**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 2**πr = 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 **2**πr = 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

= **2**πr = 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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