Problem 1 :
What must be the radius of a circular running path, around which an athlete must run 5 times in order to describe 1 km?
Solution :
Distance covered by athlete = 1 km
5 ⋅ 2πr = 1 km
5 ⋅ 2 ⋅ (3.14) ⋅ r = 1000 m
r = 31.84 m
Problem 2 :
In a circle of diameter 40 cm, a chord is of length 20 cm. Find the length of the minor arc of the chord.
Solution :
Let arc AB = s.
It is given that OA = 20 cm and chord AB = 20 cm.
Hence triangle OAB is an equilateral triangle.
<AOB = 60 degree (or) (π/3) radians
θ = arc length/radius
(π/3) = s/20
s = 20π/3
π = 3.14
s = 20(3.14)/3 = 20.93
So, the required arc length is 20.93 cm.
Problem 3 :
Find the degree measure of the angle subtended at the center of circle of radius 100 cm by an arc of length 22 cm.
Solution :
Radius of the circle = 100 cm
Length of arc = 22 cm
θ = arc length/radius
= 22/100
Radian measure = (π/180) Degreee measure
22/100 = (22/7)/180 ⋅ Degree measure
Degree measure = (22 ⋅ 7 ⋅ 180)/(22 ⋅ 100)
= 126/10
12◦ (6/10)
Multiplying 6/10 by 60, we get 36'
So, the required angle is 12◦ 36'
Problem 4 :
What is the length of the arc intercepted by a central angle of measure 41◦ in a circle of radius 10 ft?
Solution :
arc length = (θ/360) ⋅ 2 ⋅ π ⋅ r
= (41/360) ⋅ 2 ⋅ (3.14) ⋅ r
= 7.16 feet
So, the required arc length is 7.16 feet.
Problem 5 :
If in two circles, arcs of the same length subtend angles 60◦ and 75◦ at the center, find the ratio of their radii.
Solution :
Let r1 and r2 be the radii of the given circles and let their arcs of same length subtend angles of 60 degree and 75 degree at their centers.
60◦ = 60 ⋅ (π /180) = π/3 θ = arc length/radius π/3 = s/r1 s = πr1 /3 -----(1) |
75◦ = 75 ⋅ (π /180) = 5π/12 θ = arc length/radius 5π/12 = s/r2 s = 5πr2/12 -----(2) |
(1) = (2)
πr1 /3 = 5πr2 /12
r1/r2 = (5π/π)(3/12)
r1/r2 = 5/4
r1 : r2 = 5 : 4
Problem 6 :
A circle has a radius of 6 inches. Find the length of the arc intercepted by a central angle of 45◦ .
Solution :
Length of arc S = rθ
radius = 6 inches
θ = 45◦
Here the angle is given as degree measure, to convert into radian measure, we have to multiply by π/180
= 45 x (π/180)
= π/4
Applying the radian measure in the formula, we get
S = 6(π/4)
= (6/4) π
= 1.5π inches
Problem 7 :
A circle has a radius of 5cm. Find the length of the arc intercepted by a central angle of 38◦
Solution :
Length of arc S = rθ
radius = 5 cm
θ = 38◦
Here the angle is given as degree measure, to convert into radian measure, we have to multiply by π/180
= 38 x (π/180)
= 38π/180
= 19π/90
Applying the radian measure in the formula, we get
S = 5(19π/90)
= (95/90) π
= 1.05 π inches
Problem 8 :
A circle has a radius of 10 ft. Find the length of the arc intercepted by a central angle of 5π/12
Solution :
Length of arc S = rθ
radius = 10 ft
θ = 5π/12
Applying these values in the formula, we get
S = 10(5π/12)
= 50π/12
= 4.16 π ft.
Problem 9 :
Find the degree measure to the nearest tenth of the central angle that has an arc length of 87 and a radius of 16 cm.
Solution :
Length of arc S = rθ
Length of arc S = 87
radius = 16 cm
θ = ? (in radian)
Applying these values in the formula, we get
87 = 16(θ)
θ = 87/16
Converting in to degree
= (87/16) x (180/π)
= 311.7
Approximately 312.
Problem 10 :
Find the degree measure to the nearest tenth of the central angle that has an arc length of 5.6 and a radius of 12 cm.
Solution :
Length of arc S = rθ
Length of arc S = 5.6
radius = 12 cm
θ = ? (in radian)
Applying these values in the formula, we get
5.6 = 12(θ)
θ = 5.6/12
Converting in to degree
= (5.6/12) x (180/π)
= 26.75
Approximately 26.75
Problem 11 :
A sector has arc length 12 cm and a central angle measuring 1.25 radians. Find the radius and the area of the sector.
Solution :
Length of arc S = rθ
Length of arc S = 12 cm
radius = ?
θ = 1.25 radians
Applying these values in the formula, we get
12 = r(1.25)
r = 12/1.25
r = 9.6 cm
Area of sector = (1/2)r2 θ
= (1/2) (9.6)2 (1.25)
= 57.6 cm2
Problem 12 :
Find the area of the sector of the circle that has a central angle measure of π/6 and a radius of 14 cm.
Solution :
radius = 14 cm
θ = π/6
Area of sector = (1/2)r2 θ
= (1/2) (14)2 (π/6)
= 16.3π cm2
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
May 03, 25 08:08 AM
May 03, 25 07:47 AM
Apr 30, 25 11:12 PM