CIRCLE WORD PROBLEMS WORKSHEET

Worksheet given in this section will be much useful for the students who would like to practice solving word problems on circles. 

Circle Word Problems Worksheet

Problem 1 :

The diameter of a cart wheel is 2.1 m. Find the distance traveled when it completes 100 revolutions.

Problem 2 :

The diameter of a circular park is 98 m. Find the cost of fencing it at $4 per meter.

Problem 3 : 

A wheel makes 20 revolutions to cover a distance of 66 m. Find the diameter of the wheel.

Problem 4 :

The radius of a cycle wheel is 35 cm. How many revolutions does it make to cover a distance of 81.40 m ?

Problem 5 :

The radius of a circular park is 63 m. Find the cost of fencing it at $12 per metre.

Problem 6 :

A goat is tethered by a rope 3.5 m long. Find the maximum area that the goat can graze.

Example 7 :

The circumference of a circular park is 176 m. Find the area of the park.

Problem 8 :

A silver wire when bent in the form of a square encloses an area of 121 sq. cm. If the same wire is bent in the form of a circle. Find the area of the circle.

Circle Word Problems Worksheet - Solutions

Problem 1 :

The diameter of a cart wheel is 2.1 m. Find the distance traveled when it completes 100 revolutions.

Solution :

In order to find the distance covered in one revolution, we have to find the circumference of the circle.

Radius of wheel is 

=  2.1 / 2

=  1.05 m

Distance traveled when 100 revolutions are completed : 

=  100 x circumference of the wheel

=  100 x 2 x (22/7) x (1.05)

=  100 x 2 x 22 x 0.15

=  660 m

Problem 2 :

The diameter of a circular park is 98 m. Find the cost of fencing it at $4 per meter.

Solution :

Diameter of the circular park is 98 m.

Radius : 

=  98/2

=  49 m

Cost of fencing is $4 per meter. 

Length covered by fencing is equal circumference of the circular park.

Circumference of the park is 

=  2πr

=  2 (22/7) x 49

=  2 x 22 x 7

=  308 m

Cost for fencing is

=  308 x 4

=  $1232

Problem 3 : 

A wheel makes 20 revolutions to cover a distance of 66 m. Find the diameter of the wheel.

Solution :

Given :

A wheel makes 20 revolutions to cover a distance of 66 m. 

Then, the distance covered in one revolution is

=  66/20

=  33/10 m

The distance covered in one revolution is equal to circumference of the circle. 

Then,

2πr  =  33/10

2 x (22/7) x  r  =  33/10

(44/7) x r  =  33/10

Multiply each side by 44/7.

r  =  (33/10) x (7/44)

r  =  0.525

Diameter : 

=  2 x radius

=  2 x 0.525

=  1.05 m

Problem 4 :

The radius of a cycle wheel is 35 cm. How many revolutions does it make to cover a distance of 81.40 m ?

Solution :

Radius is given in centimeters and the distance is given in meters.

To have all the measures in meters, we can convert the given radius measure to meters.

Radius :

=  35 cm

=  35/100 m

=  0.35 m

Let 'n' be the number of revolutions required to cover the distance of 81.40 m. 

n x one revolution of cycle wheel  =  81.40

n x 2πr  =  81.40

n x 2 x (22/7) x 0.35  =  81.40

n x 2.2  =  81.40

Divide each side by 2.2

n  =  81.40/2.2

n  =  37

So, the cycle wheel has to revolve 37 times to cover the distance.

Problem 5 :

The radius of a circular park is 63 m. Find the cost of fencing it at $12 per metre.

Solution :

Radius of the circular park is 63 m.

Length of fencing is equal to circumference of the circular park.  

Circumference of the cicular park is 

=  2πr

=  2 x (22/7) x 63

=  2 x 22 x 9

=  396 m

Cost of fencing is $12 per meter. 

Total cost for fencing the park is

=  396 x 12

=  $4752

Problem 6 :

A goat is tethered by a rope 3.5 m long. Find the maximum area that the goat can graze.

Solution :

Radius of the circle is equal to length of the rope. 

Then, the radius is 

=  3.5 m

=  7/2 m

Area grazed by the goat is equal to the area of the circle with radius 7/2 m. 

Maximum area grazed by the goat is 

=  πr2

=  (22/7) x (7/2) x (7/2)

=  77/2

  =  38.5 sq. m

Example 7 :

The circumference of a circular park is 176 m. Find the area of the park.

Solution :

Circumference of the circular park is 176 m (given). 

Then, 

2πr  =  176

2 x (22/7) x r  =  176

(44/7) x  r  =  176

Multiply each side by 7/44.

r  =  176 x (7/44)

r  =  28 m

Area of the circular park is 

=  πr2

=  (22/7) x 28 x 28

= 22 x 4 x 28

= 2464 sq. m.

Problem 8 :

A silver wire when bent in the form of a square encloses an area of 121 sq. cm. If the same wire is bent in the form of a circle. Find the area of the circle.

Solution :

Area of the square is 121 sq. cm. (given)

Let a be the length of each side of the square.

Then, 

a2  =  121

a2  =  112

a  =  11

Then, perimeter of the square is 

=  4a

= 4 x 11

= 44 cm

Length of the wire is equal to perimeter of the square. 

So, length of the wire is 44 cm.

If thee wire is bent in the form of a circle, then the circumference of the circle is equal to length of the wire

Circumference of a circle  =  44 cm

2πr  =  44

2 x (22/7) x r  =  44

(44/7) x r  =  44

Multiply each side by 7/44.

r  =  44 x (7/44)

r  =  7 cm

Area of the circle is

=  πr2

=  (22/7) x 7 x 7

=  154 sq. cm

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