NCERT Solutions for Class 7 Maths Exercise 11.3 Chapter 11 Perimeter and Area in simple PDF are available here. This exercise of NCERT Solutions for Class 7 Maths Chapter 11 contains topics related to the circumference of a circle and area of a circle. We at BYJUâ€™S have prepared the solutions step by step containing neat descriptions. Students can score more marks in Maths by practising NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area.

## Download the PDF of NCERT Solutions For Class 7 Maths Chapter 11 Perimeter and Area â€“ Exercise 11.3

Â

Â

### Access Other Exercises of NCERT Solutions For Class 7 Maths Chapter 11 â€“ Perimeter and Area

### Access Answers to NCERT Class 7 Maths Chapter 11 â€“ Perimeter and Area Exercise 11.3

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

**(a) 14 cm**

**Solution:-**

Given, radius of circle = 14 cm

Circumference of the circle = 2Ï€r

= 2 Ã— (22/7) Ã— 14

= 2 Ã— 22 Ã— 2

= 88 cm

**(b) 28 cm**

**Solution:-**

Given, radius of circle = 28 cm

Circumference of the circle = 2Ï€r

= 2 Ã— (22/7) Ã— 28

= 2 Ã— 22 Ã— 4

= 176 cm

**(c) 21 cm**

**Solution:-**

Given, radius of circle = 21 cm

Circumference of the circle = 2Ï€r

= 2 Ã— (22/7) Ã— 21

= 2 Ã— 22 Ã— 3

= 132 cm

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

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

**Solution:**

Given, radius of circle = 14 mm

Then,

Area of the circle = Ï€r^{2}

= 22/7 Ã— 14^{2}

= 22/7 Ã— 196

= 22 Ã— 28

= 616 mm^{2}

**(b) Diameter = 49 m**

**Solution:**

Given, diameter of circle (d) = 49 m

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

= 49/2

= 24.5 m

Then,

Area of the circle = Ï€r^{2}

= 22/7 Ã— (24.5)^{2}

= 22/7 Ã— 600.25

= 22 Ã— 85.75

= 1886.5 m^{2}

**(c) Radius = 5 cm**

**Solution:**

Given, radius of circle = 5 cm

Then,

Area of the circle = Ï€r^{2}

= 22/7 Ã— 5^{2}

= 22/7 Ã— 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 Ï€ = 22/7)**

**Solution:-**

From the question it is given that,

Circumference of the circle = 154 m

Then,

We know that, Circumference of the circle = 2Ï€r

154 = 2 Ã— (22/7) Ã— r

154 = 44/7 Ã— r

r = (154 Ã— 7)/44

r = (14 Ã— 7)/4

r = (7 Ã— 7)/2

r = 49/2

r = 24.5 m

Now,

Area of the circle = Ï€r^{2}

= 22/7 Ã— (24.5)^{2}

= 22/7 Ã— 600.25

= 22 Ã— 85.75

= 1886.5 m^{2}

So, the radius of circle is 24.5 and area of circle is 1886.5.

**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 â‚¹ 4 per meter. (Take Ï€ = 22/7)**

**Solution:-**

From the question it is given that,

Diameter of the circular garden = 21 m

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

= 21/2

= 10.5 m

Then,

Circumference of the circle = 2Ï€r

= 2 Ã— (22/7) Ã— 10.5

= 462/7

= 66 m

So, the length of rope required = 2 Ã— 66 = 132 m

Cost of 1 m rope = â‚¹ 4 [given]

Cost of 132 m rope = â‚¹ 4 Ã— 132

= â‚¹ 528

**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 Ï€ = 3.14)**

**Solution:-**

From the question it is give that,

Radius of circular sheet R = 4 cm

A circle of radius to be removed r = 3 cm

Then,

The area of the remaining sheet = Ï€R^{2 }â€“ Ï€r^{2}

= Ï€ (R^{2} â€“ r^{2})

= 3.14 (4^{2} â€“ 3^{2})

= 3.14 (16 â€“ 9)

= 3.14 Ã— 7

= 21.98 cm^{2}

So, 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 â‚¹ 15. (Take Ï€ = 3.14)**

**Solution:-**

From the question it is given that,

Diameter of the circular table = 1.5 m

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

= 1.5/2

= 0.75 m

Then,

Circumference of the circle = 2Ï€r

= 2 Ã— 3.14 Ã— 0.75

= 4.71 m

So, the length of lace = 4.71 m

Cost of 1 m lace = â‚¹ 15 [given]

Cost of 4.71 m lace = â‚¹ 15 Ã— 4.71

= â‚¹ 70.65

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

**Solution:-**

From the question it is given that,

Diameter of semi-circle = 10 cm

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

= 10/2

= 5 cm

Then,

Circumference of the semi-circle = Ï€r

= (22/7) Ã— 5

= 110/7

= 15.71 cm

Now,

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 â‚¹15/m ^{2}. (Take Ï€ = 3.14)**

**Solution:-**

From the question it is given that,

Diameter of the circular table-top = 1.6 m

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

= 1.6/2

= 0.8 m

Then,

Area of the circular table-top = Ï€r^{2}

= 3.14 Ã— 0.8^{2}

= 3.14 Ã— 0.8 Ã—0.8

= 2.0096 m^{2}

Cost for polishing 1 m^{2} area = â‚¹ 15 [given]

Cost for polishing 2.0096 m^{2} area = â‚¹ 15 Ã— 2.0096

= â‚¹ 30.144

Hence, the Cost for polishing 2.0096 m^{2} area is â‚¹ 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 Ï€ = 22/7)**

**Solution:-**

From the question it is given that,

Length of wire that Shazli took =44 cm

Then,

If the wire is bent into a circle,

We know that, circumference of the circle = 2Ï€r

44 = 2 Ã— (22/7) Ã— r

44 = 44/7 Ã— r

(44 Ã— 7)/44 = r

r = 7 cm

Area of the circle = Ï€r^{2}

= 22/7 Ã— 7^{2}

= 22/7 Ã— 7 Ã—7

= 22 Ã— 7

= 154 cm^{2}

Now,

If the wire is bent into a square,

The length of the each side of square = 44/4

= 11 cm

Area of the square = length of the side of square^{2}

= 11^{2}

= 121 cm^{2}

By comparing the two areas of the square and circle,

Clearly, circle encloses more area.

**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 Ï€ = 22/7)**

**Solution:-**

From the question it is given that,

Radius of the circular card sheet = 14 cm

Radius of the two small circle = 3.5 cm

Length of the rectangle = 3 cm

Breadth of the rectangle = 1 cm

First we have to find out the area of circular card sheet, two circles and rectangle to find out the remaining area.

Now,

Area of the circular card sheet = Ï€r^{2}

= 22/7 Ã— 14^{2}

= 22/7 Ã— 14 Ã— 14

= 22 Ã— 2 Ã— 14

= 616 cm^{2}

Area of the 2 small circles = 2 Ã— Ï€r^{2}

= 2 Ã— (22/7 Ã— 3.5^{2})

= 2 Ã— (22/7 Ã— 3.5 Ã— 3.5)

= 2 Ã— ((22/7) Ã— 12.25)

= 2 Ã— 38.5

= 77 cm^{2}

Area of the rectangle = Length Ã— Breadth

= 3 Ã— 1

= 3 cm^{2}

Now,

The area of the remaining part = Card sheet area â€“ (area of two small circles + rectangle

area)

= 616 â€“ (77 + 3)

= 616 â€“ 80

= 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 Ï€ = 3.14)**

**Solution:-**

From the question it is given that,

Radius of circle = 2 cm

Square sheet side = 6 cm

First we have to find out the area of square aluminium sheet and circle to find out the remaining area.

Now,

Area of the square = side^{2}

= 6^{2}

= 36 cm^{2}

Area of the circle = Ï€r^{2}

= 3.14 Ã— 2^{2}

= 3.14 Ã— 2 Ã— 2

= 3.14 Ã— 4

= 12.56 cm^{2}

Now,

The area of the remaining part = Area of aluminum square sheet â€“ area of circle

= 36 â€“ 12.56

= 23.44 cm^{2}

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

**Solution:-**

From the question it is given that,

Circumference of a circle = 31.4 cm

We know that,

Circumference of a circle = 2Ï€r

31.4 = 2 Ã— 3.14 Ã— r

31.4 = 6.28 Ã— r

31.4/6.28 = r

r = 5 cm

Then,

Area of the circle = Ï€r^{2}

= 3.14 Ã— 5^{2}

= 3. 14 Ã— 25

= 78.5 cm

**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? (Ï€ = 3.14)**

**Solution:-**

From the question it is given that,

Diameter of the flower bed = 66 m

Then,

Radius of the flower bed = d/2

= 66/2

= 33 m

Area of flower bed = Ï€r^{2}

= 3.14 Ã— 33^{2}

= 3.14 Ã— 1089

= 3419.46 m

Now 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 = Ï€r^{2}

= 3.14 Ã— 37^{2}

= 3.14 Ã— 1369

= 4298.66 m

Finally,

Area of the path = Area of the flower bed and path together â€“ Area of flower bed

= 4298.66 â€“ 3419.46

= 879.20 m^{2}

**14. A circular flower garden has an area of 314 m ^{2}. 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 Ï€ = 3.14)**

**Solution:-**

From the question it is given that,

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 = Ï€r^{2}

314 = 3.14 Ã— r^{2}

314/3.14 = r^{2}

r^{2} = 100

r = âˆš100

r = 10 m

âˆ´Radius of the circular flower garden is 10 m.

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

Hence, the sprinkler will water the whole garden.

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

**Solution:-**

From the figure,

Radius of inner circle = outer circle radius â€“ 10

= 19 â€“ 10

= 9 m

Circumference of the inner circle = 2Ï€r

= 2 Ã— 3.14 Ã— 9

= 56.52 m

Then,

Radius of outer circle = 19 m

Circumference of the outer circle = 2Ï€r

= 2 Ã— 3.14 Ã— 19

= 119.32 m

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

**Solution:-**

From the question it is given that,

Radius of the wheel = 28 cm

Circumference of the wheel = 2Ï€r

= 2 Ã— 22/7 Ã— 28

= 2 Ã— 22 Ã— 4

= 176 cm

Now we have to find the number of rotation of the wheel,

= Total distance to be covered/ circumference of wheel

= 352 m/176 cm

= 35200 cm/ 176 cm

= 200

**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 Ï€ = 3.14)**

**Solution:-**

From the question it is given that,

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

Then,

Distance travelled by the tip of minute hand in 1 hour = circumference of the clock

= 2Ï€r

= 2 Ã— 3.14 Ã— 15

= 94.2 cm

Very helpful thanks a lot