NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

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

Answer: 150000 m2

Explanation: Area of the rectangular land = Length × Breadth

Length of the rectangular land = 500m

Breadth of the rectangular land = 300m

Area of the rectangular land = 500m × 300m

= 150000 m2

(ii) The cost of the land, if 1 m2 of the land costs ₹ 10,000.

Answer: ₹ 1,500,000,000

Explanation: Area of the rectangular land = 500m × 300m

= 150000 m2

Cost of the land = Area × Cost of the land per 1m2

Cost of the land = 150000 m2 × ₹ 10,000

Cost of the land = ₹ 1,500,000,000

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

Answer: 6400 m2

Explanation: Perimeter is the sum of all the sides of the given closed polygon.

Perimeter of a square park = 4 × Side of the square park

320 m = 4 × Side of the square park

Side of the square park = 320/4

= 80 m

Area of the square park = Side × Side or (Side)2

Area of the square park = 80 m × 80 m

Area of the square park = 6400 m2

3. Find the breadth of a rectangular plot of land, if its area is 440 m2 and the length is 22 m. Also find its perimeter.

Answer: 20 m, 84 m

Explanation: Length of the rectangular land = 22 m

Area of the rectangular land = Length × Breadth

440 m2= 22 m × Breadth

Breadth = 440 m2/22m

Breadth = 20 m

Perimeter of the rectangular land = 2 × (Length + Breadth)

Perimeter of the rectangular land = 2 × (22 + 20) m

Perimeter of the rectangular land = 2 × 42 m

Perimeter of the rectangular land = 84 m

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

Answer: 15 cm, 525 cm2

Explanation: Perimeter of the rectangular sheet = 2 × (Length + Breadth)

100 cm = 2 × (35 + Breadth) cm

(35 + Breadth) = 100/2

(35 + Breadth) = 50

Breadth = 50 - 35

Breadth = 15 cm

Area of the rectangular sheet = Length × Breadth

Area of the rectangular sheet = 35 cm × 15 cm

Area of the rectangular sheet = 525 cm2

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.

Answer: 40 m

Explanation: Length of the rectangular park = 90 m

Area of the rectangular park = Length × Breadth

Area of the rectangular park = 90 m × Breadth

Side of the square park = 60 m

Area of the square park = Side × Side

Area of the square park = 60 m × 60 m = 3600 m2

Area of the square park = Area of the rectangular park

3600 m2 = 90 m × Breadth

Breadth of the rectangular park = 3600 m2/ 90 m

Breadth of the rectangular park = 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?

Answer: 31 cm; Square

Explanation: Length = 40 cm

Breadth = 22 cm

Perimeter of the rectangle = 2 × (Length + Breadth)

Perimeter of the rectangle = 2 × (40 + 22) cm

Perimeter of the rectangle = 2 × 62 cm

Perimeter of the rectangle = 124 cm

Note: When a wire is bent in the form of any shape, its perimeter is always considered. Perimeter is the sum of all sides of a given polygon.

Perimeter of a square = 4 × Side

The same wire when bent in the form of a square will have the same perimeter.

124 cm = 4 × Side

Side = 124 cm/4

Side = 31 cm

The side of the square is 31 cm.

Area of the rectangle = Length × Breadth

Area of the rectangle = 40 cm × 22 cm

Area of the rectangle = 880 cm2

Area of the square = Side × Side

Area of the square = 31 cm × 31 cm = 961 cm2

Area of square > Area of the rectangle

Hence, square encloses more area.

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.

Answer: 43 cm; 1050 cm2

Explanation: Perimeter of the rectangle = 130 cm

Length =?

Breadth = 30 cm

Perimeter of the rectangle = 2 × (Length + Breadth)

130 cm = 2 × (Length + 30) cm

(Length + 30) cm = 130/2

(Length + 30) cm = 65 cm

Length = 65 cm - 30 cm

Length = 35 cm

Area of the rectangle = Length × Breadth

Area of the rectangle = 30 cm × 35 cm

Area of the rectangle = 1050 cm2

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 (Fig11.6). Find the cost of white washing the wall, if the rate of white washing the wall is ₹ 20 per m2?

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: ₹ 284

Explanation: Length of the door = 2 m

Breadth of the door = 1 m

Area of the door = Length × Breadth

Area of the door = 2 m × 1 m

Area of the door = 2 m2

Length of the wall = 4.5 m

Breadth of the wall = 3.6 m

Area of the wall = Length × Breadth

Area of the wall = 4.5 m × 3.6 m

Area of the wall = 16.2 m2

Area of whitewashing the wall = Area of the wall - Area of the door

Area of whitewashing the wall = 16.2 m2 - 2 m2 = 14. 2 m2

Cost of whitewashing the wall = Area of whitewashing the wall × Rate per m2

Cost of whitewashing the wall = 14. 2 m2 × ₹ 20

Cost of whitewashing the wall = ₹ 284

Exercise 11.2

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

(a)

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: 28 cm2

Explanation: Area of the parallelogram = Base × Height

Base of the parallelogram = 7 cm

Height of the parallelogram = 4 cm

Area of the parallelogram = 7 cm × 4 cm

Area of the parallelogram = 28 cm2

(b)

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: 15 cm2

Explanation: Area of the parallelogram = Base × Height

Base of the parallelogram = 5 cm

Height of the parallelogram = 3 cm

Area of the parallelogram = 5 cm × 3 cm

Area of the parallelogram = 15 cm2

(c)

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: 8.75 cm2

Explanation: Area of the parallelogram = Base × Height

Base of the parallelogram = 2.5 cm

Height of the parallelogram = 3.5 cm

Area of the parallelogram = 2.5 cm × 3.5 cm

Area of the parallelogram = 8.75 cm2

(d)

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: 24 cm2

Explanation: Area of the parallelogram = Base × Height

Base of the parallelogram = 5 cm

Height of the parallelogram = 4.8 cm

Area of the parallelogram = 5 cm × 4.8 cm

Area of the parallelogram = 24 cm2

(e)

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: 8.8 cm2

Explanation: Area of the parallelogram = Base × Height

Base of the parallelogram = 2 cm

Height of the parallelogram = 4.4 cm

Area of the parallelogram = 2 cm × 4.4 cm

Area of the parallelogram = 8.8 cm2

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

(a)

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: 6 cm2

Explanation: Area of triangle = ½ × Base × Height

Base of the triangle = 4 cm

Height of the triangle = 3 cm

Area of triangle = ½ × 4 cm × 3 cm

Area of triangle = ½ × 12 cm2

Area of triangle = 6 cm2

(b)

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: 8 cm2

Explanation: Area of triangle = ½ × Base × Height

Base of the triangle = 5 cm

Height of the triangle = 3.2 cm

Area of triangle = ½ × 5 cm × 3.2 cm

Area of triangle = ½ × 16 cm2

Area of triangle = 8 cm2

(c)

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: 6 cm2

Explanation: Area of triangle = ½ × Base × Height

Base of the triangle = 3 cm

Height of the triangle = 4 cm

Area of triangle = ½ × 3 cm × 4 cm

Area of triangle = ½ × 12 cm2

Area of triangle = 6 cm2

(d)

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: 3 cm2

Explanation: Area of triangle = ½ × Base × Height

Base of the triangle = 3 cm

Height of the triangle = 2 cm

Area of triangle = ½ × 3 cm × 2 cm

Area of triangle = ½ × 6 cm2

Area of triangle = 3 cm2

3. Find the missing values:

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

(a)

Answer:

S No.BaseHeightArea of the Parallelogram
a.20 cm12.3 cm246 cm2

Explanation:

Area of the parallelogram = Base × Height

Base of the parallelogram = 20 cm

Height of the parallelogram =?

Area of the parallelogram = 246 cm2

Area of the parallelogram = Base × Height

246 cm2 = 20 cm × Height

Height = 246 cm2/20 cm

Height = 12.3 cm

(b)

Answer:

S No.BaseHeightArea of the Parallelogram
b.10.3 cm15 cm154.5 cm2

Explanation:

Area of the parallelogram = Base × Height

Base of the parallelogram =?

Height of the parallelogram = 15 cm

Area of the parallelogram = 154.5 cm2

Area of the parallelogram = Base × Height

154.5 cm2 = Base × 15 cm

Base = 154.5 cm2/15 cm

Height = 10.3 cm

(c)

Answer:

S No.BaseHeightArea of the Parallelogram
c.5.8 cm8.4 cm48.72 cm2

Explanation:

Area of the parallelogram = Base × Height

Base of the parallelogram =?

Height of the parallelogram = 8.4 cm

Area of the parallelogram = 48.72 cm2

Area of the parallelogram = Base × Height

48.72 cm2 = Base × 8.4 cm

Base = 48.72 cm2/8.4 cm

Height = 5.8 cm

(d)

Answer:

S No.BaseHeightArea of the Parallelogram
d.15.6 cm1.05 cm16.38 cm2

Explanation:

Area of the parallelogram = Base × Height

Base of the parallelogram = 15.6 cm

Height of the parallelogram =?

Area of the parallelogram = 16.38 cm2

Area of the parallelogram = Base × Height

16.38 cm2 = 15.6 cm × Height

Height = 16.38 cm2/15.6 cm

Height = 1.05 cm

4. Find the missing values:

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

(a)

Answer:

BaseHeightArea of Triangle
15 cm11.6 cm87 cm2

Explanation:

Area of triangle = ½ × Base × Height

Base of the triangle = 15 cm

Height of the triangle =?

Area of triangle = 87 cm2

87 cm2 = ½ × 15 cm × Height

Height = (87 cm2 × 2)/ 15 cm

Height = 174 cm2/ 15 cm

Height = 11.6 cm

(b)

Answer:

BaseHeightArea of Triangle
80 mm31.4 mm1256 mm2

Explanation:

Area of triangle = ½ × Base × Height

Base of the triangle =?

Height of the triangle = 31.4 mm

Area of triangle = 1256 mm2

1256 mm2 = ½ × Base × 31.4 mm

Base = (1256 mm2× 2)/ 31.4 mm

Base = 2512 mm2/ 31.4 mm

Base = 80 mm

(c)

Answer:

BaseHeightArea of Triangle
22 cm15.5 cm170.5 cm2

Explanation:

Area of triangle = ½ × Base × Height

Base of the triangle = 22 cm

Height of the triangle =?

Area of triangle = 170.5 cm2

170.5 cm2 = ½ × 22 cm × Height

Height = (170.5 cm2 × 2)/ 22 cm

Height = 341 cm2/ 22 cm

Height = 15.5 cm

5. PQRS is a parallelogram (Fig 11.23). 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.

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Find:

(a) The area of the parallelogram PQRS

Answer: 91.2 cm2

Explanation: Area of the parallelogram = Base × Height

Base of the parallelogram = SR = 12 cm

Height of the parallelogram = QM = 7.6 cm

Area of the parallelogram = 12 cm × 7.6 cm

Area of the parallelogram = 91.2 cm2

(b) QN, if PS = 8 cm

Answer: 11.4 cm

Explanation: QM and QN are the perpendiculars or the height of the parallelogram.

Area of the parallelogram = Base × Height

Base of the parallelogram = PS = 8 cm

Height of the parallelogram = QN =?

Area of the parallelogram = 8 cm × Height

91.2 cm2 = 8 cm × QN

QN = 91.2 cm2 /8 cm

QN = 11.4 cm

6. DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (Fig 11.24). If the area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL.

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: BM = 30 cm, DL = 42 cm

Explanation: Area of the parallelogram = Base × Height

Base of the parallelogram = AB and AD

Let's first find the length of the height DL.

Base of the parallelogram = AB = 35 cm

Height of the parallelogram = DL =?

Area of the parallelogram = 35 cm × Height

1470 cm2 = 35 cm × DL

DL = 1470 cm2 /35 cm

DL = 42 cm

Base of the parallelogram = AD = 49 cm

Height of the parallelogram = BM =?

Area of the parallelogram = 49 cm × Height

1470 cm2 = 49 cm × BM

BM = 1470 cm2 /49 cm

BM = 30 cm

7. ∆ABC is right angled at A (Fig 11.25). 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.

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: Area of triangle = 30 cm2, Height AD = 60/13 cm

Explanation:∆ABC is right angled at A and D. It means that AD and AB are the height of the triangle.

Area of triangle = ½ × Base × Height

Base of the triangle = AB = 12 cm

Height of the triangle = AB = 5 cm

Area of triangle = ½ × 12 cm × 5 cm

Area of triangle = ½ × 60 cm2

Area of triangle = 30 cm2

Let's find the height AD.

Height of the triangle = AD

Base of the triangle = BC = 13 cm

Area of triangle = 30 cm2

30 cm2 = ½ × 13 cm × Height

AD = (30 cm2 × 2)/ 13 cm

AD = 60 cm2/ 13 cm

AD = 60/13 cm

8. ∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 11.26). 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?

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: Area of triangle = 27 cm2, CE = 7.2 cm

Explanation: AD and CE are both the heights of the triangle ABC.

Area of triangle = ½ × Base × Height

Base of the triangle = BC = 9 cm

Height of the triangle = AD = 6 cm

Area of triangle = ½ × 9 cm × 6 cm

Area of triangle = ½ × 54 cm2

Area of triangle = 27 cm2

Let's find the height CE.

Height of the triangle = CE

Base of the triangle = AB = 7.5 cm

Area of triangle = 27 cm2

27 cm2 = ½ × 7.5 cm × Height

CE = (27 cm2 × 2)/ 7.5 cm

CE = 54 cm2/ 7.5 cm

CE = 7.2 cm

Exercise 11.3

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

(a) 14 cm

Answer: 88 cm

Explanation: Circumference of the circle = 2πr

Where,

π = 22/7

r = Radius of the circle

r = 14 cm

Circumference of the circle = 2 × π × r

Circumference of the circle = 2 × 22/7 × 14

Circumference of the circle = 88 cm

(b) 28 mm

Answer: 176 mm

Explanation: Circumference of the circle = 2πr

Where,

π = 22/7

r = Radius of the circle

r = 28 mm

Circumference of the circle = 2 × π × r

Circumference of the circle = 2 × 22/7 × 28

Circumference of the circle = 176 mm

(c) 21 cm

Answer: 132 cm

Explanation: Circumference of the circle = 2πr

Where,

π = 22/7

r = Radius of the circle

r = 21 cm

Circumference of the circle = 2 × π × r

Circumference of the circle = 2 × 22/7 × 21

Circumference of the circle = 132 cm

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

(a) Radius = 14 mm (Take π = 22/7)

Answer: 616 mm2

Explanation: Area of the circle = πr2

Where,

π = 22/7

r = Radius of the circle

r = 14 mm

Area of the circle = π × r × r

Area of the circle = 22/7 × 14 mm × 14 mm

Area of the circle = 616 mm2

(b) Diameter = 49 m

Answer: 1886.5 m2

Explanation: Area of the circle = πr2

Where,

π = 22/7

r = Radius of the circle

D = Diameter of the circle

D = 2r

49 m = 2r

r = 49/2 m

Area of the circle = π × r × r

Area of the circle = 22/7 × 49/2 m × 49/2 m

Area of the circle = (22 × 49 × 49)/ (7 × 2 × 2)

Area of the circle = 1886.5 m2

(c) Radius = 5 cm

Answer: 550/7 cm2

Explanation: Area of the circle = πr2

Where,

π = 22/7

r = Radius of the circle

r = 5 cm

Area of the circle = π × r × r

Area of the circle = 22/7 × 5 cm × 5 cm

Area of the circle = 550/7 cm2

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)

Answer: r = 24.5 m, Area = 1886.5 m2

Explanation: Circumference of the circle = 2πr

Where,

π = 22/7

r = Radius of the circle

r =?

Circumference of the circle = 2 × π × r

154 m = 2 × 22/7 × r

r = (154 m × 7)/ (2 × 22)

r = 49/2 m

r = 24.5 m

Area of the circle = πr2

Area of the circle = π × r × r

Area of the circle = 22/7 × 49/2 m × 49/2 m

Area of the circle = (22 × 49 × 49)/ (7 × 2 × 2)

Area of the circle = 1886.5 m2

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)

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: 132 m, ₹ 528

Explanation: Diameter = 21 m

D = 2r

21 = 2r

r = 21/2 m

Length of the rope to fence the circular garden = Circumference of the garden

Circumference of the circle = 2πr

Circumference of the circle = 2 × π × r

Circumference of the circle = 2 × 22/7 × 21/2

Circumference of the circle = 66 m

Length of the rope to fence 1 round = 66 m

Length of the rope to fence 2 rounds = 2 × 66 = 132 m

Cost of the rope = Cost per meter × Total length of the rope

Cost of the rope = ₹ 4 × 132 m

Cost of the rope = ₹ 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)

Answer: 21.98 cm2

Explanation: Area of the circle = πr2

Area of the circle = π × r × r

Area of the circle with radius 4 cm = 3.14 × 4 cm × 4 cm

Area of the circle with radius 4 cm = 50.26 cm2

Area of the circle with radius 3 cm = 3.14 × 3 cm × 3 cm

Area of the circle with radius 3 cm = 28.27 cm2

Remaining area of the sheet = Area of the sheet with radius 4 cm - Area of the sheet with radius 3 cm

Remaining area of the sheet = 50.26 cm2 - 28.27 cm2

Remaining area of the sheet = 21.98 cm2

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)

Answer: 4.71 m; ₹ 70.65

Explanation: Diameter = 1.5 m

d = 2r

1.5 = 2r

r = 1.5/2 m

Length of the lace to fence the circular table = Circumference of the table

Circumference of the circle = π d

Circumference of the circle = π × d

Circumference of the circle = 3.14 × 1.5

Circumference of the circle = 4.71 m

Length of the lace required to cover the edge of the table = 4.71 m

Cost of the lace = Cost per meter × Total length of the lace

Cost of the lace = ₹ 15 × 4.71 m

Cost of the lace = ₹ 70.65

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

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: 25.70 cm

Explanation: Perimeter of the semi-circle = Circumference of the full circle/2

Circumference of the circle = 2πr

Circumference of the semi-circle = 2πr/2

= πr

Diameter = 10 cm

D = 2r

2r = 10

r = 10/2 = 5 cm

Radius of the circle = 5 cm

Circumference of the semi-circle = π × 5

Circumference of the semi-circle = 3.141 × 5

Circumference of the semi-circle = 15.70 cm

Perimeter of the semi-circle including its diameter = Circumference of the semi-circle + Diameter

= 15.70 cm + 10 cm

= 25.70 cm

8. Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹ 15/m2. (Take π = 3.14)

Answer: ₹ 30.14

Explanation: Diameter = 1.6 m

2r = 1.6 m

r = 1.6/2

r = 0.8 m

Area of the circular table-top = πr2

= π × r × r

Area of the circular table-top = 3.14 × 0.8 m × 0.8 m

Area of the circular table-top = 2.0096 m2

Cost of polishing the circular table-top = Area of the table-top × Cost per meter square

Cost of polishing the circular table-top = 2.0096 m2 × ₹ 15/m2

Cost of polishing the circular table-top = = ₹ 30.14

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)

Answer: 7 cm; 154 cm2; 11 cm; circle

Explanation: Length of the wire bent into the shape of a circle = Perimeter of the circle

Perimeter of the circle = 44 cm

44 cm = 2πr

Where,

r = radius of the circle

π = 22/7

44 cm = 2 × 22/7 × r

r = (44 × 7)/ (2 × 22)

r = 7 cm

Thus, the radius of the circle is 7 cm.

Area of the circle = πr2

= π × r × r

= 22/7 × 7 × 7

Area of the circle = 154 cm2

Length of the wire bent into the shape of a square = Perimeter of the square

Perimeter of the square = 44 cm

44 cm = 4 × (Side of a square)

Side of a square = 44/4 cm

Side of a square = 11 cm

Area of the square = Side × Side

Area of the square = 11 cm × 11cm

Area of the square = 121 cm2

Area of the circle > Area of the square

154 cm2 > 121 cm2

Thus, 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 = 22/7)

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: 536 cm2

Explanation: Radius of card sheet = 14 cm

Area of the card sheet = πr2

= π × r × r

= 22/7 × 14 × 14

Area of the circle = 616 cm2

Radius of smaller circle = 3.5 cm

Area of the smaller circle = πr2

= π × r × r

= 22/7 × 3.5 cm × 3.5 cm

Area of the smaller circle = 38.5 cm2

Area of rectangle = Length × breadth

Length of the rectangle = 3 cm

Breadth of the rectangle = 1 cm

Area of the rectangle = 3 cm × 1 cm

Area of the rectangle = 3 cm2

Area of the remaining sheet = Area of the card sheet - Area of the first circle - Area of the second circle - Area of the rectangle

Area of the remaining sheet = 616 cm2 - 38.5 cm2 - 38.5 cm2 - 3 cm2

Area of the remaining sheet = 536 cm2

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)

Answer: 23.44 cm2

Explanation: Side of the square = 6 cm

Area of the square = Side × Side

Area of the square = 6 cm × 6 cm

Area of the square = 36 cm2

Radius of the circle = 2 cm

Area of the circle = πr2

= π × r × r

Area of the circle = 3.14 × 2 cm × 2 cm

Area of the circle = 12.56 cm2

Area of the leftover aluminium sheet = Area of the square - Area of the cut out circle

Area of the leftover aluminium sheet = 36 cm2 - 12.56 cm2

Area of the leftover aluminium sheet = 23.44 cm2

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

Answer: 5 cm, 78.5 cm2

Explanation: Circumference of the circle = 31.4 cm

Circumference of the circle = 2πr

2πr = 31.4 cm

2 × 3.14 × r = 31.4 cm

r = (31.4)/ (2 × 3.14)

r = 10/2

r = 5 cm

Area of the circle = πr2

= π × r × r

Area of the circle = 3.14 × 5 cm × 5 cm

Area of the circle = 78.5 cm2

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)

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer:879.20 m2

Explanation: Diameter of the flower bed = 66 m

Radius of the flower bed = 66/2 = 33 m

The width of the surrounded path = 4 m

Area of the path = Area of the flower bed with path width - Area of the flower bed without the path width

Diameter of the flower bed with path width = 4 + 66 + 4 = 74 m

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Radius of the flower bed with path width = 74/2 = 37 m

Area of the circle = πr2

= π × r × r

Area of the path = Area of the flower bed with path width - Area of the flower bed without the path width

Area of the path = 3.14 × 37 × 37 - 3.14 × 33 × 33

Area of the path = 3.14 × (37 × 37 - 33 × 33)

Area of the path = 3.14 × (1369 - 1089)

Area of the path = 3.14 × 280

Area of the path = 879.20 m2

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)

Answer: Yes

Explanation: Area of the circular flower garden = 314 m2

Radius of the circle that a sprinkler can cover = 12 m

Area of the circle = πr2

= π × r × r

Area of the circle = 3.14 × 12 × 12

Area of the circle = 452.16 m2

Area of the circle that a sprinkler can cover is greater than the area of the circular flower garden. Hence, the sprinkler can easily water the entire garden.

The answer is 'Yes'.

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

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: 119.32 m, 56.52 m

Explanation: Circumference of the circle = 2πr

Radius of the outer circle = 19 m

Width of the outer circle = 10 m

Radius of the inner circle = Radius of the outer circle - Width of the outer circle

Radius of the inner circle = 19 m - 10 m

= 9 m

Circumference of the inner circle = 2× 3.14 × 9

Circumference of the inner circle = 56.52 m

Circumference of the outer circle = 2× 3.14 × 19

Circumference of the outer circle = 119.32 m

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

Answer: 200 times

Explanation: 1 m = 100 cm

28 cm = 28/100 = 0.28 m

Radius of the wheel = 0.28 m

Circumference of the wheel = 2πr

Circumference of the wheel = 2× 22/7 × 0.28

Circumference of the wheel = 1.76 m

The number of times a wheel can rotate = 352 m/ Circumference of the wheel

The number of times a wheel can rotate = 352 m/1.76 m

The number of times a wheel can rotate = 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)

Answer: 94.2 cm

Explanation: Radius of the minute hand of a clock = 15 cm

In 1 hour, the minute hand covers the full 12 clock hours, i.e., the full circle.

The length of the minute hand moved in hour = Circumference of the clock

Circumference of the clock = 2πr

Circumference of the clock = 2× 3.14 × 15

Circumference of the clock = 94.2 cm

Thus, the length of the minute hand moved in hour is 94.2 cm.

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.

Answer: 1750 m2; 0.675 hectare

Explanation: Length of the garden = 90 m

Breadth of the garden = 75 m

Area of the rectangular garden = Length × Breadth

Area of the rectangular garden = 90 m × 75 m

Area of the rectangular garden = 6750 m2

Path to be built outside the garden = 5 m

Length of the garden with the width = 90 + 5 + 5 = 100 m

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Breadth of the garden with the width = 75 + 5 + 5 = 85 m

Area of the rectangular garden = Length × Breadth

Area of the rectangular garden = 100 m × 85 m

Area of the rectangular garden = 8500 m2

Area of the path = Area of the rectangular garden with outside width - Area of the inside rectangular garden

Area of the path = 8500 m2 - 6750 m2

Area of the path = 1750 m2

Area of the garden = 6750 m2

1 square meter = 0.0001 hectare

= 6750 m2 = 0.0001 × 6750

= 0.675 hectare

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?

Answer: 1176 m2

Explanation: Length of the park =125 m

Breadth of the park = 65 m

Area of the rectangular park = Length × Breadth

Area of the rectangular park = 125 m × 65 m

Area of the rectangular park = 8125 m2

Path to be built outside the park = 3 m

Length of the park with the width = 125 + 3 + 3 = 131 m

Breadth of the park with the width = 65 + 3 + 3 = 71 m

Area of the rectangular park = Length × Breadth

Area of the rectangular park = 131 m × 71 m

Area of the rectangular park = 9301 m2

Area of the path = Area of the rectangular park with outside width - Area of the inside rectangular park

Area of the path = 9301 m2 - 8125 m2

Area of the path = 1176 m2

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.

Answer: 30 cm2

Explanation: Length of the cardboard = 8 cm

Width of the cardboard = 5 cm

Area of the cardboard = Length × Width

Area of the cardboard = 8 cm × 5 cm

Area of the cardboard = 40 cm2

Margin along the sides of the cardboard = 1.5 cm

Length of the painting = 8 cm - 1.5 cm - 1.5 cm = 5 cm

Width of the painting = 5 cm - 1.5 cm - 1.5 cm = 2 cm

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Area of the painting = Length × Width

Area = 5 cm × 2 cm

Area = 10 cm2

Area of the margin = Area of the cardboard- Area of the painting

Area of the margin = 40 cm2 - 10 cm2

Area of the margin = 30 cm2

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 the verandah.

Answer: 63 m2

Explanation: Length of the room = 5.5 m

Width of the room = 4m

Area of the room = Length × Width

Area of the room = 5.5 m × 4 m

Area of the room = 22 m2

Width of the verandah = 2.25 m

Length of the room with verandah = 5.5 m + 2.25 m + 2.25 m = 10 m

Width of the room with verandah = 4 m + 2.25 m + 2.25 m = 8.5 m

Area of the room with verandah = Length × Width

= 10 m × 8.5 m

= 85 m2

Area of the verandah = Area of the room with verandah - Area of the room

Area of the verandah = 85 m2 - 22 m2

Area of the verandah = 63 m2

(ii) the cost of cementing the floor of the verandah at the rate of ₹ 200 per m2 .

Answer: ₹ 12600

Explanation: Cost of cementing the floor of the verandah = Area of the verandah × Rate per meter square

Cost of cementing the floor of the verandah = 63 m2 × ₹ 200

Cost of cementing the floor of the verandah = ₹ 12600

Note: When the width is added outside any area, it is added twice to the length and the breadth of the given area. Similarly, if the width lies inside any area, it is subtracted twice from the length and the breadth.

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

Answer: 116 m2

Explanation: Side of the garden = 30 m

Area of the garden = Side × Side

Area of the garden = 30 m × 30 m

Area of the garden = 900 m2

Width of the path built inside the square garden = 1 m

Side of the square built inside the garden = 30 m - 1 m - 1 m

= 28 m

Area of the inside garden = Side × Side

Area of the inside garden = 28 m × 28 m

Area of the inside garden = 784 m2

Area of the path = Area of the garden - Area of the inside garden

Area of the path = 900 m2 - 784 m2

Area of the path = 116 m2

(ii) The cost of planting grass in the remaining portion of the garden at the rate of ₹ 40 per m2

Answer: ₹ 31360

Explanation: Cost of planting trees in the remaining portion of the garden = Area of the inside × Rate per meter square

Cost of planting trees in the remaining portion of the garden = 784 m2 × ₹ 40

Cost of planting trees in the remaining portion of the garden = ₹ 31360

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.

Answer: 0.99 hectare, 20.01 hectares

Explanation: Length of the rectangular park = 700 m

Breadth of the rectangular park = 300 m

The width of the cross road = 10 m

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Area of the rectangular park = Length × Width

Area of the rectangular park = 700 m × 300 m

Area of the rectangular park = 210000 m2

Length of one side of the cross road = 300 m

Width = 10 m

Area = Length × Width

Area = 300 m × 10 m

Area = 3000 m2

Length of the other side of the cross road = 10 m

Width = 700 m

Area = Length × Width

Area = 10 m × 700 m

Area = 7000 m2

Area of the centre region (common area of the intersection of two crossroads) = 10 m × 10 m

Area = 100 m2

Total area of the two cross roads = 3000 m2 + 7000 m2 - 100 m2

Total area of the two cross roads = 9900 m2

1 square meter = 0.0001 hectare

= 9900 m2 = 0.0001 × 9900

Total area of the two cross roads = 0.99 hectare

Area of the park excluding the cross road = Total area of the rectangular park - Area of the cross roads

Area of the park excluding the cross road = 210000 m2 - 9900 m2

Area of the park excluding the cross road = 200100 m2

1 square meter = 0.0001 hectare

= 200100 m2 = 0.0001 × 200100

Area of the park excluding the cross road = 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.

Answer: 441 m2

Explanation: Length of the rectangular field = 90 m

Breadth of the rectangular field = 60 m

The width of the cross road = 3 m

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Area of the rectangular park = Length × Width

Area of the rectangular park = 90 m × 60 m

Area of the rectangular park = 5400 m2

Length of one side of the cross road = 60 m

Width = 3 m

Area = Length × Width

Area = 60 m × 3 m

Area = 180 m2

Length of the other side of the cross road = 3 m

Width = 90 m

Area = Length × Width

Area = 3 m × 90 m

Area = 270 m2

Area of the centre region (common area of the intersection of two crossroads) = 3 m × 3 m

Area = 9 m2

Total area of the two cross roads = 180 m2 + 270 m2 - 9 m2

Total area of the two cross roads = 441 m2

(ii) The cost of constructing the roads at the rate of ₹ 110 per m2.

Answer: ₹ 48,510

Explanation: Cost of constructing the roads = Area of the cross roads × Rate per meter square

Cost of constructing the roads = 441 m2 × ₹ 110

Cost of constructing the roads = ₹ 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)

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: Yes, 9. 12 cm cord length is left

Explanation: Radius of the circle = 4 cm

Side of the square = 4 cm

The length of the cord required to cover a circular pipe = Perimeter of the circle

The length of the cord required to cover a square = Perimeter of the square

Perimeter of the circular pipe = 2 π r = Circumference of the circle

Perimeter of the circular pipe = 2 × 3.14 × 4

Perimeter of the circular pipe = 25.12 cm

So, length of the cord required to cover a circular pipe is 25.12 cm.

Perimeter of the square = 4 × Side

Perimeter of the square = 4 × 4

Perimeter of the square = 16 cm

Perimeter of the square < Perimeter of the circle

Chord left = Length of the cord required to cover the circle - Length of the cord required to cover the square

Chord left = 25.12 cm - 16 cm

Chord left = 9. 12 cm

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

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Find:

(i) The area of the whole land

Answer: 62.56 m2

Explanation: Area of the whole land = Area of the rectangular lawn

Length of the rectangle = 10 m

Width of the rectangle = 5 m

Area of the rectangular lawn = Length × Width

Area of the rectangular lawn = 10 m × 5 m

Area of the rectangular lawn = 50 m2

(ii) The area of the flower bed

Answer: 12.56 m2

Explanation: Radius of the circular flower bed = 2 m

Area of the circular flower bed = π r2

Area of the circular flower bed = 3.14 × 2 × 2

Area of the circular flower bed = 12.56 m2

(iii) The area of the lawn excluding the area of the flower bed

Answer: 37.44 m2

Explanation: Area of the lawn excluding the area of the flower bed = Area of the rectangular lawn - Area of the circular flower bed

Area of the rectangular lawn = 50 m2

Area of the circular flower bed = 12.56 m2

Area of the lawn excluding the area of the flower bed = 50 m2- 12.56 m2

Area of the whole land = 37.44 m2

(iv) The circumference of the flower bed.

Answer: 12.56 m

Explanation: Radius of the flower bed = 2 m

Circumference of the flower bed = 2 π r

Circumference of the flower bed = 2 × 3.14 × 2

Circumference of the flower bed =12.56 m

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

(i)

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: 110 cm2

Explanation: Area of the shaded portion = Area of the rectangle ABCD - Area of the triangle AEF - Area of the triangle BEC

Length of the rectangle = 18 cm

Width of the rectangle = 10 cm

Area of the rectangle = Length × Width

Area of the rectangle = 18 cm × 10 cm

Area of the rectangle ABCD = 180 cm2

Area of the triangle AEF = ½ × Base × Height

Base of the triangle AEF = 10 cm

Height of the triangle AEF = 6 cm

Area of the triangle AEF = ½ × 10 cm × 6 cm

Area of the triangle AEF = 30 cm2

Area of the triangle BEC = ½ × Base × Height

Base of the triangle BEC = 8 cm

Height of the triangle BEC =10 cm

Area of the triangle BEC = ½ × 8 cm × 10 cm

Area of the triangle BEC = 40 cm2

Area of the shaded portion = Area of the rectangle ABCD - Area of the triangle AEF - Area of the triangle BEC

Area of the shaded portion = 180 cm2 - 30 cm2 - 40 cm2

Area of the shaded portion = 110 cm2

(ii)

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: 150 cm2

Explanation: Area of the shaded portion = Area of the rectangle PQRS - Area of the triangle PQT - Area of the triangle TUS - Area of the triangle QUR

Length of the rectangle = 20 cm

Width of the rectangle = 20 cm

Area of the rectangle = Length × Width

Area of the rectangle = 20 cm × 20 cm

Area of the rectangle PQRS = 400 cm2

Area of the triangle PQT = ½ × Base × Height

Base of the triangle PQT = 20 cm

Height of the triangle PQT =10 cm

Area of the triangle PQT = ½ × 20 cm × 10 cm

Area of the triangle PQT = 100 cm2

Base of the triangle TUS = 10 cm

Height of the triangle TUS =10 cm

Area of the triangle TUS = ½ × 10 cm × 10 cm

Area of the triangle TUS = 50 cm2

Base of the triangle QUR = 10 cm

Height of the triangle QUR =20 cm

Area of the triangle QUR = ½ × 10 cm × 20 cm

Area of the triangle QUR = 100 cm2

Area of the shaded portion = Area of the rectangle PQRS - Area of the triangle PQT - Area of the triangle TUS - Area of the triangle QUR

Area of the shaded portion = 400 cm2 - 100 cm2 - 50 cm2 - 100 cm2

Area of the shaded portion = 150 cm2

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

NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area

Answer: 66 cm2

Explanation: Area of the quadrilateral ABCD = Area of the triangle ABC + Area of the triangle ACD

Area of the triangle ABC = ½ × Base × Height of the perpendicular BM

Base of the triangle ABC = 22 cm

Height of the triangle ABC =3 cm

Area of the triangle ABC = ½ × 22 cm × 3 cm

Area of the triangle ABC = 33 cm2

Area of the triangle ACD = ½ × Base × Height of the perpendicular DN

Base of the triangle ACD = 22 cm

Height of the triangle ACD =3 cm

Area of the triangle ACD = ½ × 22 cm × 3 cm

Area of the triangle ACD = 33 cm2

Area of the quadrilateral ABCD = Area of the triangle ABC + Area of the triangle ACD

Area of the quadrilateral ABCD = 33 cm2 + 33 cm2

Area of the quadrilateral ABCD = 66 cm2






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