CIRCLES CALCULATE AREA CIRCUMFERENCE RADIUS AND DIAMETER WORKSHEET

Problem 1 :

Find the circumference of a circle whose diameter is 21 cm.

Solution :

Diameter  =  21 cm

Radius  =  21/2 = 10.5 cm

Circumference of circle  =  2 π r

  =  2 (22/7) x 10.5

  =  44 x 1.5

  =  66 cm

Hence the circumference of circle is 66 cm.

Problem 2 :

A wire of length 88 cm is bent as a circle. What is the radius of the circle?

Solution :

Length of the wire  =  circumference of the circle

2 π r  =  88 cm

2 x (22/7) x r  =  88

r  =  (88 x 7)/(2 x 22)

r  =  14 cm

Hence the radius of the circle is 14 cm

Problem 3 :

The diameter of a bicycle wheel is 63 cm. How much distance will it cover in 20 revolutions?

Solution :

When a wheel makes one complete revolutions,

Distance covered in one rotation = Circumference of wheel

Circumference of the wheel = 2πr units 

=  2 x (22/7) x (63/2) cm

=  22 x 9  =  198 cm

For one revolution, the distance covered = 198 cm

For 20 revolutions, the distance covered = 20 × 198 cm

=  3960 cm

=  39 m 60 cm [100 cm = 1 m]

Hence the distance covered in 20 revolution is 39 m 60 cm

Problem 4 :

A scooter wheel makes 50 revolutions to cover a distance of 8800 cm. Find the radius of the wheel.

Solution :

Distance travelled 

  =  Number of revolutions/Circumference

2πr  =  8800/50

2πr  =  176

2 x (22/7) x r  =  176

r  =  176 x (7/22) x (1/2)

r  =  28 cm

Hence the radius of the wheel is 28 cm

Problem 5 :

The radius of a cart wheel is 70 cm. How many revolution does it make in travelling a distance of 132 m.

Solution :

Given: r = 70 cm, Distance travelled = 132 m.

Circumference of a cart wheel = 2πr

= 2 x (22/7) x 70

= 440 cm

Distance travelled

= Number of revolutions x  Circumference

Number of revolutions

  = Distance travelled/Circumference

  =  132 m/440 cm

  =  13200/440

  =  30

Hence the number of revolution is 30.

Problem 6 :

The circumference of the roll of caution tape decreases 10.5 inches after a construction worker uses some of the tape. Which is the best estimate of the diameter of the roll after the decrease?

a) 5 inches     b) 7 inches    c) 10 inches    d) 12 inches

Solution :

Circumference of the circle = 31.4 inches

Let R be the radius of the roll before it uses.

2πR = 31.4

2 x 3.14 x R = 31.4

6.28 x R = 31.4

R = 31.4 / 6.28

R = 5 inches

Diameter = 10 inches

The circumference decreases to 10.5 inches after it is being used.

Let r be the radius of the roll after it uses.

2πr = 10.5

2 x 3.14 x r = 10.5

r = 10.5/6.28

r = 1.67

Diameter = 3.34 inches

Difference between the diameter = 10 - 3.34

= 6.66

approximately 7 inches. Option b is correct.

Problem 7 :

A semicircle is one-half of a circle. Find the perimeter of the semicircular region

Solution :

Diameter of semicircle = 6 m, radius = 3 m

Area of semicircle = (1/2) πr²

= (1/2) x 3.14 x 3²

= 14.13 m²

Problem 8 :

A circular sinkhole has a circumference of 75.36 meters. A week later, it has a circumference of 150.42 meters.

a. Estimate the diameter of the sinkhole each week.

b. How many times greater is the diameter of the sinkhole now compared to the previous week?

Solution :

Circumference of sinkhole = 75.36 meters

Circumference of sinkhole a week later = 150.42 meter

Radius of circle before leakage = r, after leakage = R

2πr = 75.36

r = 75.36/2π

= 75.36 / 6.28

r = 12 m, diameter = 24 m

2πR = 150.42

R = 150.42 / 6.28

R = 23.9 m, diameter = 47.8 m

a) The difference between the diameters = 47.8 - 24

= 23.8 m

b) Number of times = 47.8/24

= 1.99

Approximately 2 times.

Problem 9 :

Bicycles in the late 1800s looked very different than they do today.

a. How many rotations does each tire make after traveling 600 feet? Round your answers to the nearest whole number.

b. Would you rather ride a bicycle made with two large wheels or two small wheels? Explain.

Solution :

a)  Distance covered in one rotation by the large wheel :

R = 30 inches

n is the number of times the wheel to be rotated.

n (2πR) = 600

n(2 x 3.14 x 30) = 600

n(188.4) = 600

n = 600/188.4

n = 3.18

Approximately 4 times

Distance covered in one rotation by the small wheel :

r = 9 inches

n (2πr) = 600

n(2 x 3.14 x 9) = 600

n(56.52) = 600

n = 600/56.52

n = 10.61

Approximately 11 times.

b) To cover the given distance quickly, travelling with large wheel is preferable.

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