Concepts to be covered:-
Introduction,Perimeter and Area of a Circle ,Areas of Sector and Segment of a Circle
Introduction:-
In this chapter we are going to learn areas related to circles.Firstly let us know about circle.
Circle:-
Circle is set of all points in a plane which are at the fixed distance from a fixed point.
Here,the fixed point is known as centre.
Centre:-
Mid-Point of a circle is called centre of a circle.
Radius:-
The distance between the centre of a circle to any point in the circle.
It is denoted by 'r' or 'R'.
Chord:-
A line segment a which joins the two points in the circle is called chord.
Here AB is the chord.
Diameter:-
It is the longest chord which passes through the centre of a circle
It is denoted by 'd' or 'D'.
Diameter=2*radius
Radius=Diameter/2
PERIMETER AND AREA OF A CIRCLE:-
Perimeter of a circle:-
It is known as the circumference of the circle.
The distance covered by travelling once around a circle is known as perimeter.
(or)
The distance covered around the edge of a circle.
circumference =π ×radius
As we are calculating the circumference for circle.
circumference=π × 2r
=2πr
where 'r' is the radius of the circle.
The great Mathematician Aryabhata gave an approximate value of π.
π=22/7 or 3.14 approximately
Area of a Circle:-
The area of a circle is the region occupied by circle in the two-dimensional space.
Area of circle=πr^2
where 'r' is the radius of the circle.
Example:- The cost of fencing a circular field at the rate of ₹5280.The field is to be ploughed at the rate of ₹0.50 per m^2.Find the cost of ploughing the field.(Take π=22/7)
Area of the field=πr^2
In order to calculate radius r
Length of the fence=(Total cost)/(Rate)=5280/24=220m
Circumference of the fence=2πr=220
2×22/7×r=220
r=(220×7)/(2×22)
r=35cm
Radius of the field=35cm
Area of the field=πr^2=(22/7)×35×35m^2
=22×5×35m^2
=3850m^2
Cost of ploughing 1 m^2 of the field=₹0.50
Total cost of ploughing the field=3850×0.50
=₹1925
Example:-The radii of two circles are 19cm and 9cm respectively.Find the radius of the circle which has circumference equal to the two circles.
Given,
Radius of the first circle=19cm
Circumference of first circle=2πr1
=2×π×19
=38π
Radius of the second circle=9cm
Circumference of the second circle=2πr2
=2π×9
=18πcm
Circumference of required circle=38π+18π
=56πcm
Radius of required circle=R
2πR=56π
R=56π/2π
R=28cm
Sector:-
The portion of circular region enclosed by two radii and the corresponding arc is called sector of the circle.
Segment:-
The portion or part of a circular region enclosed between a chord and the corresponding arc is called a segment of a circle.
Area of the sector :-
The area enclosed by the sector is called the area of the sector.
Let OAPB be a sector with radius r an ∠AOB be θ
Area of the the circle is πr^2
When the degree measure of the angle at the centre is θ,then
area of the sector=(πr^2/360°)×θ
where'r' is the radius of the circle and θ is the angle of sector in degrees
Length of the arc of a sector of angle θ=( θ /360° )×2πr
From figure
Area of major sector OAQB=πr^2 -Area of minor segment APB
Example:-Find the area of the sector of a circle with radius 4 cm and angle 30°.Also,find the area of the corresponding major sector.
Given sector is OAPB
Area of the sector=(θ/360°)×πr^2
=(30°/360°)×3.14×4×4 cm^2
=(12.56/3)cm^2
=4.19cm^2(approx)
Area of corresponding major sector=πr^2-area of sector OAPB
=3.14×16-4.19cm^2
=46.05cm^2
=46.1(approx)
We can also calculate area of major segment in another way
Area of major segment=(360°-θ/360°)πr^2
=(360°-30°/360°)3.14×16 cm^2
=(330°/360°)3.14×16 cm^2
=46.05 cm^2
46.1 cm^2(approx)
Example:-Find the area of the segment AYB shown in figure,if radius of the circle is 21 cm and ∠A0B=120°
Area of the segment AYB=Area of sector OAYB-Area of △OAB
Area of sector OAYB=(120°/360°)×(22/7)×21×21 cm^2
=462cm^2
To find the area of △OAB,
draw OM⊥AB
M is the midpoint of AB and
∠AOM=∠BOM=(1/2)×120°=60°
Let OM=x cm
from △OMA=OM/OA=cos 60°
x/21=1/2
x=21/2
OM=21/2 cm
AM/OA=sin 60°=√3/2
AM=21√3/2 cm
AB=2AM=2×21√3/2 cm
=21√3 cm
Area of △OAB=1/2 AB×OM
=1/2×21√3×21/2 cm^2
=(441/4)√3 cm^2
Area of segment AYB=[462-(441/4)√3] cm^2
=(21/4)(88-21√3) cm^2
Areas of Combinations of Plane Figures:-
A combination of plane figures forms various exciting designs like a square drawn with a semicircle on each side with the same diameter as the square's side.
Example:-In figure,two circular flower beds have been shown on 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.
Area of the square lawn ABCD=56×56 m^2
OA=OB=x m
x^2+x^2=(56)^2 m
2x^2=56×56
x^2=28×56
Area of sector OAB=(90°/360°)×πr^2
=(1/4)×πx^2
=(1/4)×(22/7)×28×56 m^2
Area of △OAB=(1/4)×56×56 m^2
Area of flower bed AB=[(1/4)×(22/7)×28×56-(1/4)×56×56] m^2
=(1/4)×28×56×((22/7)-2) m^2
=(1/4)×28×56×(8/7) m^2
Similarly area of other flower bed=(1/4)×28×56×(8/7) m^2
Total area=[56×56+(1/4)×28×56×(8/7)+(1/4)×28×56×(8/7)] m^2
=28×56[2+(2/7)+(2/7)] m^2
=28×56×(18/7) m^2
=4032 m^2
Alternative Solution:-
Total area=Area of sector OAB + Area of sector +Area of △OAD + Area of △OBC
=[(90°/360°)×(22/7)×28×56+(90°/360°)×(22/7)×28×56+(1/4)×56×56+(1/4)×56×56] m^2
=(1/4)×28×56[(22/7)+(22/7)+2+2] m^2
=[(7×56)/7](22+22+14+14) m^2
=56×72 m^2
=4032 m^2
Example:-Find the area of the shaded region in the figure where ABCD is a square of side 14 cm.
Area of square ABCD=14×14 cm^2
=196 cm^2
Diameter of each circle=14/2 cm=7 cm
Radius of each circle=7/2 cm
Area of one circle=πr^2=(22/7)×7/2×7/2 cm^2
=154/4 cm^2
=77/2 cm^2
Area of four circles=4×77/2 cm^2
=154 cm^2
Area of shaded region=(196-154)cm^2
=42cm^2
Example:-Find the area of the shaded design in figure where ABCD is a square of side 10cm and semicircles are drawn with each side of the square as diameter (Use π=3.14)
Let us mark the four unshaded regions as I,II,III,IV
Area of I + Area of III =Area of ABCD - Areas of two semicircles of each of radius 5cm
=[10×10-2×(1/2)×π×(5)^2] cm^2
=(100-3.14×25) cm^2
=(100-78.5) cm^2
=21.5 cm^2
Similarly,
Area of II + Area of IV=21.5 cm^2
so area of shaded region=Area of ABCD-Area of (I+II+III+IV)
=(100-2×21.5) cm^2
=(100-43) cm^2
=57 cm^2
Example:-If ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter.Find the area of the shaded region.
Area of semicircle=(1/2)πr^2
=1/2×22/7×14×14
=22×2×7
=308 cm^2
Area of sector=(θ/360°)×π(14)^2
=(90°/360°)×22/7×14×14
=(1/4)×22×2×14
=11×14
=154 cm^2
Area of △ABC=(1/2)×b×h
=(1/2)×14×14
=7×14
=98 cm^2
Area of segment BOC=Area of sector-Area of △ABC
=154-98
=56 cm^2
Area of shaded region=Area of semicircle-Area of segment BOC
=308-56
=252 cm^2
Example:-Calculate the area of the designed region in the figure common between the two quadrants of circles of radius 8 cm each.
Given
Square of side=8 cm
Area of square=s^2
=8×8
=64 cm^2
Area of sector I =(θ/360°)×πr^2
=(90°/360°)×22/7×8×8
=(1/4)×22/7×64
=16π cm^2
Area of △ADC=(1/2)×b×h
=(1/2)×8×8
=36 cm^2
Area of region a=Area of sector I - Area of △ADC
=16π - 36
=16×3.14-36
=50.24-36
=14.24 cm^2
Area of sector II =(θ/360°)×πr^2
=(90°/360°)×22/7×8×8
=50.24 cm^2
Area of △ADC=(1/2)×b×h
=(1/2)×8×8
=36 cm^2
Area of region b=Area of sector II - Area of region b
=50.24-36
=14.24 cm^2
Area of shaded region= Area of region a + area of region b
=14.24+14.24
=28.48 cm^2
EXERCISE
1.If the difference between the circumference and radius of a circle is 37cm then using π=22/7 the circumference(in cm)of the circle is
Difference between circumference and radius of the circle=37 cm
2πr-r=37
(2π- 1)r=37
r=37/(2π- 1)
r=37/(2×(22/7)-1)
r=37×7/37
r=7 cm
Circumference=37+7=44cm
2.If the circumference of a circle increases from 4π to 8π, then its area is
we know,
Circumference of a circle=4π
Radius(r)=circumference/2π
=4π /2π
Radius(r)=2
Area=πr^2=π2^2=4π
In second case,c=8π
Radius(R)=circumference/2π
=8π/2π
=4
Area=πR^2=π4^2=16π
16π is four times of 4π
4.If π is taken as 22/7,the distance (m)covered by a wheel of diameter 35cm,in one revolution is
Diameter of a wheel=35cm=35/100 m
Circumference of the wheel=d
3.If the area of a square is same as the area of a circle then ratio of their perimeters is
Let the side of square =a
Perimeter=4a
Area=a^2
Area of circle=a^2
Radius=√(Area)/π
=√(a^2)/π
=a/√ π
Circumference=2πr
=2π×a/√ π
=2a√ π
Ratio=4a:2√ π
=2:√ π
4.If π is taken as 22/7,the distance(m) covered by a wheel of diameter 35cm,in one revolution is
Diameter of a wheel=35cm=35/100 m
Circumference of the wheel=πd
=(22/7)×(35/100)
=110/100
=1.10
=1.1 m
Distance in one revolution=1.1m
Multiple choice Questions
1.The perimeter of a circle having radius 5 cm is equal to [ c ]
a)30 cm b)3.14 cm c)3.14 cm d)40 cm
Explanation:-The perimeter of a circle is equal to the circumference of a circle.
Circumference=2πr=2×3.14×5=31.4 cm
2.Area of the circle with radius 5 cm is equal to [ a ]
a)60 sq cm b)75.5 sq cm c)78.5 sq cm d)10.5 sq cm
Explanation:-Radius=5 cm
Area=πr^2=3.14×5×5=78.5 sq cm
3.The largest triangle inscribed in a semi-circle of radius r,then the area of that triangle is
a)r^2 b)1/2 r^2 c)2 r^2 d)√2 r^2 [ a ]
Explanation:-The height of the largest triangle inscribed will be equal to the radius of the semicircle and base will be equal to the diameter of the semi-circle
Area of the triangle=1/2×base×height
=1/2×2 r×r=r^2
4.If the perimeter of the circle and square are equal,then the ratio of their areas will be equal to
a)14:11 b)22:7 c)7:22 d)11:14 [ a ]
Explanation:-The perimeter of circle=Perimeter of square
2πr=4a
a=πr/2
a^2=(πr/2)^2
Area of circle/Area of square=πr^2/(πr/2)^2
=πr^2/(πr^2/4)
=πr^2×(4/π^2 r^2)
=4/π
=4/(22/7)
=28/22
=14/11
5.The area of the circle that can be inscribed in a square of side 8 cm is [ b ]
a)36πcm^2 b)16πcm^2 c)12πcm^2 d)9πcm^2
Explanation:-Side of square=8 cm
Diameter of circle=Side of square=8 cm
Radius of circle=4 cm
Area of circle=π(4)^2=16πcm^2
6.The area of the square that can be inscribed in a circle of radius 8 cm is [ d ]
a)256 cm^2 b)64 cm^2 c)642 cm^2 d)128 cm^2
Explanation:-Radius of circle=8 cm
Diameter of circle=16 cm=diagonal of the square
Let 'a' be the triangle side,and hypotenuse is 16 cm
By Pythagorean theorem,
(16)^2=a^2 +a^2
a^2=256/2
a^2=128 cm^2
7.The area of a sector of a circle with radius 6 cm if the angle of the sector is 60° [ c ]
a)142/7 b)152/7 c)132/7 d)122/7
Explanation:-Angle of the sector=60°
Area of the sector=(θ/360°)×πr^2
Area of sector with angle 60°=(60°/360°)×πr^2 cm^2
=(36/6)π cm^2
=6×(22/7) cm^2
=132/7 cm^2
8.In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre.The length of the arc is
a)20 cm b)21 cm c)22 cm d)25 cm [ c ]
Explanation:-Length of an arc=(θ/360°)×2πr
Length of the arc AB= 60°/360°×2×(22/7)×21
=1/6×2×(22/7)×21
Arc AB length=22 cm
9.In a circle of radius 21 cm,an arc subtends an angle of 60° at the centre.the area of the sector formed by the arc is [ c ]
a)200 cm^2 b)220 cm^2 c)231 cm^2 d)250 cm^2
Explanation:-The angle subtended by an arc=60°
Area of the sector=(60°/360°)×πr^2 cm^2
=441/6×(22/7) cm^2
=231 cm^2
10.Area of the sector of angle P(in degrees)of a circle with radius R is [ d ]
a)p/180×2πR b)p/180×πR^2 c)p/360×2πR d)p/720×2πR^2
Explanation:-The area of the sector=(θ/360°)×πr^2
Given θ=p
So area of sector=p/360°×πR^2
Multiplying and dividing by 2 simultaneously
=[(p/360°)/(πR^2)]×[2/2]
=(p/720)×2πR^2
11.If the area of a circle is 154 cm^2,then its perimeter is [ c ]
a)11 cm b)22 cm c)44 cm d)55 cm
Explanation:-Area of the circle=154 cm^2
πr^2=154
(22/7)×r^2=154
r^2=(154×7)/22
r^2=7×7
r=7 cm
Perimeter of circle=2πr
=2×(22/7)×7
=44 cm
12.If the sum of the areas of two circles with radii R 1 and R 2 is equal to the area of a circle of radius R,then [ b ]
a)R 1+ R 2 =R b)R 1^2 +R 2^2=R^2 c)R 1 +R 2 <R d)R 1 ^2+R 2 ^2<R ^2
Explanation:-According to the given problem
πR 1^2+πR 2^2=πR^2
π(R 1^2+R 2^2)=πR^2
R 1^2+R 2^2=R^2
13.If θ is the angle (in degrees) of a sector of a circle of radius r, then the length of arc is [ a ]
a)(πr^2θ)/360 b)(πr^2θ)/180 c)2πrθ/360 d)2πrθ/180
Explanation:-If θ is the angle(in degrees) of a sector of a circle of radius r,then the area of sector is 2πrθ/360
14.It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality.The radius of new park would be [ a ]
a)10 m b)15 m c)20 m d)24 m
Explanation:-
Radii of two circular parks will be
R 1=16/2=8 m
R 2=12/2=6 m
Let R be the radius of the new circular park.If the areas of two circles with radii R 1 and R 2 is equal to the area of circle with radius R,then
R^2=R 1^2 +R 2^2
=8^2 + 6^2
=64+36
=100
R=10 m
15.The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36 cm and 20 cm is [ c ]
a)56 cm b)42 cm c)28 cm d)16 cm
Explanation:-Given R=R 1+R 2
Here,
R 1=36/2=18 cm
R 2=20/2=10 cm
R=R 1+R 2=18+10=28 cm
The radius of required circle is 28 cm
16.Find the area of a sector of circle of radius 21 cm and central angle 120° [ b ]
a)441 cm^2 b)462 cm^2 c)386 cm^2 d)512 cm^2
Explanation:- Given
radius(r)=21 cm
central angle=120°
area of sector=πr^2θ/360°=(22/7)×21×21×120/360
=22×21
=462 cm^2
17.The wheel of a motor cycle is of radius 35 cm.The number of revolutions per minute must the wheel make so as to keep a speed of 66 km/hr will be [ c ]
a)50 b)100 c)500 d)1000
Explanation:-Circumference of the wheel=2πr
=2×(22/7)×35
=220 cm
Speed of the wheel=66 km/hr
=(66×1000)/60 m/min
=1100×100 cm/min
=110000 cm/min
Number of revolutions in 1 min=110000/220
=500
18.If the perimeter and the area of a circle are numerically equal,then the radius of the circle is
a)2 units b)π units c)4 units d)7 units [ a ]
Explanation:-According to the given,
Perimeter of circle=Area of circle
2πr=πr^2
r=2
Radius=2 units
19.The area of a quadrant of a circle with circumference of 22 cm is [ b ]
a)77 cm^2 b)77/8 cm^2 c)35.5 cm^2 d)77/2 cm^2
Explanation:-Circumference=22 cm
2πr=22
2×(22/7)πr=22
r=7/2 cm
Area of quadrant of a circle=(1/4)πr^2
=(1/4)×(22/7)×(7/2)×(7/2)
=77/8 cm^2
20.In a circle of radius 14 cm,an arc subtends an angle of 30° at the centre,the length of the arc is
a)44 cm b)28 cm c)11 cm d)22/3 cm [ d ]
Explanation:-Radius(r)=14 cm
Length of arc=2πrθ/360
=2×(22/7)×14 ×(30/360)
=22/3 cm