**1. Find the circumference of the circle with the following radius: (Take π = 22/7)**

**(a) 14 cm**

**Solution:**

Given,

Radius of circle = 14 cm

Now,

Circumference of the circle = 2πr

= 2 × (22/7) × 14

= 2 × 22 × 2

= 88 cm

**(b) 28 cm**

**Solution:**

Given,

Radius of circle = 28 cm

Now,

Circumference of the circle = 2πr

= 2 × (22/7) × 28

= 2 × 22 × 4

= 176 cm

**(c) 21 cm**

**Solution:**

Given,

Radius of circle = 21 cm

Now,

Circumference of the circle = 2πr

= 2 × (22/7) × 21

= 2 × 22 × 3

= 132 cm

**2. Find the area of the following circles, given that:**

**(a) Radius = 14 mm (Take π = 22/7)**

**Solution:**

Given,

Radius of circle = 14 mm

Now,

Area of the circle = πr^{2}

= 22/7 × 14^{2}

= 22/7 × 196

= 22 × 28

= 616 mm^{2}

**(b) Diameter = 49 m**

**Solution:**

Given,

Diameter of circle (d) = 49 m

We know that, radius (r) is half of diameter = d/2

= 49/2

= 24.5 m

Now,

Area of the circle = πr^{2}

= 22/7 × (24.5)^{ 2}

= 22/7 × 600.25

= 22 × 85.75

= 1886.5 m^{2}

**(c) Radius = 5 cm**

**Solution:**

Given,

Radius of circle = 5 cm

Now,

Area of the circle = πr^{2}

= 22/7 × 5^{2}

= 22/7 × 25

= 550/7

= 78.57 cm^{2}

**3. If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π = 22/7)**

**Solution:**

Given,

Circumference of the circle = 154 m

Now,

Circumference of the circle = 2πr

So,

154 = 2 × (22/7) × r

154 = 44/7 × r

r = (154 × 7)/44

r = (14 × 7)/4

r = (7 × 7)/2

r = 49/2

r = 24.5 m

Now,

Area of the circle = πr^{2}

22/7 × (24.5)2

= 22/7 × 600.25

= 22 × 85.75

= 1886.5 m^{2}

Hence, the radius of the circle is 24.5 m, and the area of the circle is 1886.5 m^{2}.

**4. A gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs ₹ 4 per meter. (Take π = 22/7)**.

**Solution:**

Given,

Diameter (d) of the circular garden = 21 m

We know that, radius (r) = d/2

= 21/2

= 10.5 m

Now,

Circumference of the circle = 2πr

= 2 × (22/7) × 10.5

= 462/7

= 66 m

The length of rope required = 2 × 66 = 132 m

Now,

Cost of 1 m rope = ₹ 4 (given)

Cost of 132 m rope = ₹ 4 × 132

= ₹ 528

**5. From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)**

**Solution:**

Given,

Total radius of circular sheet (R) = 4 cm

A circle of radius to be removed (r) = 3 cm

Now,

The area of the remaining sheet = πR^{2} – πr^{2}

= π (R^{2} – r^{2})

= 3.14 (42 – 32)

= 3.14 (16 – 9)

= 3.14 × 7

= 21.98 cm^{2}

Hence, the area of the remaining sheet is 21.98 cm^{2}.

**6. Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs ₹ 15. (Take π = 3.14)**

**Solution:**

Given,

Diameter of the circular table cover = 1.5 m

So, radius (r) = d/2

= 1.5/2

= 0.75 m

Now,

Circumference of the circular table cover = 2πr

= 2 × 3.14 × 0.75

= 4.71 m

Hence, the length of the lace = 4.71 m

Now,

Cost of 1 m lace = ₹ 15 [given]

Cost of 4.71 m lace = ₹ 15 × 4.71

= ₹ 70.65

**7. Find the perimeter of the adjoining figure, which is a semicircle including its diameter.**

**Solution:**

Given,

Diameter of semi-circle = 10 cm

So, radius (r) = d/2

= 10/2

= 5 cm

Now,

Circumference of the semi-circle = πr

= (22/7) × 5

= 110/7

= 15.71 cm

Hence,

Perimeter of the given figure = Circumference of the semi-circle + semi-circle diameter

= 15.71 + 10

= 25.71 cm

**8. Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹15/m ^{2}. (Take π = 3.14)**

**Solution:**

Given,

Diameter of the circular table-top = 1.6 m

So, radius (r) = d/2

= 1.6/2

= 0.8 m

Now,

Area of the circular table-top = πr^{2}

= 3.14 × (0.8)^{ 2}

= 2.0096 m^{2}

Cost of polishing 1 m^{2} area = ₹ 15 [given]

Cost of polishing 2.0096 m^{2} area = ₹ 15 × 2.0096

= ₹ 30.144

Hence, the cost of polishing 2.0096 m^{2} area is ₹ 30.144.

**9. Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take π = 22/7)**

**Solution:**

Given,

Length of wire =44 cm

If the wire is bent into a circle,

So, the circumference of the circle = 2πr

44 = 2 × (22/7) × r

44 = 44/7 × r

(44 × 7)/44 = r

r = 7 cm

Then,

Area of the circle = πr^{2}

= 22/7 × 7^{2}

= 22 × 7

= 154 cm^{2}

Now,

If the wire is bent into a square,

Length of wire = perimeter of the square

So, the perimeter of square = 4 x side

44 = 4 x side

44 = 4S

S = 44/4

S = 11cm

Area of square = (side)2 = 11^{2}

= 121 cm^{2}

Hence, the circle has more area than square

**10. From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet. (Take π = 22/7)**.

**Solution:**

Given,

The radius of the circular card sheet = 14 cm

The radius of the two small circles = 3.5 cm

Length of the rectangle = 3 cm

Breadth of the rectangle = 1 cm

First, we have to find out the area of the circular card sheet,

Area of the circular card sheet = πr^{2}

= 22/7 × 14^{2}

= 22 × 2 × 14

= 616 cm^{2}

Now, we find the area of 2 small circles,

Area of the 2 small circles = 2 × πr^{2}

= 2 × (22/7 × 3.5^{2})

= 2 × ((22/7) × 12.25)

= 2 × 38.5

= 77 cm^{2}

Now, we find the area of the rectangle,

Area of the rectangle = Length × Breadth

= 3 × 1

= 3 cm^{2}

Now,

Area of the remaining sheet = Area of circular card sheet – (Area of two small circles + Area of the rectangle)

= 616 – (77 + 3)

= 616 – 80

= 536 cm^{2}

Hence, the area of the remaining sheet is 536 cm^{2}

**11. A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take π = 3.14)**

**Solution:**

Given,

Radius of circle = 2 cm

Side of square sheet = 6 cm

First, we have to find out the area of the square aluminium sheet,

Area of the square = side^{2}

So, the area of the square aluminium sheet = 62 = 36 cm^{2}

Now,

Area of the circle = πr^{2}

= 3.14 × 2^{2}

= 3.14 × 4

= 12.56 cm^{2}

Now,

Area of the aluminium sheet left = Area of the square aluminium sheet – Area of the circle

= 36 – 12.56

= 23.44 cm^{2}

Hence, the area of the aluminium sheet left is 23.44 cm^{2}

**12. The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take π = 3.14)**

**Solution:**

Given,

Circumference of a circle = 31.4 cm

So, the circumference of a circle = 2πr

31.4 = 2 × 3.14 × r

31.4 = 6.28 × r

31.4/6.28 = r

r = 5 cm

Then,

Area of the circle = πr^{2}

= 3.14 × 5^{ 2}

= 3. 14 × 25

= 78.5 cm^{2}

Hence, radius of the circle is 5 cm and area of the circle is 78.5 cm^{2}.

**13. A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)**.

**Solution:**

Given,

Diameter of the flower bed = 66 m

Radius of the flower bed = d/2

= 66/2

= 33 m

Now,

Area of flower bed = πr^{2}

= 3.14 × 33^{2}

= 3.14 × 1089

= 3419.46 m

We have to find area of the flower bed and path together

So, radius of flower bed and path together = 33 + 4 = 37 m

Area of the flower bed and path together = πr^{2}

= 3.14 × 37^{2}

= 3.14 × 1369

= 4298.66 m^{2}

Now,

Area of the path = Area of the flower bed and path together – Area of flower bed

= 4298.66 – 3419.46

= 879.2 m^{2}

Hence, the area of the path is 879.2 m^{2}

**14. A circular flower garden has an area of 314 m2. A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden? (Take π = 3.14)**

**Solution:**

Given,

Area of the circular flower garden = 314 m^{2}

Sprinkler at the centre of the garden can cover an area that has a radius = 12 m

Area of the circular flower garden = πr^{2}

314 = 3.14 × r^{2}

314/3.14 = r^{2}

r^{2} = 100

r = √100

r = 10 m

Hence, the radius of the circular flower garden is 10 m.

Since the sprinkler can cover an area of a radius of 12 m

Yes, the sprinkler will water the whole garden.

**15. Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)**.

**Solution:**

According to the figure,

The radius of inner circle = outer circle radius – 10

= 19 – 10

= 9 m

Circumference of the inner circle = 2πr

= 2 × 3.14 × 9

= 56.52 m

Now,

The radius of outer circle = 19 m

Circumference of the outer circle = 2πr

= 2 × 3.14 × 19

= 119.32 m

Hence, the circumference of the inner circle is 56.52 m and the circumference of the outer circle is 119.32 m

**16. How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = 22/7).**

**Solution:**

Given,

Radius of the wheel = 28 cm

Total distance = 352 m = 35200 cm

Now,

Circumference of the wheel = 2πr

= 2 × 22/7 × 28

= 2 × 22 × 4

= 176 cm

We have to find the number of rotations of the wheel,

Total number of times the wheel should rotate = Total distance covered by wheel / Circumference of the wheel

= 352 m/176 cm

= 35200 cm/ 176 cm

= 200

Hence, the wheel rotates 200 times

**17. The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour. (Take π = 3.14)**

**Solution:**

Given,

Length of the minute hand of the circular clock = 15 cm

So, distance travelled by the tip of a minute hand in 1 hour = circumference of the clock

= 2πr

= 2 × 3.14 × 15

= 94.2 cm

Hence, the minute hand moves 94.2 cm in 1 hour

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