ncert solutions for class 9 maths

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AREAS RELATED TO CIRCLES
S.No
.
Name Figure Perimeter Area
Nomenclatur
e
2
1. Circle 2r or d r
r = radius
2r = d =
diameter
2. Semicircl
e
r + 2r
1 r
2
2
r = radius
A
3. Quadrant
r
r
2r
+
2
1 r
4
2
r = radius
O r B
4. Ring
(shaded
region)
R
r
2R + 2r
2
(
R − r
2
)
or
(R + r)(R – r)
R = outer
radius
r = inner
radius
5. Sector of
a circle
6.
Segment
of a
circle
For Example:
A
r
C
O
D
r
B
r
l + 2 r = + 2r
180
r
+ 2r
180
sin
2
2
r
360
or
1
2
lr
Area of the minor
segment (ACB)
2
= r
360º
area of AOB
=
2
r
1
− r
360 2
2
–
sin
Area of the major
segment (ADB)
= area of the
circle – area of
the minor
segment
r
2
= – area of
the minor
segment
= angle of
the sector
r = radius of
sector
l = length of
arc
r = radius
= angle of
the related
sector
In a circle of radius 21 cm, an arc subtends an angle of 60º at the centre. Find
(i)
the length of the arc,
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(ii)
(iii)
(iv)
the area of the sector,
the area of the minor segment, and
the area of the major segment.
Let ACB be the given arc subtending an angle of 60º at the
centre. Then, r = 21 cm and = 60º
D
2r
(i) Length of the arc ACB = cm
360
22 60
= 2 21
cm
7 360
= 22 cm
A
O
C
B
(ii) Area of the sector OACBO =
2
r
cm
360
2
22 60 2
= 21
21
cm
7 360
= 231 cm 2
(iii)
Area of the minor segment ACBA
= (area of the sector OACB) – (area of the OAB)
=
=
1 2 2 1
231
− r
2
sin cm
= 231
− 21
21
sin60º
2
1
3
2
2
231 − 21
21
cm
= (231 − 190.953)cm
2
2
cm
2
= 40.047 cm
2
(iv)
Area of the major segment BDAB
= (area of the circle) – (area of the minor segment)
=
22
21
21 − 40.047 7
2
cm
= (1386 – 40.047) cm 2
= 1345.953 cm 2
• Distance covered by a wheel in 1 revolution = Circumference of the wheel.
• Number of revolutions in 1 minute =
distance covered in1minute
circumference
.
For example:
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The diameter of the wheels of a bus is 140 cm. How many revolutions per minute must
a wheel make in order to move at a speed of 66 km per hour?
Distance covered by a wheel in 1 minute
66 1000
100
cm = 110000 cm
60
Circumference of a wheel
22
= 2 70 cm = 440 cm
7
Number of revolutions in 1 min
=
110000
= 250
440
.
• Angle subtended by the minute hand of a clock in 60 minutes = 360º.
• Angle subtended by minute hand (of clock) in 1 minute =
360º =
60
6º
.
• Angle described by the hour hand of a clock in 24 hours = 360º.
For example:
The minute hand of a clock is 12 cm long. Find the area of the face of the clock
described by the minute hand in 35 minutes.
Angle described by the minute hand in 60 minutes = 360º.
Angle described by the minute hand in 35 minutes
º
360
= 35 = 210º
60
= 210º and r = 12 cm
Area swept by the minute hand in 35 minutes
2
r
2 22 210 2
2
= cm
= 12
12
cm = 264 cm
360
7
360
B
O
A
210º
• Areas of combinations of plane figures
For example:
In the figure, two circular flowerbeds have been shown on two
sides of a square lawn ABCD of side 56 m. If the centre of each
circular flowerbed is the point of intersection O of the diagonals
of the square lawn, find the sum of the areas of the lawn and
the flowerbeds.
Area of the square lawn ABCD = (side) 2 = 56 × 56
2
m
... (i)
A
D
O
B
56m
C
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Let
OA = OB = x metres
In AOB, AOB = 90 [Diagonals of square bisect at 90]
So,
or,
2
2
2
x + x = 56
[By Pythagoras theorem]
2x 2 = 56 56
or,
x 2
= 28 56
Now, area of sector OAB =
Also, area of OAB =
So,
=
90
360
2 1
x = x
4
1 22
28 56
4 7
1
1
OAOB
= x
2
2
1
= 28 56
2
area of flower bed AB = Area of sector OAB − area of OAB
=
1 22
1 2
28 56 − 28 56 m
4 7
2
2
2
... (ii)
[From (ii)] ... (iii)
[AOB = 90º] ... (iv)
=
1 22 2
28 56 − 2 m
4 7
[From (iii) and (iv)]
Similarly, area of the other flower bed
Therefore, totals area =
=
=
1
28 56
4
1
28 56
4
8 2
m
7
8 2
m
7
... (v)
... (vi)
1 8 1 8 2
56 56 + 28 56 + 28 56 m
4 7 4 7
=
=
2 2 2
28 56 2
+ + m
7 7
18 2
28 56 m = 4032 m
7
2
[From (i), (v) and (vi)]
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