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

**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.

**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,**

**2**πr = 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 **2**πr = 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

= **2**πr

= 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

= πr^{2}

= (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

= πr^{2}

= (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,

a^{2} = 121

a^{2} = 11^{2}

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

= πr^{2}

= (22/7) x 7 x 7

= 154 sq. cm

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