = 24.5 m.
Area of the path
= 22 )21()5.24( m2
= 22 )21()5.24( m2
= 215.24215.24 m2
= 5.35.45 m2
=
5.35.457
22
m2
= 500.5 m2
Hence, cost of gravelling the path
= Rs (500.5 × 4) = Rs 2002.
Ex. 85: A p a p er i s i n t h e f or m o f a r e c t a n gl e
ABCD i n w h i ch AB = 20 cm and BC = 14
cm . A sem i c i r c u l a r p o r t i o n w i t h BC a s
d i a m e t er i s cu t o f f . F i n d t h e a r e a o f
t h e r em a i n i n g p a r t .
S o l n : Length of the rectangle ABCD
= AB = 20 cm
Breadth of the rectangle ABCD= BC = 14 cm
Area of rectangle ABCD
= (20 × 14) cm2 = 280 cm2.
Diameter of the semi-circle
= BC = 14 cm
Radius of the semi-circle = 7 cm.Area of the semi-circular portion cut off
from the rectangle ABCD
=
22 7
7
22
2
1
2
1r cm2 = 77 cm2
Area of the remaining part = Area of
rectangle ABCD - Area of semi-circle
= (280 - 77) cm2 = 203 cm2
Ex. 86: T h e c i r c um f e r e n ce s o f t w o ci r c l e s a r e
i n t h e r a t i o 2 : 3 . Fi n d t h e r a t i o o f t h e i r
a r e a s .
S o l n : Let r 1 and r
2 be the radii of two given circles
and C 1 and C
2 be their circumference. Then,
C1 = 2r
1 and C
2 = 2r
2
Now, C1 : C
2 = 2 : 3
or, 3
2
C
C
2
1 or 3
2
2
2
2
1 r
r or 3
2
2
1 r
r
Let A1 and A
2 be the areas of two circles.
Th en ,
A1 = 2
1r and A2 = 2
2r
9
4
A
A22
21
22
21
2
1
r
r
r
r
9
4
3
222
21
2
1
r
r
r
r
A1 : A
2 = 4 : 9
Hence, the areas of two given circles are
in the ratio 4 : 9.
Ex. 87: T h e a r e a s o f t w o c i r c l es a r e i n t h e
r a t i o 1 6 : 2 5 . Fi n d t h e r a t i o of t h e i r
c i r c um f e r e n c e s .
Soln: Let r 1 and r
2 be the radii of two circles and
let their areas be A1 and A
2 respectively.
Th en ,
211A r and 2
22A r
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531Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
Now, A1 : A
2 = 16 : 25 [Given]
or, 22
21 : r r = 16 : 25
or, 25
1622
2
1
r
r or 2
2
22
2
15
4
r
r
5
4
2
1 r
r ..... (i)
[Taking square root of both sides]
Let C1 and C
2 be the circumferences of
two circles. Then,
C1 = 2r
1 and C
2 = 2r
2.
2
1
2
1
2
1
2
2
C
C
r
r
r
r
or, 5
4
C
C
2
1 .... [Using (i)]
C1 : C2 = 4 : 5Hence, the circumferences of the two
circles are in the ratio 4 : 5.
Ex. 88: A squ a r e p a r k h a s ea c h s i d e of 1 0 0 m .
A t e a ch c or n e r o f t h e p a r k , t h e r e i s a
f l o w e r b ed i n t h e f or m o f a q u a d r a n t
o f r a d i u s 1 4 m a s sh ow n i n t h e f i g u r e
g i v e n b el o w . F i n d t h e a r e a o f t h e
r e m a i n i n g pa r t o f t h e p a r k .
(Tak e7
2 2 ).
Soln:
Area of each quadrant of radius 14 m
= 14147
22
4
1
4
1 2 r [ r = 14]
= 154 m2
Area of 4 quadrants = (4 × 154) m2
= 616 m2.
Area of square park having side 100 m
long = (100 × 100) m2 = 10,000 m2
Hence, area of the remaining part of the
park = (10000 - 616 =) 9384 m2.Ex. 89: Fou r e q u a l c i r c l e s a r e d e s cr i b ed a b o u t
t h e f o u r c or n e r s o f a s q u a r e s o t h a t
e a ch t o u c h e s t w o o f t h e o t h e r s a s
s h ow n i n f i g u r e . F i n d t h e a r ea o f t h e
shaded r eg i on , each si de o f t he squa r e
m e a su r i n g 1 4 c m .
Soln: Let ABCD be the given square each side
of which is 14 cm long. Clearly, the radius
of each circle is 7 cm.
We have:
Area of the square of side 14 cm long
= (14 × 14) cm2 = 196 cm2
Area of each quadrant of a circle of radius
7 cm
=
22 7
7
22
4
1
4
1r cm2
= 38.5 cm
2
Area of 4 quadrants
= 4 × 38.5 cm2 = 154 cm2.
Hence, area of the shaded region
= Area of the square ABCD – Area of 4
quadrants
= (196 – 154) cm2 = 42 cm2.
Ex. 90: An a r c sub t ends an an g l e o f 36º a t t he
cen t r e of a c i r c l e o f r a d i u s 3 . 6 cm , f i n d
t h e l e n g t h o f t h e a r c .
Soln: We know that the length of an arc of a
circle of radius r is given by
ncecircumfere
360
D = r 2
360
D
Here, D = 36° and r = 3.6 cm
Length of the arc
=
6.3
7
222
360
36cm = 2.26 cm.
Ex. 91: A se ct o r i s cu t f r om a c i r c l e o f r a d i u s
21 cm. T he an g le o f t h e sec to r i s 15 0°.
F i n d t h e l en g t h o f i t s a r c a n d a r e a .
Soln: The arc length l and area A of a sector of
angle D° in a circle of radius r are given
by l = r 360
D and
A =2
360
Dr respectively.
Here, r = 21 cm and D = 150°
l =
217
222360
150 cm = 55 cm
And, A =
2
217
22
360
150cm2
=
2
1155 cm2 = 577.5 cm2
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532 Concep t o f A r i t hm e t i c
K KUNDAN
Ex. 92: The a r ea o f a sect o r o f a c i r c l e i s t h 1 0
1
o f t h e a r ea o f t h e c i r c l e. F i n d t h e a n g l e
o f t h e s ec t o r .S o l n : Let the radius of the circle be r cm and the
sector angle be of x°. Then,
Area of the sector =
2
º360r
x cm2
and, Area of the circle = r 2 cm2
It is given that:
Area of the sector =10
1× Area of the circle
or,22
10
1
º360
ºr r
x
or, x = 2
2 360
10
1
r r
or, x = 36
Hence, the sector angle is of 36°.
Ex. 93: A 36°sec t o r o f a c i r c l e h a s a r e a 3 . 8 5
cm 2 . W h a t i s t h e l e n g t h o f t h e a r c o f
t h e s ec t o r ?
Soln: Let r cm be the radius of the circle. We
have, sector angle = 36° and area of the
sector = 3.85 cm2.
We have,
Area of the sector
=
circletheof Area
360º
angleSector
or, 3.85 =2
7
22
º360
º36r
or,2236
736085.32
r
or, r 2 = 12.25
or, r = 25.12 cm = 3.5 cm
Now,
Area of the sector
=2
1× length of the arc × radius
or, 3.85 = 3.5arctheof length2
1
or, length of the arc
= 5.3
285.3
cm = 2.2 cm
Ex. 94: I n t h e f i g u r e g i v en b e l o w a r e s h o w n
s ec t o r s o f t w o c o n c en t r i c c i r c l e s o f
r a d i i 7 cm a n d 3 . 5 cm . F i n d t h e a r e a
o f t h e sh a d e d r e gi o n .
(Use7
2 2 )
Soln: Let A1 and A
2 be the areas of sectors OAB
and OCD respectively. Then,
A1 = Area of a sector of angle 30° in a
circle of radius 7 cm.
=
27
7
22
360
30 cm2
2
360
D Using A r
=
6
77cm2
Area of a sector of angle 30° in a circle of
radius 3.5 cm.
=
2
5.37
22
360
30cm2
=
2
7
2
7
7
22
12
1cm2 =
24
77cm2
Area of the shaded region = A1 – A
2
=
24
77
6
77cm2 = 14
24
77 cm2
=8
77cm2 = 9.625 cm2
Ex. 95: The pe r im e t er o f a sec t o r o f a c i r c l e o f r a d i u s 5 . 2 cm i s 1 6 . 4 cm . F i n d t h e a r ea
o f t h e s ec t o r .
Soln: Let OAB be the given sector. Then
Perimeter of sector OAB = 16.4 cm
or, OA + OB + arc AB = 16.4 cm
or, 5.2 + 5.2 + arc AB = 16.4or, arc AB = 6 cm
Area of sector OAB
=2
1 × length of the arc × radius
=
2.56
2
1cm2 = 15.6 cm2
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533Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
Ex. 96: T h e a r ea o f a n e qu i l a t e r a l t r i a n g l e i s
3 4 9 cm 2 . T a k i n g e a ch a n g u l a r p o i n t
a s c en t r e , a c i r c l e i s d e s cr i b ed w i t h
r a d i u s eq u a l t o h a l f t h e l en g t h o f t h e
s i d e o f t h e t r i a n g l e a s s h own i n f i g u r e .
F i n d t h e a r e a o f t h e t r i a n g l e n o t
i n c l u d e d i n t h e c i r c l e.
Soln: Let each side of the triangle be a cm. Then,
Area = 349 cm2 or 349
4
3 2 a
2
side4
3 Area
or, a 2 = 49 × 4
a = 14 cm
Thus, radius of each circle is 7 cm.
Now, required area
= Area of ABC – 3 × (Area of a sector of
angle 60° in a circle of radius 7 cm)
=
27
7
22
360
603349 cm2
= ]77349[ cm2
= [49 × 1.73 – 77] cm2 = 7.77 cm2
Ex. 97: The l en g t h o f m i n u t e-h a n d o f a c l o c k i s 1 4 c m . F i n d t h e a r e a s w ep t b y t h e
m i n u t e-h a n d i n o n e m i n u t e.
(Use7
2 2 ).
Soln: Clearly, minute-hand of a clock describes
a circle of radius equal to its length ie 14
cm.
Since the minute-hand rotates through 6°
in one minute, therefore, area swept by
the minute hand in one minute is the area
of a sector of angle 6° in a circle of radius
14 cm.
Hence, required area =2
360
anglesectorr
=
214
7
22
360
6cm2
=
1414
7
22
60
1cm2
=15
154cm2 = 10.26 cm2
Ex. 98: T h e m i n u t e-h a n d o f a c l o ck i s 1 0 c m
l o n g . F i n d t h e a r e a o n t h e f a c e o f t h e
c l o c k d e s cr i b ed b y t h e m i n u t e -h a n d
b et w e en 9 am a n d 9 : 3 5 am .
Soln: Angle described by the minute-hand inone minute = 6º.
So, angle described by the minute-hand
in 35 minutes = (6 × 35)° = 210°
Area swept by the minute-hand in 35
minutes = Area of a sector of angle 210°
in a circle of radius 10 cm
=
2
107
22
360
210cm2
= 183.3 cm2
Ex. 99: T h e s h o r t a n d l o n g h a n d s o f a c l o ck
a r e 4 cm a n d 6 cm l o n g r e s p ec t i v e l y .
F i n d t h e s um o f d i s t a n c es t r a v el l e d b y
t h e i r t i p s i n 2 d a y s .
(Tak e 7
2 2 ) .
Soln: In 2 days, the short hand will complete 4
rounds.
Distance moved by its tip = 4
(circumference of a circle of radius 4 cm)
=
4
7
2224 cm =
7
704cm
In 2 days, the long hand will complete 48
rounds.
Distance moved by its tip
= 48 (circumference of a circle of
radius 6 cm)
=
6
7
22248 cm =
2
12672cm
Hence, the sum of the distances moved
by the tips of two hands of the clock
=
7
12672
7
704cm = 1910.57 cm
Ex.100:F i n d t h e a r e a o f a s egmen t o f a c i r c l e
o f r a d i u s 2 1 c m i f t h e a r c of t h e
segm ent h as m easu r e 60°.
Soln: Let O be the centre of the circle and PXQ
the arc of the segment such that m(PXQ)
= 60º
POQ = 60º
Now,
Area of sector POQ
=
21217
22
º360
º60cm2
=
21322
3
1cm2
= 231 cm2
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534 Concep t o f A r i t hm e t i c
K KUNDAN
Since OP = OQ and POQ = 60º.
Therefore, OPQ is an equilateral triangle.
Therefore,
Area of OPQ = 2side4
3
=
2121
4
3cm2
=4
44173.1 cm2 = 190.73 cm2
Area of segment PXQ
= Area of sector OPQ – Area of OPQ
= (231 – 190.73) cm2 = 40.27 cm2.
Ex.101: I f t h e a r c o f a s egmen t o f a c i r c l e h a s
m easure 12 0°. I f t h e ra d i us o f t he c i r c l e
i s 6 cm , f i n d t h e a r e a o f t h e se gmen t .
Soln: Let O be the centre of the circle and PXQ
the arc of the segment such that m (PXQ)
= 120°.
POQ = 120°
Now, Area of sector POQ =2
º360
º120r
=
66
7
22
º360
º120cm2
=
667
22
3
1
cm2
=7
264cm2
Let ONPQ. Then OPN is right angled
triangle, right angled at N and OPN =
30°.
In PNO,
sin 30° =OP
ON
ON = sin 30° × OP
ON =2
1(Hypotenuse)
[since, sin 30° =2
1]
=
6
2
1OP
2
1cm = 3 cm
In OPN, we have
222 PNONOP
222 ONOPPN
or, PN = 22 PNOP
or, PN = 936 = 27 cm = 33 cm
PQ = 2PN = 332 cm = 36 cm
Area of OPQ =2
1× PQ × ON
=22 cm39cm336
2
1
Hence,
Area of segment PXQ = Area of sector OPQ
– Area of OPQ
=
39
7
264cm2
= (38.142 – 15.588) cm2
= 22.554 cm2 ]732.13[
Ex.102:A ho r s e i s p l a c e d f o r g r a z i n g i n s i d e a
r e ct a n g u l a r f i e l d 7 0 m b y 5 2 m a n d i s
t e t h e r e d t o o n e co r n e r b y a r o p e 2 1 m
l o n g . On how m uch a r e a ca n i t g r a z e ?
Soln: Shaded portion indicates the area which
the horse can graze. Clearly, shaded area
is the area of a quadrant of a circle of
radius r = 21 m.
Hence, required area =2
4
1r
=
2
217
22
4
1cm2
=2
693cm2 = 346.5 cm2
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535Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
Ex.103: PQRS i s a d i am e t er o f a c i r c l e o f r a d i u s
6 cm . T h e l e n g t h s PQ , QR and RS a r e
equa l . Sem i -c i r c l es a r e d r aw n on PQ and
QS a s d i am et e r s a s s h own i n t h e f i g u r e
g i v e n be l ow . Fi n d t h e p er i m e t e r o f t h e s h a d e d r eg i o n .
Soln: We have PS = diameter of a circle of radius
6 cm = 12 cm
cm43
12RSQRPQ
QS = QR + RS = (4 4) cm = 8 cm
Hence, required perimeter
= Arc of semi-circle of radius 6 cm + Arc
of semi-circle of radius 4 cm + Arc of semi-circle of radius 2 cm
cm12cm)246(
Ex.104: I n t h e f i g u r e g i v en b e l o w , A O B CA
r e p r e sen t s a q u a d r a n t o f a c i r c l e o f
r a d i u s 3 . 5 cm w i t h c en t r e O . Ca l c u l a t e
t h e a r e a o f t h e s h a d e d p o r t i o n .
(Take7
2 2 ).
Soln: Area of quadrant AOBCA =2
4
1r
= 25.37
22
4
1
=8
77
2
7
2
7
7
22
4
1 cm2 = 9.625 cm2
Area of AOD = HeightBase2
1
= OBOA2
1
= 2cm25.32
1 = 3.5 cm2
Hence, area of the shaded portion
= Area of quadrant – Area of AOD
= (9.625 – 3.5) cm2 = 6.125 cm2
Ex.105: F i n d t h e a r e a o f t h e s h a d e d r eg i o n i n
t h e f i g u r e g i v e n b el ow .
Soln: Clearly, radius of the bigger semi-circle
= 14 cm
Area of the bigger semi-circles
= 222 cm147
22
2
1
2
1r = 308 cm2
Radius of each of the smaller circle
= 7 cm
Area of 2 smaller semi-circles
=222 cm154cm7
7
22
2
12
Hence, required area
= (308 + 154) cm2 = 462 cm2
Ex.106: ABCD i s a f l owe r bed . I f OA = 21 m and
OC = 14 m , f i n d t h e a r ea o f t he bed .
(Take7
2 2 ).
S o l n : We have OA = R = 21 m and
OC = r = 14 m.
Area of the flower bed = area of a
quadrant of a circle of radius R – Area of a
quadrant of a circle of radius r
2222
44
1
4
1r R r R
222 cm14217
22
4
1
m14andm21 r R
2m142114217
22
4
1
22 m5.192m7357
22
4
1
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536 Concep t o f A r i t hm e t i c
K KUNDAN
Ex.107: ABCP i s a q u a d r a n t o f a c i r c l e o f r a d i u s
1 4 c m . W i t h A C a s d i a m et e r , a s em i -
c i r c l e i s d r a w n . F i n d t h e a r e a o f t h e
s h a d e d p o r t i o n .
Soln: In the right-angled triangle ABC,
We have AC2 = AB2 + BC2
or, AC2 = 142 + 142
or, 214142AC 2 cm
or, 272
214
2
AC
cm
Now, required area = Area APCQA
= area ACQA – Area ACPA
= Area ACQA – (Area ABCPA –
Area of ABC)
= (Area of semi-circle with AC as
diameter) – [Area of a quadrant of
a circle with AB as radius – Area of
ABC]
14142
114
7
22
4
1
277
22
2
1
2
2
cm2
1414
2
1
1414
7
22
4
1249
7
22
2
1
cm2
22 cm98cm 98154154
Ex.108: I n a n e q u i l a t e r a l t r i a n g l e of s i d e 2 4 cm ,
a c i r c l e i s i n s c r i b e d t o u c h i n g i t s s i d e s.
F i n d t h e a r e a o f t h e r ema i n i n g p o r t i o n
o f t h e t r i a n g l e.
(Tak e 7 3 2 .1 3 ).
Soln: Let ABC be an equilateral triangle of side
24 cm, and let AD be perpendicular from
A on BC. Since the triangle is equilateral,
so D bisects BC.
BD = CD = 12 cm
The centr e of the inscr ibed ci rcle wi ll
coincide with the centroid of ABC
Therefore, AD31OD
In ABD, we have
222 BDADAB [Using Pythagoras Theorem]
or, 222 12AD24
or, 22 1224AD
12241224
3121236 cm
OD = AD3
1 cm
= 343123
1
Area of the incircle
= 222 cm347
22OD
22 cm85.150cm487
22
Area of the triangle ABC
= 2224
4
3
4
3 side
= 279.4 cm2
Area of the remaining portion of thetriangle
2cm85.1504.249
2cm55.98
Ex.109: Two c i r c l e s t o u c h e x t e r n a l l y . T h e s um
o f t h e i r a r ea s i s 1 3 0 s q cm a n d t h e
d i s t ance be tween t h e i r cen t r e s i s 14 cm .
F i n d t h e r a d i i o f t h e c i r c l es .
Soln: Note that if two circles touch externally,
the distance between their centres is equal
to the sum of their radii.
Let the radii of the two circles be r 1 cm
and r 2 cm respectively.
Let C1 and C
2 be the centres of the given
circles. Then,2121CC r r
or, 2114 r r (given)cm14CC 21
or, 1421 r r .... (i)
It is given that the sum of the areas of two
circles is equal to 130 cm2.
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537Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
13022
21 r r
or, 13022
21 r r .... (ii)
Now, 212
22
12
21 2 r r r r r r
or, 212 213014 r r [Using (i) and (ii)]
or, 212130196 r r
3321 r r ... (iii)
Now, 212
22
12
21 2 r r r r r r
or, 332130221 r r
[Using (ii) and (iii)]
or, 64221 r r
821 r r ... (iv)
Solving (i) and (iv), we get
r 1 =11 cm and r
2 = 3 cm
Hence, radii of the two circles are 11 cm
and 3 cm.
Ex.110: I n t h e g i v e n f i g u r e , a s em i c i r c l e i s
d r a w n w i t h s egm e n t P R a s d i a m et e r .
Q i s t h e m i d -p o i n t o f s egmen t PR , t w o
sem i c i r c l e s w i t h segmen t PQ and QR as
d i am et e r s a r e d r awn . A ci r c l e i s d r awn
wh i c h t o u c h e s t h e t h r ee sem i c i r c l e s . If PR = 28 cm , f i n d t he a r ea o f t h e shad ed
r e g i o n .
[ = 3.14]
Soln: Let A be the centre of the circle touching
the three semicircles at points D, E and F
respectively.
Let r be the radius of the circle
Then PR6
1
r [From Geometry]
286
1r cm =
3
14cm
Now area of shaded region
= Area of semi-circle PDR – Area of semi-
circle PEQ – Area of QFR – Area of circle
with centre A
= 2
222
3
147
2
17
2
114
2
1
=
2
196
2
49
2
4998
=
9
19649
=9
770
9
3522
9
245
7
22
= 85.56 cm2.
Ex.111: F i n d t h e a r e a o f a r i g h t -a n g l e d t r i a n g l e,
i f t h e r a d i u s o f i t s c i r c u m -c i r l c e i s 5
c m a n d t h e a l t i t u d e d r a w n t o t h e
h y p o t e n u es i s 4 cm .
Soln: We know that the circumcentre of a right-
angled triangle is the mid-point of its
hypotenues and the circum-radius is half
of the hypotenues.
Let ABC be the given triangle with right-
angle at B. Let O be the mid-point of
hypotenues AC. Let BD be the
perpendicular from B on AC. Then,
AC = 2(OA) = 2 × 5 = 10 cm
[OA=radius of the circumcirlce=5 cm]
and, BD = 4 cm (given)
Area of ABC =2
1(Base × Height)
= BDAC2
1
= )410(2
1 cm2
= 20 cm2.
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538 Concep t o f A r i t hm e t i c
K KUNDAN
1. Calculate the area of a rectangle 23 metres 7
decimetres long and 14 metres 4 decimetres
8 centimetres wide.
2. Find the diagonal of a rectangle whose sides
are 12 metres and 5 metres.
3. How many metres of carpet 75 cm wide will
be required to cover the floor of a room which
is 20 metres long and 12 metres broad?
4. How many paving stones, each measuring2
12
metres by 2 metres, are required to pave a
rectangular courtyard 30 metres long and
2
116 metres broad?
5. A hall room, 39 m 10 cm long and 35 m 70
cm broad, is to be paved with equal square
tiles. Find the largest tile which will exactly
fit and the number required.
6. A wire is in the shape of a square of side 10
cm. If the wire is rebent into a rectangle of
length 12 cm, find the breadth. Which
encloses more area, the square or the
rectangle?
7. The area of a square and a rectangle are equal.
If the side of the square is 40 cm and the
breadth of the rectangle is 25 cm, find the
length of the rectangle. Also find the perimeter
of the rectangle.8. A map is drawn to a scale of 120 cm to the
km. How many square cm on the map will
represent a hectare of ground?
9. Find the width of a roller which traverses
128 sq km while cutting 6.4 hectares of grass.
10. The diagonal of a rectangular field is 15 m
and its area is 108 m2. What will be the cost
of fencing this field if the cost of fencing for
one metre is Rs 5.
11. A strip of paper 2.2 km long and .075 mm
thick is rolled up into a solid cylinder.
Assuming the area of a circle to be7
13 times
the square of its radius, find approximately
the radius of the circular ends of the cylinder.
12. A square field containing 31684 square metresis to be enclosed with wire placed at heights
1, 2, 3 and 4 metres above the ground. What
length of the wire will be needed, if the length
required for each circuit is 5% greater than
the perimeter of the field?
13. The area of a rectangular field is 27000 square
metres and the ratio between its length and
breadth is 6 : 5. Find the cost of the wire
required to go four times round the field at Rs
740 per kilometre of length of the wire.
14. A rectangular park is 100 metres long and 80
metres wide. There are two paths, each 5
metres wide, in the middle of the park
running one parallel to the length and the
other parallel to the width of the park. Find,
(i ) the area of the paths,
( ii) the expenditure involved in constructing
the paths at 25 paise per square metre,
and
( iii) the expenditure involved in laying grass
in the remaining portion of the park at5 paise per square metre.
15. A rectangular field 150 metres long and 100
metres wide, has within it a 10 metres wide
uniform path running round it. Find,
(i) the area of the path, and
( ii) the cost of cultivating the remaining part
of the field at Rs 1.50 per square metre.
16. A school hall 20 m long and 15 m broad is
surrounded by a verandah of uniform width
of2
12 m. Find the cost of flooring the
verandah at Rs 2.50 per square metre.
17. A room is 8 m long and 6 m wide. It is
surrounded by a verandah. Find the width of
the verandah if it occupies 72 sq metres.18. A path 2 m wide, running all around outside
a square garden occupies 204 sq metres. Find,
(i) the length of the square garden.
( ii) the area of the part of the garden enclosed
by the path.
19. A square carpet is spread in the centre of a
hall 9 m square leaving some margin of equal
width all around. The total cost of carpeting
at Rs 2.50 per sq m and decorating the margin
at 20 paise per sq m is Rs 163.40. Find the
width of the margin.
20. A rectangular field is 200 metres long and
121 metres broad. It is planted with trees in
rows perpendicular to the length, one metre
from row-to-row, and one metre from tree-to-
tree in the same row. If the width of a metreall-round the field remains unplanted, find
the number of trees.
21. A path 2 metres wide running all-round a
square garden has an area of 9680 sq metres.
Find the area of the part of the garden enclosed
by the path.
22. A marginal walk all-round the inside of a
Practice Exercise
Exercise–1
(Rectangle, Square and Area of Path)
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rectangular space 37 m by 30 m occupies 570
sq metres. Find the width of the walk.
23. A garden, whose length is 22 metres, has a
path 1.5 metres wide on the two sides and at
one end. If it costs Rs 2460 to turf theremainder at Rs 15 a sq metre, what is the
width of the garden?
24. In the centre of a room 10 metres square, there
is a square of turkey carpet, and the rest of
the floor is covered with oilcloth. The carpet
and the oilcloth cost respectively Rs 150 and
Rs 65 per square metre, and the total cost of the carpet and the oilcloth is Rs 13385. Find
the width of the oilcloth border.
Exercise–2(Triangle)
7. Find the area of an isosceles right triangle,
the length of whose each side containing the
right angle is 15 cm.
8. The sides of an equilateral triangle is 8 cm.
Find its area and the height.
9. From a point in the interior of an equilateral
triangle, perpendiculars drawn to the three
sides, are 16 cm, 20 cm and 22 cm
respectively. Find the area of the triangle.10. A triangular park ABC has sides 120 m, 80 m
and 50 m (see figure). A gardener Dhania has
to put a fence all round it and also plant grass
inside. How much area does he need to plant?
Find the cost of fencing it with barbed wire
at the rate of Rs 20 per metre leaving a space
3 m wide for a gate on one side.
1. The base of triangular field is 880 metres and
its height 550 metres. Find the area of the
field. Also calculate the charges for supplying
water to the field at the rate of Rs 242.50 per
sq hectometre.
2. The base of a triangular field is three times
its height. If the cost of cultivating the field
at Rs 1505.52 per hectare is Rs 20324.52,
find its base and height.3. Find the area of 1 triangular field whose sides
are 50 metres, 78 metres, 112 metres
respectively and also find the perpendicular
from the opposite angle on the side 112
metres. If it is lent at Rs 10000 per hectare,
find the rent of the field.
4. X is a point on side CD of a square ABCD
such that CX = 5 cm. If area of the triangle
ADX is 43 cm2, find the length of the side of
the square.
5. Find the area of a triangle, one of whose angles
is 90°, hypotenuse is 12.5 cm and the base is
7.5 cm.
6. The area of a tri angle equals the area of a
square whose side is 45 m. Find the length
of the side of the triangle which is 75 m from
the opposite vertex.
3. Find the area of a quadrilateral piece of ground
ABCD in which AB = 85 metres, BC = 143
metres, CD = 165 metres, DA = 85 metres
and DB = 154 metres.
Exercise–3(Quadrilateral, Parallelogram, Rhombus, Trapezium and Regular Polygon)
1. Find the area of the quadrilateral ABCD in
which the diagonal DB = 10 m and the
perpendiculars AL and CM drawn on it from
A and C are respectively 4 m and 6 m. (These
perpendiculars are called offsets).
2. The sides of a parallelogram are 40 m and 30
m respectively and its diagonal is 50 m. Find
its area.
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540 Concep t o f A r i t hm e t i c
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Exercise–4
(Circle)
10. Find the ratio of area of a square inscribed in
a semi-circle of radius r to the area of another
square inscribed in the entire circle of radius
r .
11. Find to the three places of decimals the radius
of the circle whose area is the sum of the
areas of two triangles whose sides are 35,
53, 66 and 33, 56, 65 measured in
centimetres.
(Take = 22/7).
12. PQRS is a diameter of a circle of radius 6 cm.
The length PQ, QR and RS are equal. Semi-
circles are drawn on PQ and QS as diameters
as shown in the figure below. Find the area
of the shaded region.
13. A square water tank has its side equal to 40
m. There are four semi-circular grassy plots
all round it. Find the cost of turfing the plots
at Rs 1.25 per sq cm.
(Use = 3.14)
14. A rectangular park is 100 m by 50 m. It is
surrounded by semi-circular flower beds all
around. Find the cost of levelling the semi-
circular flower beds at 60 per sq metre.
(Use = 3.14)
15. A park is in the form of a rectangle 120 × 100
m. At the centre of the park there is a circular
lawn. The area of the park excluding the lawn
is 8700 m2. Find the radius of the circular
lawn.
(Use = 22/7)
1. The radius of a circular wheel is4
31 m. How
many revolutions will it make in travelling
11 km?
2. The circumference of a circular garden is 1012
metres. Find the area. Outside the garden a
road of 3.5 metres width runs round it.
Calculate the area of this road and find the
cost of gravelling the road at Rs 32 per 100 sq
metres.
3. A bicycle wheel makes 5000 revolutions in
moving 11 km. Find the diameter of the wheel.
4. A boy is cycling such that the wheels of the
cycle are making 140 revolutions per minute.
If the diameter of the wheel is 60 cm, calculate
the speed per hour with which the boy is
cycling.
5. The diameter of the wheel of a bus is 140 cm.
How many revolutions per minute must the
wheel make in order to keep a speed of 66
km per hour?
6. A copper wire, when bent in the form of a
square, encloses an area of 484 cm2. If the
same wire is bent in the form of a circle, find
the area enclosed by it.
(Use = 22/7)
7. A wire is looped in the form of a circle of
radius 28 cm. It is re-bent into a square form.
Determine the length of the side of the square.
8. A bucket is raised from a well by means of a
rope which revolves round a wheel of diametre
75 cm. If the bucket ascends in 1 minute 12
seconds with a uniform speed of 1.3 m per
second, find the number of complete
revolutions made by the wheel in raising the
bucket.
9. The radius of circle is 20 cm. Three more
concentric circles are drawn inside it in such
a manner that it is divided into 4 equal parts.
Find the radius of the smallest circles?
4. The perimeter of a rhombus is 146 cm and
one of its diagonals is 55 cm. Find the other
diagonal and the area of the rhombus.
5. The parallel sides of a field, which is in the
shape of a trapezium, are 20 m and 41 m andthe remaining two sides are 10 m and 17 m.
Find the cost of levelling the field at the rate
of Rs 30 per sq metre.
6. Ratio between the parallel sides of the
trapezium is 1:3, while ratio between
unparallel sides of the trapezium is 2:3. Ratio
between bigger parallel and unparallel sides
is 2:1. If height of the trapezium is4
1515,
then find the area of the trapezium?
7. A regular hexagon of side 6 cm is inscribedin a circle. Find the area of the region in the
circle which is outside the hexagon.
[Use = 3.14, 732.13 ]
8. The diagonals of a rhombus are 8 cm and 6
cm, find the sides and the area.
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541Mensu r a t i o n I (A rea an d Per i me t e r )
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in the figure. Find the area of the shaded
region.
[ = 22/7, 3 = 1.73]
21. In a circle of radius 21 cm an arc subtends
an angle of 60º at the centre. Find
(i) the length of the arc,
(ii) the area of the sector bounded by the arc
and
( iii) the area of the segment made by this arc.
22. A circular swimming pool is surrounded by a
circular path which is 4 m wide. If the area
of the path is25
11th part of the area of the
swimming pool, then find the radius of the
swimming pool (in metres).
23. If the circumference of a circle is 80 cm, then
find the side of a square inscribed in the circle.
24. In the figure given below, square OABC is
inscribed in a sector OPBQ. If OC = 20 cm,
find the area of the shaded region.
(use = 3.14)
16. In the figure given below, ABCD is a rectangle.
The radius of the semic ircles drawn on AD
and BC as diameters and radius of circle drawn
in between is the same. If BC = 7 cm, find
the area of the shaded region.
17. An athletic track 14 m wide consists of two
straight sections 120 m long joining semi-
circular ends whose inner radius is 35 m.
Calculate the area of the shaded region.
18. In the given figure, the centre of the circle is
A and ABCDEF is a regular hexagon of side 6
cm. Find the following:
(i) Area of segment BPF
(ii) Area of the shaded portion.
19. Quadrilateral ABCD is a rectangle. Sectors
with centre C and D are drawn as shown in
the figure. If AB = 21 cm, CB = 14 cm, find
the area of the shaded portion.
20. An equilateral ΔABC has each of its sides
14 cm with each of its vertices as centres,
and radius as 7 cm, arcs are drawn as shown
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542 Concep t o f A r i t hm e t i c
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25. In the given figure, cresent is formed by two
circles which touch at the point A. O is the
centre of the bigger circle. If CB = 9 cm and
DE = 5 cm, find the area of the shaded
portion.
26. Two circles touch internally. The sum of their
areas is 116 sq cm and the distance between
their centres is 6 cm. Find the radii of the
circle.
27. Find the area of the shaded portion of the
given diagram. Give your answer correct to
three significant figures.
28. Find the area of the shaded portion in the
given figure, where the arcs are quadrants of
a circle.
29. Find the area of the shaded portion in the
figure given below:
[use = 22/7]
30. In a right ABC, A = 90°, AB = 4 cm, AC = 3
cm. On its three sides as diameters, three
semi-circles are drawn as shown in the figure
given below. Find area of shaded parts.
31. The area of the shaded circular ring is 770
sq cm and the difference between the radii of
the two circles is 7 cm. Find the area of the
unshaded region.
32. In the figure given below, ABC is an
equilateral triangle of side 7 cm and segment
BC is diameter of the semicircle. Find the
area of the shaded region.
( 3 = 1.732)
33. In the given diagram AC is a diameter of a
circle with radius 5 cm. If AB = BC and CD =
8 cm, calculate area of the shaded region.
34. In the figure as mentioned below, POQ and
ROS are diameters of a circle with centre O
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543Mensu r a t i o n I (A rea an d Per i me t e r )
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42. Find the area of the shaded region in the
given figure, where ABCD is a square of side
14 cm.
43. Find the area of the shaded region in the
given figure, where ABCD is a square of side
10 cm and semi-circle are drawn with each
side of the square as diameter.
(use = 3.14)
44. The circumference of a circle is 16.8 cm more
than its diameter. What will be the radius of
the circle?
(use = 3.14)
45. A square has been inscribed in a circle. What
will be the ratio of the areas of circle and the
square?
46. The length of sides AB, BC and CA of a ABC
are 4 cm, 6 cm and 8 cm respectively. Three
sectors of circles drawn with centres A, B, Cand each with one centimetre of radius,
starting and terminating with the sides of
the triangle are cut off. Find the area of the
remaining position of the triangle.
47. In a circle of radius 28 cm, an arc subtends
an angle of 72° at the centre. Find the length
of an arc and the area of the sector so formed.
48. The radii of three concentric circles are in
the ratio 1 : 2 : 3. Find the ratio of the area
between the two inner circles to that between
the two outer circles.
and radius 14 cm. Find the area of shaded
region.
(use = 22/7).
35. A rectangular field is surrounded by four
semicircular flower-beds. If the length and
the breadth of the field are 6 m and 4 m
respectively, find the cost of raising the
flower-beds at the rate of Rs 8 per m2.
(Take = 3.14)
36. The length of the side of a square is 14 cm.
Taking ver tices of the square as centres, 4circles are drawn each with a radius of 7 cm.
Find the area of the region of the square that
remains outside the region of any of the
circles.
37. A brick, 5 cm thick (high), is placed against a
wheel to act for a stop. The horizontal distance
of the face of the brick stopping the wheel
from the point where the wheel touches the
ground is 15 cm. What is the radius of the
wheel?
38. The length of the minute-hand of a clock is
10 cm. What is the area swept by the minute-
hand in one minute?
(Use = 3.14)
39. A chord AB of a circle of radius 10 cm makes
a right angle at the centre of the circle. Findthe area of the major and minor segments.
(Take = 3.14).
40. A chord AB of a circle of radius 15 cm makes
an angle of 60° at the centre of the circle.
Find the area of the major and minor segment.
(Take = 3.14, 3 = 1.732).
41. In the given figure, two circular flower beds
have been shown in two sides of a square
lawn ABCD of side 56 m. If the centre of each
circular flower bed is the point of intersection
O of the diagonals of the square lawn, find
the sum of the areas of the lawn and the
flower beds.
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1. Length = 23.70 metres
Breadth = 14.48 metres
Area = (23.70 × 14.48) square metres
= 343.176 square metres
2. Length of the diagonal
= 22 512 metres = 169 metres
= 13 metres
3. Area of carpet = Area of the floor
= (20 × 12) sq metres
Width of carpet = 75 cm =4
3metre
Length of carpet =
4
3
1220 = 320 metres
4. Area of courtyard =2
11630 sq m
= 495 sq metres
Area of each paving stone =
2
2
12 sq m
Number of stones required =5
495 = 99
5. 39 m 10 cm = 3910 cm
35 m 70 cm = 3570 cm
The side of the largest square tile
= HCF of 3910 and 3570= 170 cm
= 1 m 70 cm
Number of tiles =170170
35703910
= 483
6. Side of the square = 10 cm
Length of the wire = Perimeter of the square
= 4 × side = (4 × 10 =) 40 cm
Length of the rectangle = l = 12 cm. Let b be
the breadth of the rectangle.
Perimeter of the rectangle = Length of wire
= 40 cm
Perimeter of rectangle = 2 (12 + b )
Thus, 40 = 2 (12 + b )
or, 20 = 12 + b
b = (20 – 12 =) 8 cm The breadth of the rectangle = 8 cm.
Area of the square = (Side)2
= 10 cm × 10 cm = 100 cm2
Area of the rectangle = l × b = 12 cm × 8 cm
= 96 cm2
So, the square encloses more area even though
its perimeter is the same as that of the rectangle.
7. Area of the square = (Side)2
= 40 cm × 40 cm = 1600 cm2
It is given that, the area of the rectangle
= The area of the square
Area of the rectangle = 1600 cm2
Breadth of the rectangle = 25 cm
Area of the the rectangle = l × b
or, 1600 = l × 25
l =25
1600 = 64 cm
Hence, the length of rectangle is 64 cm
Perimeter of the rectangle = 2 (l + b )
= 2 (64 + 25) cm
= 178 cm
Hence, the perimeter of the rectangle is 178
cm.
8. 1 hectare = 10000 sq m
1000 metres are represented by 120 cm.
(1000 × 1000) sq m are represented by
(120 × 120) sq m.
1 sq m is represented by10001000
120120
sq cm
10000 sq m are represented by
10001000
10000120120sq cm = 144 sq cm
9. 128 km = 128000 m
6.4 hectare = (6.4 × 10000) sq m
Imagining the grass area to be 128000 m long,
and as wide as the roller, we have
Width required =128000
100004.6 m =
2
1m = 50 cm
10. Let the length of the rectangle be x metres
and breadth be y metres.
Area of rectangular field = x × y = 108 m2 ... (i)
Area of rectangle = length × breadth
And 15
2
= x
2
+ y
2
or, 225 = x 2 + y 2 .... (ii)
In a right-angled triangle
Hypotenuse2 = Base2 + Height2
From equations (i) and (ii)
(x + y )2 = x 2 + y 2 + 2xy
= 225 + 2 × 108 = 441 = (21)2
or, x + y = 21
Answers and explanations
Exercise–1
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Now, the perimeter of the field
= 2(length + breadth)
Perimeter of field = 2(x + y)
= 2 × 21 = 42 metres
Cost of fencing the field = (42 × 5 =) Rs 21011. By rolling up a rectangular piece of paper it
will be seen that the area of the longer edge
of the paper rolled into solid cylinder is equal
to that of the circular end of the cylinder.
2.2 km = 2.2 × 1000 × 100 cm
.075 mm = .0075 cm
(radius)2 ×7
13
= 2.2 × 1000 × 100 × 0.0075 sq cm
(radius)2 =
22
7
1000010
75100100022 sq cm
radius = 22.91 cm (Approx.)
12. Area of the field = 31684 sq m
perimeter = 431684 metres
= 178 × 4 metres
length of each circuit = 178 × 4 ×100
105 m
Since the wire goes round 4 times.
total length of wire required
= 178 × 4 ×100
105× 4 m = 2990.4 m
13. Let the length be 6x m and the breadth be 5x m.
area = )56( x x sq m = 27000 sq m
or, 30x 2 = 27000
or, x 2 = 900
x = 30
Hence length = 180 m and breadth = 150 mLength of wire required to go round the field
four times = [4× 2(180 + 150)] m = 2.64 km
required cost = Rs (2.64 × 740)
= Rs 1953.60
14. (i ) The area of the path HG
= (100 × 5 =) 500 sq m
The area of the path EF
= (80 × 5 =) 400 sq m
Area of the shaded portion
= (5 × 5 =) 25 sq m
The shaded port ion is common to both
the roads, so while finding the actual area
of the paths, we should subtract this
common area from the sum of the areas of
the two paths.
Area of the paths
= (500 + 400 – 25 =) 875 sq units.
( ii) The expenditure involved in constructingthe paths
= Rs 875100
25 = Rs
4
875 = Rs 218.75
(iii) Area of the park = 100 × 80 = 8000 sq m
Area of the remaining portion of the park
= (8000 – 875 =) 7125 sq m
The expenditure of laying grass in the
remaining portion of the park
= Rs 7125100
5 = Rs 356.25
15. (i ) Area of the field = 150 × 100 = 15000 sq m
Length of the inner rectangle
= {150 – (10 + 10) =} 130 m
Width of the inner rectangle
= {100 – (10 + 10) =} 80 m
Area of the inner rectangle
= (130 × 80 =) 10400 sq m
Area of the path
= (15000 – 10400 =) 4600 sq m( ii) The area of the remaining part of the field
to be cultivated = 10400 sq m
Cost of cultivating the remaining part of
the field = Rs 104002
3 = Rs 15600
16. The area of the school hall ABCD
= (15 × 20 =) 300 sq m
The length of the rectangular region PQRS
=
2
12
2
1220 25 m
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The width of the rectangular region PQRS
=
2
12
2
1215 m = 20 m
Area of the rectangular region PQRS
= (25 × 20 =) 500 sq m Area of the verandah
= (500 – 300) sq m = 200 sq m
Cost of flooring the verandah
= Rs 200 ×2
5 = Rs 500
17. Let the width of the verandah be x metres.
Area of the room ABCD = (8 × 6 =) 48 sq m
Length of PQRS = (8 + x + x ) m = (8 + 2x ) m
Width of PQRS = (6 + x + x ) m = (6 + 2x ) m
Area of PQRS = (8 + 2x ) × (6 + 2x ) sq m
= (48 + 28x + 4x 2) sq m
Area of the room + area of the verandah
= (48 + 28x + 4x 2) sq m
2428487248 x x
or, 72284 2 x x
or, 1872 x x
or, 01872 x x
or, 018292 x x x
or, 0)9(2)9( x x x
or, 0)2)(9( x x
Either x = –9 or x = 2
x cannot be –9 because width of the verandah
cannot be negative
x = 2
Width of the verandah = 2 m
18. (i ) Let the length of the square garden ABCD
be x metres.
Area of the square garden ABCD
= (x × x =) x 2 sq m
Length of square PQRS
= )4(m)22( x x m
Area of square PQRS = (x + 4)2 sq m
(x + 4)2 = x 2 + 204
or, x 2 + 8x + 16 = x 2 + 204
or, 8x = 204 – 16
or,2
123
2
47
8
188 x m
The length of the square garden ABCD
=2
123 m
( ii) Area of the part of the garden enclosed by
the path, ie, of ABCD
=
4
2209
2
47
2
47 = 552.25 sq m
19. Let the width of the margin be x m.
Area of the square hall ABCD = (9 × 9 =) 81 sq m
Area of the square hall PQRS = (9 – 2x )2 sq m
= 81 – 36x + 4x 2 sq m
Area of the margin = 81 – (81 – 36x + 4x 2) sq m
= 81 – 81 + 36x – 4x 2 sq m
= 36x – 4x 2 sq m
Cost of the carpet = Rs 2
543681 2 x x
= Rs2
20180405 2x x
Cost of decorating the margin
= Rs 5
1436 2 x x = Rs
5
436 2x x
It is given that total cost of carpeting at Rs
2.50 per sq m and decorating the margin at
20 paise per sq m is Rs 163.40.
4.1635
436
2
20180405 22
x x x x
or, 10
1634
10
8721009002025 22
x x x x
or, 16348721009002025 22 x x x x
or, 039182892 2 x x
or, 03917824692 2 x x x
or, 46x (2x – 1) – 391(2x – 1) = 0
or, (46x – 391) (2x – 1) = 0
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548 Concep t o f A r i t hm e t i c
K KUNDAN
Solving equations (i) and (ii), we have
x = 81
The area of the square carpet is 81 sq
metres.
Therefore, the carpet is 9 metres in lengthand breadth. But the room is 10 metres in
length and breadth.
Hence double the width of the border is
(10 – 9 =) 1 metre
the width of the border =2
1metre = 5 dm
Alternative Method:
The area of the square room = 100 sq metres
The mean cost per sq metre = Rs
100
13385
= Rs 13.385
= 81 : 19
By the Alligation Rule, the area of the square
carpet is 81 sq metres. Therefore, the carpet
is 9 metres in length and breadth.
But room is 10 metres in length and breadth.
Hence double the width of the border is
(10 – 9 =) 1 metre
the width of the border =2
1metre = 5 dm
Exercise–2
(s – a ) = (120 – 50 =) 70 metres
(s – b ) = (120 – 78 =) 42 metres
(s – c ) = (120 – 112 =) 8 metres
Area = 84270120 = 1680 sq m
Perpendicular =Base
Area2 =
112
21680 metres
= 30 metres
Rent per hectare = Rs 10000
required rent = Rs
10000
168010000
= Rs 1680
4. Let ABCD be the given square and X is pointon side CD.
CX = 5 cm
Let the length of the side of square be x cm.
XD = (x – 5) cm
Now, ADX is a right-angled triangle
2
1 × (DX) × (AD) = 42
or, )()5(2
1x x = 42
or, x 2 – 5x – 84 = 0
or, x 2 – 12x + 7x – 84 = 0
1. Area of the field =2
HeightBase
=
2
550880 sq metres
=
100100
550440 sq hectometres
= 24.20 sq hectometres
Cost of supplying water to 1 sq hectometre
= Rs 242.50
Cost of supplying water to the whole field
= Rs 24.20 × 242.5 = Rs 5868.5
2. Area of the field =2
2752.150552.20324
hectares
Also, area of the field =2
1 × Base × Height
=2
1 × 3 × Height × Height =
2
3(Height)2
2
3(Height)2 =
2
27hectares
(Height)2 =
3
2
2
27 9 hectares
= 90000 sq metres
Height = 90000 m = 300 m
Also, Base = 3 × Height = 900 m3. Here, a = 50 metres, b = 78 metres, c = 112
metres
s = )1127850(2
1 metres
=
240
2
1metres = 120 metres
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549Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
or, x (x – 12) + 7(x – 12) = 0
or, (x – 12) (x + 7) = 0
or, x = 12 because x –7
The side of square = 12 cm
5.
Let ABC be a right-angled triangle,
whose ABC = 90°
Hypotenuse, AC = 12.5 cm
Base, BC = 7.5 cm
Perpendicular (AB) = 22 BCAC
= 22 )5.7()5.12(
= )5.75.12()5.75.12(
= 520 = 100 = 10 cm
required area =2
1 × Base × Height
= 105.72
1 = 375 sq cm
6. Area of the square = (45 × 45 =) 2025 sq m
Area of the triangle = 2025 sq m
Height of the triangle = 75 m
Required side of the triangle
=75
22025 = 54 m
7. Area of an isosceles right triangle
=2
1 × (length of one of its two equal sides)2
= 2152
1 sq cm =
2
225= 112.5 sq cm
8. Area of an equilateral triangle = 2side4
3
= 884
3 sq cm
= 1.732 × 16 = 27.712 sq cm
Height of an equilateral triangle = side2
3
= 82
3 = 1.732 × 4 = 6.928 cm
9. Let each side of ABC be x cm.
Area of BOC =2
122x = 11x sq cm
Area of AOC =2
116x = 8x sq cm
Area of AOB = x x 102
120 sq cm
Area of ABC
= area of BOC + area of AOC + area of AOB
= )10811( x x x sq cm = 29x sq cm
But area of equilateral ABC=2
4
3x sq cm
x x 294
3 2 or, 294
3x
or, 97.663
116
3
429
x cm
Area of ABC =2
4
3x
= 297.664
732.1 = 1942.0 sq cm
10. For finding area of the park, we have2s = 50 m + 80 m + 120 m = 250 m
ie s =
2
250125 m
Now, (s – a ) = (125 – 120 =) 5 m
(s – b ) = (125 – 80 =) 45 m
(s – c ) = (125 – 50 =) 75 m
Therefore, area of the park
= )()()( c s b s a s s
= 75455125 m2
= 15375 m2
Also perimeter of the park = AB + BC + CA
= 250 m
Ther efor e, length of th e wi re needed for
fencing
= 250 m – 3 m (to be left for gate)
= 247 m
And so the cost of fencing = Rs (20 × 247)
= Rs 4940
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550 Concep t o f A r i t hm e t i c
K KUNDAN
1. Here the area of the quadrilateral ABCD = the
area of the triangle ABD + the area of the
triangle BCD =2
1 × BD × AL +
2
1 BD × CM
=2
1 × BD (AL + CM)
ie, The area of a quadrilateral
=2
1 × diagonal × sum of offsets.
Ther ef ore, the requi re d ar ea of the
quadrilateral
= 64102
1 sq m
=
1010
2
1 50 sq m
2. Area of parallelogram ABCD
= area of ABC + area of ACD
= 2 area of ABC
( Each diagonal of a parallelogram bisects it)
Semi-perimeter (s) of ABC
=
2
304050 60 m
Area of ABC = )()()( c s b s a s s
= )5060()3060()4060(60
= 10302060
= 360000 = 600 sq m
Area of parallelogram ABCD = 2 × 600 sq m
= 1200 sq m
3. Area of quadrilateral ABCD
= Area of triangle ADB + Area of triangle DBC.
In triangle ADB,
2
1 × perimeter =
21548585 162 m
Area of the triangle ADB
= )154162()85162()85162(162 sq m
= 87777162
= 2772 sq m
In triangle DBC,
2
1× perimeter =
2
154143165231 m
Area of the triangle DBC
= )154231()143231()165231(231 sq m
= 778866231 = 10164 sq m Area of the quadrilateral ABCD
= (2772 + 10164) sq m
= 12936 sq m
4.
Let ABCD be the rhombus in which AC
= 55 cm Perimeter of the rhombus = 146 cm
AB =4
146= 36.5 cm
and AO =2
55= 27.5 cm
Exercise–3
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551Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
BO = 22 )5.27()5.36( cm = 24 cm
Hence the other diagonal BD = 48 cm
Area of the rhombus = 2
1
× AC × BC
=
4855
2
1
= 1320 sq cm
5.
Let ABCD be the trapezium such that AB||CD,AB = 41 m, DC = 20 m, AD = 10 m and
BC = 17 m
Draw CE||DA and CF AB.
Clearly, AECD is a parallelogram
Now, EB = AB – AE = AB – DC [AE = DC]
= (41 – 20 =) 21 m
Also, EC = AD = 10m
Thus, in ECB, we have
EB = 21 m, EC = 10 m and BC = 17 m
Let s be the semi-perimeter of the ECB. Then
s =
2
171021 24 m
Area of ECB = )()()( c s b s a s s
= )1724()1024()2124(24 sq m
= 714324 = 84 sq m ....(i)
Also, area of ECB =2
1× Base × Height
=2
1× 21 × (CF) ....(ii )
From equations (i) and (ii), we get
84CF2
21
CF =21
284 = 8 m
Area of parallelogram AECD = Base × Height
= AE × CF = (20 × 8 =) 160 sq m
Now, area of trapezium ABCD
= (Area of parallelogram AECD) + (Area of ECB)
= (160 + 84) sq m = 244 sq m
Cost of levelling the field at the rate of Rs
30 per sq metre = Rs (30 × 244) = Rs 7320
6. Let x be present in the parallel sides of the
trapezium.
parallel sides are x and 3x .
Let k be present in the non-parallel sides of
the trapezium. Non-parallel sides are 2k and 3k .
According to the question,
1
2
3
3
k
x
or, k x 2
Combining triangle AEC and BFD, we get a
triangle of base 6k – 2k = 4k and two other
sides are 2k and 3k.
Semi-perimetre of combined triangle
= x =2
324 k k k =
2
9k
Area of the triangle
= )4()3()2( k s k s k s s
=
k
k k
k k
k k 4
2
93
2
92
2
9
2
9
=
22
3
2
5
2
9 k k k k
= 3594
2
k
= 154
3 2k
Also area of the triangle
= 151524
15154
2
1 k k
or, 154
3 2k = 1515
2
k
k = 10 units
Parallel sides are 20 units and 60 units.
Area of trapezium = 4
15156020
2
1 units2
= 2units1575 .
7. Each side of the hexagon inscribed in the
circle is 6 cm, the radius of the circle is 6 m.
Area of the circle =2r
= 3.14 × (6)2 = 3.14 × 36 = 113.04 cm2
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552 Concep t o f A r i t hm e t i c
K KUNDAN
1. Distance to be travelled = 11 km = 11000 m
Radius of the wheel =4
31 m
Circumference of the wheel
=
4
31
7
222 m = 11 m
In travelling 11 m the wheel makes 1
revolution.
In travelling 11000 m the wheel makes
11
110001000 revolutions.
2. Area = r 2, circumference = 2r
2r = 1012 metres.
r = 1012 ×22
7
2
1 m = 161 m
area of the garden = r 2
= 1611617
22 sq m
= 81466 sq m
Area of the road
= area of bigger circle – area of the garden.
Radius of the bigger circle
=
2
13161 m =
2
329 m
area of bigger circle =2
329
2
329
7
22 sq m
= 2
185046 sq m
area of road = 814662
185046
=2
13580 sq m
required cost = Rs100
32
2
7161 = Rs 1145.76
3. Distance covered by the wheel in one
revolution
=srevolutionof Number
movedDistance
=
1001000
500011km
500011
cm
= 220 cm
Circumference of the wheel = 220 cm
Let the radius of the wheel be r cm. Then,
Circumference = 220 cm
or, 2r = 220 cm
Exercise–4
Area of the hexagon =2
33(side)2
=22 cm354)6(
2
33
= 54 × 1.732 = 93.53 cm2
Hence the area of the region of the circle
which is outside the hexagon
= 113.04 cm2 – 93.53 cm2 = 19.51 cm2.
8. We know that the diagonals of a rhombus
bisect one another at right angles. Therefore
from the given figure the area of the rhombus
ABCD = area of the triangle ABC + area of the
triangle ADC = 2 × area of the triangle ABC.
=2
12 × AC × OB = AC × OB
= 2
1 × AC × BD
BD2
1
OB
ie, the area of a rhombus
=2
1× product of its two diagonals.
Here, AC = 8 cm and BD = 6 cm
AO = 4 cm and BO = 3 cm
AB = 22 34 = 5 cm
Each side of the rhombus is 5 cm.
The required area of the rhombus
= 682
1 sq cm = 24 sq cm
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553Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
or, 887
222 r
or, r = 14 cm
Area of the circle
= 2222 cm616cm147
22
r
7. Length of the wire = circumference of the
circle
=
28
7
222 cm [Using C = r 2 ]
= 176 cm ... (i)
Let the side of the square be x cm. Then,
perimeter of the square = length of the wire
or, 4x = 176 [Using (i)]
or, x = 44 cm
Hence, the length of the side of the square is
44 cm.
8. Radius of the wheel =2
75cm
Now, perimeter of the wheel = r 2
=
7
7522
2
75
7
222 cm
Total time taken to pull up bucket is 1 minute
12 seconds = (60 + 12 =) 72 seconds
Distance travelled in one second = 1.3 m
Distance travelled in 72 seconds
= (1.3 × 72 =) 93.6 m or 9360 cm
Let number of revolutions be N.
According to the question,
Distance covered by bucket = perimeter of
wheel × number of revolutions made by the
wheelN × (perimeter of the wheel) = 9360 cm
or, N × 93607
7522
or, N = 70.397522
79360
Number of complete revolutions = 39.
9. Area of circle = r 2, where r is the radius.
Area of the biggest circle = 22 cm40020
Area of the smallest circle =2cm400
4
1
= 100 cm2
or, 2207
222 r
or, r = 35 cm
diameter = 2r cm = (2 × 35) cm = 70 cm
Hence, the diameter of the wheel is 70 cm.
4. Radius of the wheel = r =2
60cm = 30 cm
Circumference of the wheel
=
30
7
2222 r cm =
7
1320 cm
Distance covered in one revolution
= circumference =7
1320 cm
Distance covered in 140 revolutions
=
140
7
1320cm = (1320 × 20) cm
= 26400 cm =
100
26400
m
= 264 m =
1000
264km
It is given that the wheels are making 140
revolutions per minute. So, distance covered
in one minute = Distance covered in 140
revolutions
=
1000
264km
Distance covered in one hour
=
60
1000
264km = 15.84 km
Hence, the speed with which the boy is
cycling= 15.84 km/hr
5. Distance covered by the wheel in one minute
=
60
100100066 = 110000 cm
Circumference of the wheel
=
70
7
222 = 440 cm
Number of revolutions in one minute
=
440
110000 = 250
6. Area of the square = 484 cm2
side of the square = 484 cm = 22 cm
]Areaside)side(Area[ 2
So, Perimeter of the square
= 4(side) = (4 × 22) cm = 88 cm
Let r be the radius of the circle. Then,
circumference of the circle
= Perimeter of the square
2r = 88
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554 Concep t o f A r i t hm e t i c
K KUNDAN
Now, if the radius of the smallest circle be r .
Now, according to the question,
1002r
or, r 2 = 100
or, r = 10
radius of the smallest circle = 10 cm
10. Each side of the square inscribed in semicircle
= BC = OA = a
In right angle triangle OAB
OA2 + AB2 = OB2
or,2
22
4
r a
a
or,22
4
5r a
5
4 22 r
a
Area of the square inscribed in semicircle
=5
4 22 r
a
Diagonal of the square inscribed in a circle
= 2r
Area of this square =22 2)2(
2
1r r
Required ratio = 1:5
2
2:5
4 22
r
r
or 2 : 5
11. For the first triangle, we have
a = 35, b = 53 and c = 66
s =
2
665335
2
c b a 77 cm
1
= Area of the first triangle
= )()()( c s b s a s s
= )6677()5377()3577(77
= 11244277
= 114667117
= 2222 26117
= 7 × 11 × 6 × 2 = 924 cm2 ....(i)
For the second triangle, we have
a = 33, b = 56, c = 65
s =
2
655633
2
c b a 77 cm
2
= Area of the second triangle
= )()()( c s b s a s s
= )6577()5677()3377(77
= 12214477
= 4373114117
= 2222 34117
= 7 × 11 × 4 × 3 = 924 cm2 ....(ii)
Let r be the radius of the circle. Then,
Area of the circle = Sum of the areas of two
triangles
or, 212 r
or, 9249242 r
or, 18487
22 2 r
or, 58878422718482 r
or, 249.24588 r cm
12. For semi-circle with diameter PS,
Radius = 6 cm
Area of such semi-circle = 2
2
1r
=7
396)6(
7
22
2
1 2 cm2
Diameter PS = 6 + 6 = 12 cm
PQ = QR = RS =
3
124 cm
QS = QR + RS = 8 cm
For semi-circle with diameter QS, radius= 4 m
Its area =7
176)4(
7
22
2
1
2
1 22 r cm2
For semi-circle with diameter PQ, radius
= 2 cm
Its area = 7
442
7
22
2
1 2 cm2
Area of shaded region
=7
264
7
44
7
176
7
396 = 37.71 cm2.
13. Diameter of each of the semi-circle = 40 m.
Radius, r =2
40
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555Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
Area of each semi-circle =2
2
1r
Area of semi-circle grassy plots
=22 2
2
14 r r
= 2 × 3.14 × 20 × 20 = 2512 sq m
Cost of turfing the plots at the rate of Rs
1.25 per sq m = Rs (1.25 × 2512 =) Rs 3140.
14. Radius of flower beds I and II =
2
10050 m.
Area of each of flower beds I and II = 221 r
= 505014.32
1 sq m = 3925 sq m
Total area of flower beds I and II
= 2 × 3925 = 7850 sq m
Radius of each of flower beds III and IV
=
2
5025 m
Area of each of flower beds III and IV =2
2
1r
= 252514.3
2
1 sq m = 981.25 sq m
Total area of flower beds III and IV
= 2 × 981.25 sq m = 1962.62 sq m
Total area of 4 semi-circular beds
= (7850 + 1962.50) sq m = 9812.50 sq m
Cost of levelling the flower beds at the rate
of 60 p per sq m
= Rs 50.9812100
60 = Rs 5887.50
15. Area of rectangular park
= 120 m × 100 m = 12000 m2
Area of the park excluding the circular lawn
= 8700 m2
Area of circular lawn
= (12000 – 8700) m2 = 3300 m2
Let r be the radius of circular lawn
33002 r or 33007
22 2 r
or, r 2 =22
73300 = 150 × 7 = 1050
r = 2/1)1050(1050
Radius of the circular lawn = 32.40 m.
16. Area of rectangle ABCD
= Length × Breadth = AB × BC
= (3.5 + 7 + 3.5) × 7 = 98 cm 2
Area of unshaded region
=2222 2
2
1
2
1r r r r
= 5.35.37
44)5.3(
7
222 2 = 77 cm2
Area of the shaded region
= Area of rectangle ABCD – Area of unshaded
region
= 98 cm2 – 77 cm2 = 21 cm2.
17. Area of the shaded region
= 2(Area of rectangle with sides 120m and 14m)
+ 2[Area of the semi-circle with radius (35+14)
cm
ie 49 cm - Area of the semi-circle with radius
35 m]
= 2 (120 × 14) +
)3549(
7
22
2
12 22
= )3549()3549(7
22)1680(2
=
1484
7223360 m2
= (3360 + 22 × 84 × 2) m2
= (3360 + 3696) m2 = 7056 m2
18. (i ) Area of segment BPF
= Area of sector ABPF – Area of ABF
Draw AK BF
BAF is an angle of regular hexagon
BAF =120° . .. .(1)
The perpendicular from the centre of a
circle to a chord disects the chord
BK = KF ....(2)
AB = AF (Side of a regular hexagon)
AK = AK (common side)
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556 Concep t o f A r i t hm e t i c
K KUNDAN
AFKABK (SSS congruence rule)
FAKBAK
Also 120ºBAFFAKBAK
BAK = 60º ABK is a 30º – 60º – 90º triangle
AK=2
1hypotenuse (AB) = 36
2
1
and 3362
3)AB(
2
3BK cm
BF = 2BK = 2 × 3 363 cm.
Now Area of sector ABPF
=2
360
anglesectorr
= 68.3714.312614.3
360
120 2 cm2
Area of ABF = AKBF2
1
= 393362
1
= 9 × 1.73 = 15.57 cm2
Area of segment BPF
= Area of sector ABPF – Area of ABF
= 37.68 m2 – 15.57 cm2 = 22.11 cm2.
( ii) Area of shaded portion
= Area of hexagon ABCDEF – area of ABF)
= 57.1562
33 2
= 57.15354 = 54 × 1.73 – 15.57
= 93.42 cm2 – 15.57 cm2 = 77.85 cm2.
19. Radius of sector CXB = 14 cm
Radius of sector DXY = 21 cm – 14 cm = 7 cm
Area of sector CXB = 2147
22
360
90 = 154 cm2
Area of sector DXY = 277
22
360
90
= 38.5 cm2
Area of rectangle = ABCD = AB × BC
= 21 cm × 14 cm = 294 cm2
Area of the shaded portion
= Area of rectangle – Area of two sectors
= 294 – (154 + 38.5) = 294 – 192.5
= 101.5 cm2.
20. Area of equilateral4
3 (side)2
= 1964
732.114
4
3 2
= 1.732 × 49 = 84.868 cm2
Area of three sectors
= 22 77
22
120
60
360
anglesector3
r
= 11 × 7 = 77 cm2
Area of shaded region = Area of equilateral – Area of three sectors
= (84.868 – 77 =) 7.868 cm2.
21. The radius of circle = 21 cm
An arc ABC subtends an angle of 60º at the
centre = OA = OB = 21 cm
6060 – 1802
1OBAOAB
OAB is equilateral.
A
B
O60°
21cm
(i ) Length of the arc =
360
60 × circumference
= 217
222
360
60 cm = 22 cm
( ii) Area of the sector =360
60 × area of the circle
= 21217
22
360
60 sq cm = 231 sq cm
(iii) Area of segment
= Area of sector – Area of equilateral AOB
of side 21 cm
=
2121
4
3231 sq cm
=
4
441732.1231 sq cm
= (231 – 190.953) sq cm = 40.047 sq cm.
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557Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
22. Let the radius of the swimming pool be r m.
Width of the path = 4 m
Then area of the swimming pool = r 2 sq m
and area of the swimming pool including the
path
= msq )4( 2 r
222
25
11)4( r r r
It is given that the area of the path is2511 th
part of the area of the swimming pool.
or, 222
25
114 r r r
or, 222 1125)168(25 r r r r
or, 222 112540020025 r r r r
or, 040020011 2 r r
22
1760040000200 r
2257600200 r
= ,22
440
22
240200
22
40,20
20
40
Since radius of the swimming pool cannot be
negative. Hence radius = 20 m.
23. Circumference of circle = 80 cm
or, r 2 = 80 cm or r =
40cm
Let the side of square be 2a cm.
Since from the centre of a circle to a chord
bisects the chord,
OM = a and AM = a
Now in right-angled triangle OAM, we have
222 AMOMOA or 222 a a r
or, ,2
2
240
a
or
22 40
2
1
a
or,
40
2
1a
Hence side of square
=
402
40
2
122a cm
24. OC = CB = 20 cm
In right-angled OCB,
OB = cm 220202022
Radius of sector OPBQ = 220 cm.
Area of sector OPBQ
= 22 22014.3360
º90
360
anglesector r
= 80014.34
1 = 628 cm2
Area of square OABC = {(20)2 =} 400 cm2.
Hence the area of the shaded region
= Area of sector OPBQ – Area of square OABC
= 628 cm2 – 400 cm2 = 228 cm2.25. Let r be the radius of inner circle and R be
the radius of outer circle
Then, 9R2 r ...(i)
ΔDOC~ΔAOD
ODOC
OAOD
or, OCOAOD2 or 9RR5R 2
or, 9RR2510RR 22 or R = 25
2(25 – r ) = 9 [using (i)]
or, 2r = 50 – 9
or, 2r = 41
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558 Concep t o f A r i t hm e t i c
K KUNDAN
r =2
41 = 20.5
Area of shaded portion =22R r
= [(25)2 – (20.5)2] = 25.4206257
22
= 75.2047
22 = 643.5 cm2.
26. Let the radii of the given circles be R and r
respectively.
Sum of their areas = 116 cm2
116R 22 r
or, 116)R( 22 r
or, 116R2 r ....(i)
If O and O be the centres of the given circles,
then r ROO
6R r (Given) ....(ii)
Now, 2222 R2RR r r r
or, 22 6R r = 2 × 116 [Using (i) and (ii)]
or, 196362322 r R
14196R r ....(iii)
Solving (ii) and (iii), we get
2R = 20
R =2
20 = 10
From (iii), 10 + r = 14 or r = 14 – 10 = 4
Hence radii of the given circles are 10 cm
and 4 cm respectively.
27. Area of rectangle = (28 × 26 =) 728 m2
Area of one corner (unshaded)
= 2)10(4
1 = 10014.3
4
1 =
4
314= 78.5 cm2
Total area of 4 unshaded corners
= 78.5 × 4 = 314 m2
Area of shaded portion
= Area of rectangle – Area of 4 corners
= (728 – 314 =) 414 m2
.28. Required area
= Area of square – 4 × Area of one sector
=
27
4
141414
= 497
22
4
14196
= (196 – 154 =) 42 cm2.
29. Area of shaded region
= Area of a square
– 2(Area of a semi-circle of radius 7 cm)
=
77
7
22
2
1214
2
= (196 – 154 =) 42 cm2
.30. In ABC, A = 90°
By pythagoras theorem, we get
222 ACABBC = 25)3()4( 22
25BC cm = 5 cm
Area of shaded portion
= 222 5.22
15.1
2
12
2
1
= 25.625.247
22
2
1
= 64.197
5.1375.12
7
11 cm2.
31. Let r cm be the radius of the inner circle, then
The radius of the outer circle = (r + 7) cm.
Area of the shaded region = 770 cm2
or, 770)7( 22 r r
or, 770)4914( 22 r r r
or, 77049147
22r
or, 3572 r
or, 282 r
r = 14
Radius of outer circle = (14 + 7) cm = 21 cm.
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559Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
Side of the square = 2 × 21 cm = 42 cm
The area of the unshaded region of the
square
= (42)2 – 770 = (1764 – 770) cm2 = 994 cm2.
32.
Area of the shaded region
= Area of the equilateral + Area of the
semicircle
= 22
2
1side
4
3r =
22
2
7
7
22
2
17
4
3
[ BC = 7 cm is the diameter of semicircle
cm r 2
7
]
=4
7749
4
732.1
4
49
7
1149
4
3
= 12.25 × 1.732 + 19.25
= 21.22 + 19.25 = 40.47 cm2.
33. AC = 2 × radius = 2 × 5 cm = 10 cm.
As angle in a semi-circle is 90º.
So ADC = 90° and ABC = 90°
By pythagoras theorem;
AD2 = AC2 – CD2
= 102 – 82 = 100 – 64 = 36
AD = 36 = 6 cm
and AB2 + BC2 = AC2
or, AB2 + AB2 =100
or, 2AB2 =100
or, AB2 = 50
or, AB = 2550 cm
The area of the shaded region
= area of the circle – area of ADC – area of
ABC
= BCAB2
1CDAD
2
12 r
= 25252
186
2
15
7
22 2
[ AD = 5 cm, CD = 8 cm
AD = 25 , BC = 25 ]
= 25247
550
= 78.59 – 49 = 29.57 cm2.
34. From the given figure it can be seen that
Radius = OP = OR = OQ = OS = 14 cm
Taking the semi-c ircle POQS first
Area of shaded small circle =
2
2
OS
=4
(OS)2 =
47
141422
= 154 cm2
Now taking the semi-circle OPRQ
Area of the semi-circle OPRQ =2
(OR)2
=
27
141422
= 308 cm2
Area of triangle PRQ=2
1 × base × height
=2
1 × PQ × OR
=2
1 × 14 × 28
= 196 cm2
Area of shaded region in semi-circle OPRQ
= (308 – 196) cm2 = 112 cm2
The area of the shaded region = Area of shaded
region in semi-circle OPQR + Area of shaded
region in POQS = (154 + 112) cm2 = 266 cm2
35.
In the given figure PQRS is a rectangular field
in which SR = 6 m and PS = 4 m. We have to
find the cost of shaded portion in the given
figure. The semicircle PAS and QBR makes a
complete circle of radius
=2
4
2
PS
2
QR = 2 m
area of circle = r 2 = 3.14 × 4 m2
Similarly, PCQ and SDR also makes a complete
circle of radius
=2
6
2
SR
2
PQ = 3 m
area of circle = r 2 = 3.14 × 9 m2
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560 Concep t o f A r i t hm e t i c
K KUNDAN
Total area in which flower is raised
= 3.14 × 4 + 3.14 × 9 = 3.14 × 13 m2
total cost = (3.14 × 13 × 8 =) Rs 326.56
36.
Area of the square = (14 cm)2 = 196 cm2.
Area of circular part at one vertex
= 777
22
º360
º90 =
4
154cm2
Total area of circular parts = 44
154
= 154 cm2.
Area of the region of the square that remains
outside the region of any of the circle
= (196 – 154) cm2 = 42 cm2.
37. In the following figure, BDEF is the brick put
to stop the wheel.
Here BD = AC = EF = 5 cmAB = CD = 15 cm
OA = OD = r (say) [Radius of the wheel]
In the OCD; OC = OA – CA = (r – 5) cm
CD = 15 cm
Here, OD2 = OC2 + CD2
or, 222 15)5( r r
or, 225102522 r r r
or, 10r = 250
r = 25 cm
Therefore, the required radius of the wheel
= 25 cm.
38. Angle described by minute-hand in 60 minutes
= 360°
Angle described by minute-hand in one
minute =60
360 = 6°
We know that the area A of a sector of angle
D, in a circle of radius r is given.
A =2
360
Dr
Here, r = 10 cm and D = 6°
A =30
157101014.3
360
6
sq cm
=3075 sq cm
39. Area of the minor segment
= Area of sector OAB – Area of the right-
angled triangle OAB
=
1010
2
1101014.3
360
90cm2
= (78.5 – 50) cm2 = 28.5 cm2
Area of the major segment
= Area of the circle – Area of the minor segment
= [3.14 × 102 – 28.5] cm2
= [314 – 28.5] cm2 = 285.5 cm2
Note: We know that the area of a minor
segment of angle in a circle of radius r is
given by
sin
2
1
360A 2r [Always Remember]
Here, r = 10, = 90°
A =
90sin
2
1
360
9014.3)10( 2
cm2
=
2
1
4
14.3)10( 2
cm2
= [3.14 × 25 – 50] cm2
= (78.5 – 50)cm2 = 28.5 cm2
40. We know that the area of a minor segment of
angle in a circle of radius r is given by
A =
sin
2
1
360
2r
=
60sin
21
3606014.3)15( 2
cm2
=
4
3
6
14.3225 cm2
= 225 [0.5233 – 0.4330] cm2
= 225 × 0.0903 cm2 = 20.317 cm2
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561Mensu r a t i o n I (A rea an d Per i me t e r )
K KUNDAN
Area of the major segment
= Area of the circle – Area of the minor segment
= [3.14 × (15)2 – 20.317] cm2
= [706.5 – 20.317] cm2 = 686.183 cm2
Note: We can have another method for findingarea of the minor segment. Clearly, from the
figure, triangle OAB is an equilateral triangle
with the side 15 cm.
Now, area of the minor segment
= Area of the sector OAB – Area of the
triangle OAB
=
22 )15(
4
3)15(
360
60cm2
=
4
3
6
14.3152
cm2
= 20.317 cm2
41. Area of the square lawn ABCD = (56 × 56) sq m
Let OA = OB = x metresso, x 2 + x 2 = 562
or, 2x 2 = 56 × 56
or, x 2 = 28 × 56
Now, area of sector OAB =22
4
1
360
90x x
=
5628
7
22
4
1sq m
[Putting the value of x 2 = 28 × 56]
Also, area of OAB =
5656
4
1sq m
(AOB = 90°)
So, area of flower bed AB
=
5656
4
15628
7
22
4
1sq m
=
2
7
225628
4
1sq m
=
7
85628
4
1sq m
Similarly, area of the other flower bed
=
7
85628
4
1sq m
Therefore, total area
=
7
85628
4
1
7
85628
4
15656 sq m
=
7
2
7
225628 sq m
=
7
185628 sq m = 4032 sq m
Alternative Method:
Total area
= Area of sector OAB + Area of sector ODC +
Area of OAD + Area of OBC
=
5628
7
22
360
905628
7
22
360
90
5656
4
15656
4
1sq m
= 141422227
567
sq m
= (56 × 72 =) 4302 sq m
42. Area of square ABCD
= (14 × 14) sq cm = 196 sq cm
Diameter of each circle =
2
14 7 m
Radius of each circle = 2
7cm
Area of one circle =2r =
2
7
2
7
7
22sq cm
=2
77sq cm
Therefore, area of the four circles
=
2
774 154 sq cm
Hence, area of the shaded region
= (196 – 154 =) 42 sq cm
43.
Let us mark the four unshaded regions as I,
II, III and IV as in the figure.
Area of I + Area of III
= Area of ABCD – Areas of two semicircles
of each of radius 5 cm
=
25
2
121010 sq cm
= (100 – 3.14 × 25) sq cm
= 21.5 sq cmSimilarly, Area of II + Area of IV = 21.5 sq cm
So, area of the shaded region
= Area of ABCD – Area of (I + II + III + IV)
= (100 – 2 × 21.5) sq cm
= (100 – 43 =) 57 sq cm
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562 Concep t o f A r i t hm e t i c
K KUNDAN
44. Let the radius of the circle be r cm.
As per given information,
r r 22 = 16.8
or, )1(2 r = 16.8
or,
1
7
222r = 16.8
or,7
152 r = 16.8
or, r =152
78.16
= 3.92
Hence, radius of the circle = 3.92 cm
45.
Let the radius of circle be x .
Area of the circle =22 x r
A square has been inscribed in the circle.
Diagonal of square = Diameter of circle = 2x
Side of square =2
2x
Its area = 22
22
4
2
2
2
2x
x x x
Required ratio =22 2: x x
= 2:
7
22
= 11 : 7
46.
Let a = 6 cm, b = 8 cm and c = 4 cm
Semi-perimeter(s) =2
486
2
c b a = 9 cm
Area of the triangle = )()()( c s b s a s s
= )49()89()69(9
= 5139 = 153 sq m
Area of sectors = )QQQ( 321
2
r
= 180360
12
( Sum of all the angles of a
triangle is equal to 180°)
=7
11
27
22
sq m
Area of the remaining portion of the triangle
=
7
11153 sq cm
= (3 × 3.9 – 1.6) sq cm
= (11.7 – 1.6) sq cm
= 10.1 sq cm
47.
Area of the sector AOB
=360
AngleSector × Area of the circle
=
2828
7
22
360
72cm2
=5
2464 cm2 = 492.8 cm2
Length of the arc AB
=360
AngleSector× Circumference of the circle
=
287
22
2360
72
cm
=5
176 cm = 35.2 cm
48. Let the radii of three concentric circles be x ,
2x and 3x respectively.
required ratio
=circlesoutertwothebetweenArea
circlesinnertwothebetweenArea
= 22
22
)2()3(
)()2(
x x
x x
= 2
2
)49(
)14(
x
x
=5
3 = 3 : 5