# CIRCLES WORD PROBLEMS

## About "Circles word problems"

Circles word problems :

Here we are going to see some practical problems based on the concept circles.

Area of circle  =  πr²

Circumference of circle  =  2πr

Example 1 :

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

Solution :

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

radius of wheel  =  2.1 / 2  =  1.05 m

Distance travelled when it completes 100 revolutions  =  1000 x circumference of the wheel

=  100 x 2 x (22/7) x (1.05)

=  100 x 2 x 22 x 0.15

=  660 m²

Example 2 :

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

Solution :

Diameter of the circular park  =  98 m

radius  =  98/2  =  49 m

Cost of fencing  =  \$ 4 per meter

length covered by fencing  =  circumference of the circular park

Circumference of park  =  2πr

=  2 (22/7) x 49

=  2 x 22 x 7

=  308 m

Required cost for fencing  =  308 x 4  =  \$1232

Example 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

20 x one revolution  =  66 m

20 x 2πr  =  66

20 x 2 x (22/7) x  r  =  66

r  =  66 x (7/22) x (1/20) x (1/2)

r  =  21/40

r  =  0.525

Diameter  =  2r  =  2(0.525)  =  1.05 m

Example 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 of cycle wheel  =  35 cm  =  0.35 m

n x Revolution of cycle wheel  =  81.40

n x 2πr  =  81.40

n x 2 x (22/7) x 0.35  =  81.40

n  =  81.40/(2 x 22 x 0.05)

n  =  81.40/2.2

n  =  37

Hence the cycle wheel has to revolve 374 times to cover the distance.

Example 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  =  63 m

Length of fencing  =  Circumference of circular park

=   2πr   =  2 x (22/7) x 63

=  2 x 22 x 9

=  396 m

Cost of fencing  = \$ 12 per meter

Total cost for fencing  =  396 x 12

=  \$4752

Example 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 = Length of the rope

radius r = 3.5 m = 7/2 m

Maximum area grazed by the goat = πr²sq. units.

= (22/7) x (7/2) x (7/2)

=  77/2  =  38.5 sq. m

Maximum area grazed by the goat is 38.5 sq. m.

Example 7 :

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

Solution :

Circumference = 176 m (given)

2πr = 176

2 x (22/7) x r  =  176

44 r /7  =  176

r  =  (176 x 7)/44

r  =  28 m

Area of the park = πr²

=  (22/7) x 28 x 28

= 22 x 4 x 28

= 2464 sq. m.

Example 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 :

Let a be the side of the square

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

a² = 121, a² = 11 cm (11 x 11 = 121)

Perimeter of the square = 4a units

= 4 x 11 cm

= 44 cm

Length of the wire = Perimeter of the square

= 44 cm

The wire is bent in the form of a circle

The circumference of the circle = Length of the wire

Circumference of a circle = 44 cm

2πr = 44

2 x (22/7) x r  =  44

r = 44 x (7/22) x (1/2)

r  =  7 cm

Area of the circle = πr²

= (22/7) x 7 x 7

Area of the circle = 154 cm² After having gone through the stuff given above, we hope that the students would have understood "Circles word problems".

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