**Download NCERT Solutions for Class 7 Maths Chapter 11 – Perimeter and Area**

**Class 7 Maths NCERT Solutions Chapter 11 – Perimeter and Area comprises of the 4 Exercises**

### NCERT Solutions for Class 7 Maths Chapter 11 Exercise 11.1

**1. The length and the breadth of a rectangular piece of land are 500 m and 300 m respectively. Find
**

**i. Its area**

**ii. The cost of the land, if 1 m**

^{2}of the land costs Rs. 10,000.

**Solution:**

**i.** Length = 500 m

Breadth = 300 m

Area of rectangular piece of land = length × breadth

= 500 × 300

= 150000 m^{2}

**ii.** Cost of 1 m^{2} of land = Rs. 10,000

Cost of 150000 m^{2} of land = Rs. 10,000 × 150000

= Rs. 1500000000

**2. Find the area of a square park whose perimeter is 320 m.**

**Solution:**

Let the side of the square be s

Perimeter of the square = 320 m

4s = 320 m

s = 80 m

Area of the square = (side)^{2} = s^{2
}= 80^{2
}= 6400 m^{2}

**3. Find the breadth of a rectangular plot of land, if its area is 440 m ^{2} and the length is 22 m. Also find its perimeter.**

**Solution:**

Let the breadth of the rectangle be b

Length of the rectangle = 22 m

Area of the rectangle = 440 m^{2
}Length × Breadth = 440 m^{2
}22 × b = 440

b = 20 m

Perimeter of rectangle = 2 × (Length + Breadth)

= 2 × (22 + 20)

= 2 × 42

= 84 m^{2}

**4. The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.**

**Solution:**

Let the breadth of the rectangle be b

Length of the rectangle = 35 cm

Perimeter of the rectangle = 100 cm

2 × (Length + Breadth) = 100 cm

2 × (35 + b) = 100

35 + b = 50

b = 15 cm

Area of the rectangle = Length × Breadth

= 35 × 15

= 525 cm^{2}

**5. The area of a square park is the same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.**

**Solution:**

Let the breadth of the rectangular park be b

Length of the rectangle = 90 m

Side of the square = 60 m

Area of square = (side)^{2} = 60^{2} = 3600 m^{2
}Area of rectangle = Length × Breadth = 90 × b

According to the question,

Area of square = Area of rectangle

3600 = 90 × b

b = = 40 m

**6. A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side. Also find which shape encloses more area?**

**Solution:**

Length of rectangle = 40 cm

Breadth of rectangle = 22 cm

Let the side of the square be s

Perimeter of rectangle = 2 × (Length + Breadth) = 2 × (40 + 22) = 2 × (62) = 124 cm

Perimeter of square = 4 × (side) = 4s

Since the same wire is rebent from a rectangle to a square

Therefore, perimeter of square = perimeter of rectangle = length of the wire

4s = 124

s = 31 cm

Area of rectangle = Length × Breadth = 40 × 22 = 880 cm^{2
}Area of square = (Side)^{2} = 31^{2} = 961 cm^{2}

Hence, the side of the square is 31 cm and square encloses more area than the rectangle.

**7. The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also find the area of the rectangle.**

**Solution:**

Let the length of rectangle be l

Breadth of rectangle = 30 cm

Perimeter of rectangle = 2 × (Length + Breadth) = 130 cm

2 × (l + 30) = 130

l + 30 = 65

l = 35 cm

Area of the rectangle = Length × Breadth = 35 × 30 = 1050 cm^{2 }

**8. A door of length 2 m and breadth 1m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m. Find the cost of white washing the wall, if the rate of white washing the wall is Rs. 20 per m ^{2}.**

**Solution:**

Length of door = 2 m

Breadth of door = 1 m

Length of wall = 4.5 m

Breadth of wall = 3.6 m

Area of wall = (Length of wall × Breadth of wall) – (Length of door × Breadth of door)

= (4.5 × 3.6) – (2 × 1)

= 16.2 – 2

= 14.2 m^{2}

Cost of white washing 1 m^{2} of wall = Rs. 20

Cost of white washing 14.2 m^{2} of wall = Rs. 20 × 14.2 = Rs. 284

### NCERT Solutions for Class 7 Maths Chapter 11 Exercise 11.2

**1. Find the area of each of the following parallelograms:**

**a) **

**b)**

**c) **

**d) **

**e)**

**Solution:**

**a)** Area of parallelogram = Base × Height

= 7 × 4

= 28 cm^{2}

**b)** Area of parallelogram = Base × Height

= 5 × 3

= 15 cm^{2}

**c)** Area of parallelogram = Base × Height

= 2.5 × 3.5

= 8.75 cm^{2}

**d)** Area of parallelogram = Base × Height

= 5 × 4.8

= 24 cm^{2}

**e)** Area of parallelogram = Base × Height

= 2 × 4.4

= 8.8 cm^{2}

**2. Find the area of each of the following triangles:**

**a)**

**b)**

**c)**

**d)**

**Solution:**

**a)** Area of triangle = (Base × Height)

= × 4 × 3

= 6 cm^{2}

**b)** Area of triangle = (Base × Height)

= × 5 × 3.2

= 8 cm^{2}

**c)** Area of triangle = (Base × Height)

= × 3 × 4

= 6 cm^{2}

**d)** Area of triangle = (Base × Height)

= × 3 × 2

= 3 cm^{2}

**3. Find the missing values:**

**Solution:**

**a)** Area of parallelogram = Base × Height

20 × Height = 246

Height = cm = 12.3 cm

**b)** Area of parallelogram = Base × Height

Base × 15 = 154.5

Base = cm = 10.3 cm

**c)** Area of parallelogram = Base × Height

Base × 8.4 = 48.72

Base = cm = 5.8 cm

**d)** Area of parallelogram = Base × Height

15.6 × Height = 16.38

Height = cm = 1.05 cm

**4. Find the missing values:**

**Solution:**

**a)** Area of triangle = (Base × Height)

87 = × 15 × Height

Height = = 11.6 cm

**b)** Area of triangle = (Base × Height)

1256 = × Base × 31.4

Base = = 80 mm

**c)** Area of triangle = (Base × Height)

170.5 = × 22 × Height

Height = = 15.5 cm

**5. PQRS is a parallelogram. QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find:**

**a) The area of the parallelogram PQRS
b) QN, if PS = 8 cm**

**Solution:**

**a)** Area of parallelogram = Base × Height = SR × QM

= 12 × 7.6

= 91.2 cm^{2}

**b)** Area of parallelogram = Base × Height = PS × QN

91.2 = 8 × QN

QN = = 11.4 cm

**6. DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is 1470 cm ^{2}, AB = 35 cm and AD = 49 cm, find the length of BM and DL.**

**Solution:**

Area of ABCD = 1470 cm^{2
}Area of parallelogram = Base × Height = AB × DL = AD × BM

AB × DL = 1470 cm^{2
}35 × DL = 1470

DL = = 42 cm

AD × BM = 1470 cm^{2
}49 × BM = 1470

BM = = 30 cm

**7. ∆ABC is right angled at A. AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, Find the area of ∆ABC. Also find the length of AD.**

**Solution:**

Area of triangle = × Base × Height = × AB × AC

= × 5 × 12

= 30 cm^{2}

_{ }Area of triangle = × Base × Height = × BC × AD

× 13 × AD = 30

AD =

AD = cm

**8. ∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to BC, is 6 cm. Find the area of ∆ABC. What will be the height from C to AB i.e., CE?**

**Solution:**

Area of triangle = × Base × Height = × BC × AD

= × 9 × 6

= 27 cm^{2}

Area of triangle = × Base × Height = × AB × CE

× 7.5 × CE = 27

CE =

CE = 7.2 cm

### NCERT Solutions for Class 7 Maths Chapter 11 Exercise 11.3

**1. Find the circumference of the circles with the following radius: (Take π = )**

**a) 14 cm**

**b) 28 mm**

**c) 21 cm**

**Solution:**

**a)** Circumference of circle = 2πr

= 2 × × 14

= 88 cm

**b)** Circumference of circle = 2πr

= 2 × × 28

= 176 mm

**c)** Circumference of circle = 2πr

= 2 × × 21

= 132 cm

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

**a) Radius = 14 mm**

**b) Diameter = 49 m**

**c) Radius = 5 cm**

**a)** Area of circle = πr^{2} = × 14^{2} = 616 mm^{2}

**b)** Radius(r) = = m

Area of circle = πr^{2} = x = 1886.5 m^{2}

**c)** Area of circle = πr^{2} = × 5^{2} = 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 = )**

**Solution:**

Let the radius of the circle be r

Circumference = 2πr

2 × × r = 154

r = = = 24.5 m

Area of circle = πr^{2} = x = = 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 Rs. 4 per meter. (Take π = )**

**Solution:**

Radius of circular garden(r) = = m

Circumference of the circular garden = 2πr = 2 × × = 66 m

Since the gardener makes 2 rounds of fence

Length of rope required = 2 × Circumference = 2 × 66 m = 132 m

Cost of 1 m of rope = Rs. 4

Cost of 132 m of rope = Rs. 4 × 132 = Rs. 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:**

Radius of circular sheet (R) = 4 cm

Radius of removed circle (r) = 3 cm

Area of remaining portion = Area of sheet – Area of removed circle

= πR^{2} – πr^{2
}= 3.14 × 4^{2} – 3.14 × 3^{2
}= 3.14 × (16 – 9)

= 3.14 × 7

= 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 Rs. 15. (Take π = 3.14)**

**Solution:**

Radius of table (r) = = = 0.75 m

Circumference of table = 2πr = 2 × 3.14 × 0.75 = 4.71 m

Length of lace required = Circumference of the table = 4.71 m

Cost of 1 m of lace = Rs. 15

Cost of 4.71 m of lace = Rs. 15 × 4.71 = Rs. 70.65

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

**Solution:**

Diameter of the semicircle (d) = 10 cm

Radius of the semicircle (r) = = = 5 cm

Length of the curved part of semicircle = × Circumference of circle = × 2πr = πr

Circumference of the given semicircle = Length of curved part + diameter

= πr + d

= 3.14 × 5 + 10

= 15.7 + 10

= 25.7 cm

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

**Solution:**

Radius of table (r) = = = 0.8 m

Area of table = πr^{2} = 3.14 × (0.8)^{2} = 3.14 × 0.64 = 2.0096 m^{2}

Cost of polishing 1 m^{2} of table-top = Rs. 15

Cost of 2.0096 m of lace = Rs. 15 × 2.0096 = Rs. 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 = )**

**Solution:**

Let the radius of the wire be r

Length of the wire = Circumference of the circle = 2πr

2 × × r = 44

r = = 7 cm

Area enclosed by circle = πr^{2} = × 7^{2} = 154 cm^{2}

Let the length of the side of the square be s

Perimeter of square = 4s

Since the same wire is rebent from a circle to square

Perimeter of square = Length of the wire

4s = 44

s = 11 cm

Area enclosed by square = s^{2} = 11^{2} = 121 cm^{2}

Therefore, the 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 π = )**

**Solution:**

Radius of circular card sheet (R) = 14 cm

Radius of removed circles (r) = 3.5 cm

Length of removed rectangle (l) = 3 cm

Breadth of removed rectangle (b) = 1 cm

Area of card sheet = πR^{2} = × 14^{2} = 616 cm^{2}

Area of removed portion = Area of 2 removed circles + Area of removed rectangle

= 2 × (πr^{2}) + (l × b)

= 2 × ( × 3.5^{2}) + (3 × 1)

= 77 + 3

= 80 cm^{2}

Remaining area = Area of card sheet – Area of removed portion

= 616 – 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 π = 3.14)**

**Solution:**

Radius of removed circle (r) = 2 cm

Side of the square sheet (s) = 6 cm

Area of left-over sheet = Area of square – Area of removed circle

= s^{2} – πr^{2
}= 6^{2} – 3.14 × 2^{2
}= 36 – 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 π = 3.14)**

**Solution:**

Let the radius of the circle be r

Circumference = 2πr

2 × 3.14 × r = 31.4

r = = = 5 cm

Area of circle = πr^{2} = 3.14 × 5^{2} = 3.14 × 25 = 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)**

Radius of the flower bed (r) = = =33 m

Radius of the surrounding path (R) = r + 4 = 33 + 4 = 37 m

Area of the path = Area of outer circle – Area of flower bed

= πR^{2} – πr^{2
}= 3.14 × 37^{2} – 3.14 × 33^{2
}= 3.14 × (37^{2} – 33^{2})

= 3.14 × (1369 – 1089)

= 3.14 × 280

= 879.2 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 π = 3.14)**

**Solution:**

Area of garden = 314 m^{2
}Radius of area covered by sprinkler (r) = 12 m

Area of garden covered by the sprinkler = πr^{2} = 3.14 × 12^{2} = 452.16 m^{2
}Since the area covered by the sprinkler is greater than the area of the garden

Hence, the sprinkler will water the entire garden.

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

**Solution:**

Radius of outer circle (R) = 19 cm

Radius of inner circle (r) = 19 – 10 = 9 cm

Circumference of inner circle = 2πr = 2 × 3.14 × 9 = 56.52 cm

Circumference of outer circle = 2πR = 2 × 3.14 × 19 = 119.32 cm

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

**Solution:**

Radius of wheel (r) = 28 cm

Circumference of the wheel = 2πr = 2 × × 28 = 176 cm = 1.76 m

Distance travelled in 1 revolution = Circumference = 1.76 m

Revolution required to travel 1.76 m = 1

Revolution required to travel 1 m =

Revolution required to travel 352 m = = 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 π = 3.14)**

**Solution:**

Length of the minute hand (r) = 15 cm

Circumference of circle traversed by the tip of minute hand = 2πr = 2 × 3.14 × 15 = 94.2 cm

In 1 hour, the minute hand completes a full circle

∴ Distance travelled by the tip of minute hand = Circumference = 94.2 cm

### NCERT Solutions for Class 7 Maths Chapter 11 Exercise 11.4

**1. A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.**

**Solution:**

The dimensions of the garden and path are shown in the figure

Area of the path = Area of PQRS – Area of ABCD

= 100 × 85 – 90 × 75

= 8500 – 6750

= 1750 m^{2}

Area of garden = 90 × 75 = 6750 m^{2
}= 0.675 hectare (1 hectare = 10000 m^{2})

**2. A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.**

**Solution:**

The dimensions of the rectangular park and the path are shown in the figure

Area of the park = Area of PQRS – Area of ABCD

= 131 × 71 – 125 × 65

= 9301 – 8125

= 1176 m^{2}

**3. A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.**

**Solution:**

The dimensions of the cardboard and the picture are as shown

Area of margin = Area of PQRS – Area of ABCD

= 8 × 5 – 5 × 2

= 40 – 10

= 30 cm^{2}

**4. A verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find:**

**i. The area of verandah
ii. The cost of cementing the floor of the verandah at the rate of Rs. 200 per m ^{2}**

**Solution:**

**i.** The dimensions of the verandah and the room are shown in the figure

Area of the verandah = Area of PQRS – Area of ABCD

= 10 × 8.5 – 5.5 × 4

= 85 – 22

= 63 m^{2}

**ii.** Cost of cementing 1 m^{2} of floor = Rs. 200

Cost of cementing 63 m^{2} of floor = Rs. 200 × 63 = Rs. 12600

**5. A path 1 m wide is built along the border and inside a square garden of side 30 m. Find:**

**i. The area of the path
ii. The cost of planting grass in the remaining portion of the garden at the rate of Rs. 40 per m ^{2}**

**Solution:**

**i.** The dimensions of the garden and the path are shown in the figure

Area of the path = Area of PQRS – Area of ABCD

= 30^{2} – 28^{2
}= (30 + 28) × (30 – 28)

= 58 × 2

= 116 m^{2}

**ii.** Area of remaining portion of the garden = 28^{2} = 784 m^{2
}Cost of planting grass in 1 m^{2} of land = Rs. 40

Cost of planting grass in 784 m^{2} of land = Rs. 40 × 784 = Rs. 31,360

**6. Two cross roads, each of width 10 m, cut at right angles through the centre of a rectangular park of length 700 m and breadth 300 m and parallel to its sides. Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares.**

**Solution:**

**i.** The dimension of the park and the roads are shown in the figure

Area of roads = Area of EFGH + Area of PQRS – Area of KLMN

= 700 × 10 + 300 × 10 – 10 × 10

= 7000 + 3000 – 100

= 10000 – 100

= 9900 m^{2
}= 0.99 hectares

Area of park excluding the roads = Area of ABCD – Area of roads

= 700 × 300 – 9900

= 210000 – 9900

= 200100 m^{2
}= 20.01 hectares

**7. Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find
i. The area covered by the roads
ii. The cost of constructing the roads at the rate of Rs. 110 per m**

^{2}

**Solution:**

**i.** The dimensions of the field and the roads are shown in the figure

Area of roads = Area of EFGH + Area of PQRS – Area of KLMN

= 90 × 3 + 60 × 3 – 3 × 3

= 270 + 180 – 9

= 450 – 9

= 441 m^{2}

**ii.** Cost of constructing 1 m^{2} of road = Rs. 110

Cost of constructing 441 m^{2} of road = Rs. 110 × 441 = Rs. 48,510

**8. Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left? (π = 3.14)**

**Solution:**

Radius of the circular pipe (r) = 4 cm

Side of the square box (s) = 4 cm

Length of cord = Circumference of pipe = 2πr = 2 × 3.14 × 4 = 25.12 cm

Length of cord required to wrap around the square box = Perimeter of the box = 4s = 16 cm

Because the length of cord required to wrap around the box is less than the length of cord Pragya already have, therefore, she will have 9.12 cm (= 25.12 – 16) of cord left.

**9. The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find:**

**i. The area of whole land
**

**ii. The area of the flower bed**

iii. The area of the lawn excluding the area of the flower bed

iv. The circumference of the flower bed

iii. The area of the lawn excluding the area of the flower bed

iv. The circumference of the flower bed

**i.** Length of lawn = 10 m

Breadth of lawn = 5 m

Area of lawn = Length × Breadth = 10 × 5 = 50 m^{2}

**ii.** Radius of flower bed (r) = 2 m

Area of flower bed = πr^{2} = 3.14 × 2^{2} = 12.56 m^{2}

**iii.** Area of lawn excluding the area of the flower bed = Area of lawn – Area of flower bed

= 50 – 12.56

= 37.44 m^{2}

**iv.** Circumference of flower bed = 2πr = 2 × 3.14 × 2 = 12.56 m

**10. In the following figures, find the area of the shaded portions:**

**i.**

**ii.**

**i.** Shaded area = Area of rectangle ABCD – (Area of ΔAEF + Area of ΔBEC)

= 18 × 10 – ( × 10 × 6 + × 8 × 10)

= 180 – (30 + 40)

= 180 – 70

= 110 cm^{2}

**ii.** Shaded area = Area of square PQRS – (Area of ΔTSU + Area of ΔURQ + Area of ΔPTQ)

= 20^{2} – ( × 10 × 10 + × 10 × 20 + × 20 × 10)

= 400 – (50 + 100 + 100)

= 400 – 250

= 150 cm^{2}

**11. Find the area of the quadrilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM ⊥ AC, DN ⊥ AC**

**Solution:**

Area of quadrilateral ABCD = Area of ΔABC + Area of ΔADC

= × AC × BM + × AC × ND

= × 22 × 3 + × 22 × 3

= 33 + 33

= 66 cm^{2}

**Maths – NCERT Solutions Class 7**

**NCERT Solutions Class 7**

**This Chapter contains the Exercises relating to the following topics , which are discussed in Chapter 11 – Perimeter and Area Class 7 NCERT book : –**

- 11.1 INTRODUCTION
- 11.2 SQUARES AND RECTANGLES
- 11.2.1 Triangles as Parts of Rectangles
- 11.2.2 Generalising for other Congruent Parts of Rectangles
- 11.3 AREA OF A PARALLELOGRAM
- 11.4 AREA OF A TRIANGLE
- 11.5 CIRCLES
- 11.5.1 Circumference of a Circle
- 11.5.2 Area of Circle
- 11.6 CONVERSION OF UNITS
- 11.7 APPLICATIONS

**Download NCERT Solutions for Class 7 Maths Chapter 11 – Perimeter and Area**

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