Chapter 3:Pair of Linear Equations in Two Variables
Chapter 5: Arithmetic Progression
Chapter 8: Introduction to Trigonometry
Chapter 9:Some Applications of Trigonometry
Chapter 12:Area Related to Circles
Area Related to Circles
angle of the sector is 60°
the area of sector = (θ/360°)×πr2
∴ Area of the sector with angle 60° = (60°/360°)×πr2 cm2
= (36/6)π cm2
= 6×22/7 cm2 = 132/7 cm2
Circumference of the circle = 22 cm (given)
quadrant of a circle is a sector which is making an angle of 90°.
radius of the circle = r
here
2πr = 22,
r = 22/2π cm = 7/2 cm
∴ Area of the quadrant = (θ/360°) × πr2
Here, θ = 90°
So, A = (90°/360°) × π r2 cm2
= (49/16) π cm2
= 77/8 cm2 = 9.6 cm2
Length of minute hand = radius of the clock
Radius of the circle r= 14 cm (given)
Angle swept by minute hand in 60 minutes = 360°
The angle swept by the minute hand in 1 minutes = 360°/60 = 6o
The angle swept by the minute hand in 5 minutes =5 x 6o =30o
Area of a sector = (θ/360°) × πr2
= (30°/360°) × πr2 cm2
= (1/12) × π142
= (49/3)×(22/7) cm2
= 154/3 cm2
(i) minor segment
(ii) major sector. (Use π = 3.14)
Given radius of a circle AO=10cm
A perpendicular is drawn from the center of circle to the chord of the circle which bisect the chord
AD=DC
∠AOD=∠COD =45°
∠AOC =∠AOD +∠COD
=45° +45°=90
In ΔAOD
sin45°= AD/AO= 1/√2 =AD/10
AD= 5√2
and
cos45°= OD/AO= 1/√2 =OD/10
OD= 5√2
and
AC = 2 AD =2 x 5√2 =10√2
Also, area of ΔAOC = 1/2 × AC x OD
So, area of ΔAOB = 1/ × 10√2 × 5√2
= 50 cm2
(i) Area of minor sector = (90/360°)×πr2
= 90/360 x 3.14 x 102
= 78.5 cm2
Now, area of minor segment = area of minor sector – area of ΔAOB
= 78.5 – 50
= 28.5 cm2
(ii) Area of major sector = Area of circle – Area of minor sector
= (3.14×102)-78.5
= 235.5 cm2
(i) the length of the arc
(ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord
Given,
Radius = 21 cm
θ = 60°
(i) Length of an arc = θ/360°×2πr
∴ = (60°/360°)×2×(22/7)×21
= (1/6)×2×(22/7)×21
Or Arc AB Length = 22cm
(ii) the angle subtend at center = 60°
= (60°/360°)×π r2 cm2
= 441/6×22/7 cm2
= 231 cm2
In ΔOAB
∠OAB =∠OBA
∠OAB + ∠AOB + ∠OBA =180
2∠OAB +60 =180
∠OAB =60
So
ΔOAB is equileteral triangle
area will be √3/4×a2 sq. Units.
√3/4×212 = (441×√3)/4 cm2
(iii)Area of segment APB = Area of sector OAPB – Area of ΔOAB
Or, Area of segment APB = [231-(441×√3)/4] cm2
Given,
Radius = 15 cm
θ = 60°
So,
Area of sector OAPB = (60°/360°)×πr2 cm2
= 225/6 πcm2
In ΔOAB
∠OAB =∠OBA
∠OAB + ∠AOB + ∠OBA =180
2∠OAB =120
∠OAB =60
So
ΔOAB is equileteral triangle
So, Area of ΔAOB = (√3/4) ×a2
Or, (√3/4) ×152
∴ Area of ΔAOB = 97.31 cm2
Now, area of minor segment APB = Area of OAPB – Area of ΔAOB
Or, area of minor segment APB = ((225/6)π – 97.31) cm2 = 20.43 cm2
And,
Area of major segment = Area of circle – Area of segment APB
Or, area of major segment = (π×152) – 20.4 = 686.06 cm2
cord AB substended an angle 120° to the center
∠ AOB = 120°
draw OD perpendicular to AB
AB=BD
∠ AOD =∠ BOD= 60°
here radius r=12 cm
IN ΔAOD
sin60°=AD/AO= √3/2=AD/12
AD =6√3 =10.38
cos60°=OD/AO= 1/2=OD/12
OD =6cm
length AB = 2AD =2 x 10.38 = 20.76
So, the area of ΔAOB = 1/2 × base × height
So, area of ΔAOB = 1/2 × 20.76 × 6 = 36√3 cm = 62.28 cm2
Now, the area of the minor sector = (θ/360°)×πr2
= (120/360)×(22/7)×122
= 150.72 cm2
∴ Area of the corresponding Minor segment = Area of the Minor sector – Area of ΔAOB
= 150.72 cm2– 62.28 cm2 = 88.44 cm2
(i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)
side of square =15 m
radius =5m
Here, the length of rope will be the radius of the circle i.e. r = 5 m
(i)
Now, the area of the part of the field where the horse can graze = (θ/360°) × πr2
(90/360°) x 3.14 x 52 = 19.625 m2
(ii)If the rope is increased to 10 m,
Area of circle will be = πr2 =22/7×102 = 314 m2
Now, the area of the field graze by horse=(90/360°) x 3.14 x 102
= 314/4 = 78.5 m2
∴ Increase in the grazing area = 78.5 m2 – 19.625 m2 = 58.875 m2
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
Diameter (D) = 35 mm
number of diameters = 5
Now, length of 5 diameters = 35×5 = 175
Circumference of the circle = 2πr
C = πD = 22/7×35 = 110
Area of the circle = πr2
Or, A = (22/7)×(35/2)2 = 1925/2 mm2
(i)Total length of silver wire required = Circumference of the circle + Length of 5 diameter
= 110+175 = 185 mm
(ii)Number of sectors in the brooch = 10
angle of each sector = 360° /10 =36°
So, the area of each sector =(θ/360°) × πr2
(36/360°)× 22/7 x (35/2)2
= 22/70 x 1225/4
=26950/280
∴ = 385/4 mm2
The radius of the umbrella = 45 cm
Total number of ribs (n) = 8
The Angle between two rib = 360°/8 =45°
the area of one rib =(θ/360°) × πr2
45/360° × 22/7 x (45)2
22/7 x 8
∴ The area between the two consecutive ribs of the umbrella = A/n
6364.29/8 cm2
Or, The area between the two consecutive ribs of the umbrella = 795.5 cm2
Given,
Radius (r) = 25 cm
Sector angle (θ) = 115°
Since there are 2 blades,
The total area of the sector made by wiper = 2×(θ/360°)×π r2
= 2×(115/360)×(22/7)×252
= 2×158125/252 cm2
= 158125/126 = 1254.96 cm2
(Use π = 3.14)
Let O bet the position of Lighthouse.
Given, radius (r) = 16.5 km
Sector angle (θ) = 80°
Or, Area of sector = (θ/360°)×πr2
= (80°/360°)×πr2 km2
= 189.97 km2
Total number of equal designs = 6
AOB= 360°/6 = 60°
Radius of the cover = 28 cm
Cost of making design = ₹ 0.35 per cm2
Since the two arms of the triangle are the radii of the circle and thus are equal, and one angle is 60°, ΔAOB is an equilateral triangle. So, its area will be (√3/4)×a2 sq. units
Here, a = OA
∴ Area of equilateral ΔAOB = (√3/4)×282 = 333.2 cm2
Area of sector ACB = (60°/360°)×πr2 cm2
= 410.66 cm2
So, area of a single design = area of sector ACB – area of ΔAOB
= 410.66 cm2 – 333.2 cm2 = 77.46 cm2
∴ Area of 6 designs = 6×77.46 cm2 = 464.76 cm2
So, total cost of making design = 464.76 cm2 ×Rs.0.35 per cm2
= Rs. 162.66
Area of a sector of angle p (in degrees) of a circle with radius R is
(A) p/180 × 2πR
(B) p/180 × π R2
(C) p/360 × 2πR
(D) p/720 × 2πR2
The area of a sector = (θ/360°)×πr2
Given, θ = p
So, area of sector = p/360×πR2
Multiplying and dividing by 2 simultaneously,
= (p/360)×2/2×πR2
= (2p/720)×2πR2
So, option (D) is correct.