NCERT Solutions for Class 7 Maths Chapter 11 - Perimeter and Area

Discover our PDF solutions for NCERT Solutions for Class 7 Math Chapter 11, "Perimeter and Area." These comprehensive solutions are designed to help students navigate the intricacies of this chapter, offering step-by-step answers to all the questions found within. With a focus on clarity and thorough explanations, these solutions are invaluable for students seeking to excel in this subject. Our NCERT Solutions are thoughtfully prepared to simplify the complexities of perimeter and area problems, enabling students to understand and apply these concepts with confidence. These solutions serve as an essential study resource for students aiming for excellence in mathematics.

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Exercise 11.1

Question 1 :

A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m (Fig). Find the cost of white washing the wall, if the rate of white washing the wall is ₹ 20 per m2.

From the question it is given that,

Length of the door = 2 m

Breadth of the door = 1 m

Length of the wall = 4.5 m

Breadth of the wall = 3.6 m

Then,

Area of the door = Length × Breadth

= 2 × 1

= 2 m2

Area of the wall = Length × Breadth

= 4.5 × 3.6

= 16.2 m2

So, Area to be white washed = 16.2 – 2 = 14.2 m2

Cost of white washing 1 m2 area = ₹ 20

Hence, cost of whit washing 14.2 m2 area = 14.2 × 20

= ₹ 284

Question 2 :

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.

From the question it is given that.

Perimeter of the rectangle = 130 cm

Breadth of the rectangle = 30

We know that,

Perimeter of rectangle = 2 (Length + Breadth)

130 = 2 (length + 30)

130/2 = length + 30

Length + 30 = 65

Length = 65 – 30

Length = 35 cm

Then,

Area of the rectangle = Length × Breadth

= 35 × 30

= 1050 cm2

Question 3 :

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?

By reading the question we can conclude that, perimeter of the square is the same as perimeter of rectangle.

From the question it is given that,

Length of the rectangle = 40 cm

Breadth of the square = 22 cm

Then,

Perimeter of the rectangle = Perimeter of the Square

2 (Length + Breadth) = 4 × side

2 (40 + 22) = 4 × side

2 (62) = 4 × side

124 = 4 × side

Side = 124/4

Side = 31 cm

So, Area of the rectangle = (Length × Breadth)

= 40 × 22

= 880 cm2

Area of square = side2

= 312

= 31 × 31

= 961 cm2

∴ Square shaped wire encloses more area.

Question 4 :

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.

From the question it is given that,

Area of a square park is the same as of a rectangular park.

Side of the square park = 60 m

Length of the rectangular park = 90 m

We know that,

Area of the square park = (one of the side of square)2

= 602

= 3600 m2

Area of the rectangular park = 3600 m2 … [∵ given]

Question 5 :

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

Also find the area.

From the question it is given that,

Perimeter of the a rectangular sheet = 100 cm

Length of the rectangular sheet = 35 cm

We know that,

Perimeter of the rectangle = 2 (Length + Breadth)

100 = 2 (35 + Breadth)

Then,

Area of the rectangle = Length × Breadth

= 35 × 15

= 525 cm2

∴ Area of the rectangular sheet is 525 cm2

Question 6 :

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.

From the question it is given that,

Area of the rectangular plot = 440 m2

Length of the rectangular plot = 22 m

We know that,

Area of the rectangle = Length × Breadth

Then,

Perimeter of the rectangle = 2(Length + Breadth)

= 2 (22 + 20)

= 2(42)

= 84 m

∴ Perimeter of the rectangular plot is 84 m.

Question 7 :

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

From the question it is given that,

Perimeter of the square park = 320 m

4 × Length of the side of park = 320 m

Then,

Length of the side of park = 320/4

= 80 m

So, Area of the square park = (length of the side of park)2

= 802

= 6400 m2

Question 8 :

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 m2 of the land costs ₹ 10,000.

From the question it is given that,

Length of the rectangular piece of land = 500 m

Breadth of the rectangular piece of land = 300 m

Then,

(i) Area of rectangle = Length × Breadth

= 500 × 300

= 150000 m2

(ii) Cost of the land for 1 m2 = ₹ 10000

Cost of the land for 150000 m2 = 10000 × 150000

= ₹ 1500000000

Exercise 11.2

Question 1 :

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, find

(a) The area of the parallelogram PQRS

(b) QN, if PS = 8 cm.

Fig 11.23

From the question, it is given that

SR = 12 cm, QM = 7.6 cm

(a) We know that

Area of the parallelogram = Base × Height

= SR × QM

= 12 × 7.6

= 91.2 cm2

(b) Area of the parallelogram = Base × Height

91.2 = PS × QN

91.2 = 8 × QN

QN = 91.2/8

QN = 11.4 cm

Question 2 :

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.

Fig 11.24

From the question, it is given that

Area of the parallelogram = 1470 cm2

AB = 35 cm

Then,

We know that

Area of the parallelogram = Base × Height

1470 = AB × BM

1470 = 35 × DL

DL = 1470/35

DL = 42 cm

And,

Area of the parallelogram = Base × Height

1470 = 49 × BM

BM = 1470/49

BM = 30 cm

Question 3 :

ΔABC 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 ΔABC. Also, find the length of AD.

Fig 11.25

From the question, it is given that

AB = 5 cm, BC = 13 cm, AC = 12 cm

Then,

We know that

Area of the ΔABC = ½ × Base × Height

= ½ × AB × AC

= ½ × 5 × 12

= 1 × 5 × 6

= 30 cm2

Now,

Area of ΔABC = ½ × Base × Height

30 = ½ × AD × BC

30 = ½ × AD × 13

Question 4 :

ΔABC 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 ΔABC. What will be the height from C to AB, i.e., CE?

From the question, it is given that

AB = AC = 7.5 cm, BC = 9 cm, AD = 6cm

Then,

Area of the ΔABC = ½ × Base × Height

= ½ × BC × AD

= ½ × 9 × 6

= 1 × 9 × 3

= 27 cm2

Now,

Area of the ΔABC = ½ × Base × Height

27 = ½ × AB × CE

27 = ½ × 7.5 × CE

(27 × 2)/7.5 = CE

CE = 54/7.5

CE = 7.2 cm

Question 5 :

Find the area of each of the following parallelograms.

(a)

(b)

(c)

(d)

(e)

(a) From the figure,

Height of the parallelogram = 4 cm

Base of the parallelogram = 7 cm

Then,

Area of the parallelogram = Base × Height

= 7 × 4

= 28 cm2

(b) From the figure,

Height of the parallelogram = 3 cm

Base of the parallelogram = 5 cm

Then,

Area of the parallelogram = Base × Height

= 5 × 3

= 15 cm2

(c) From the figure,

Height of the parallelogram = 3.5 cm

Base of the parallelogram = 2.5 cm

Then,

Area of the parallelogram = Base × Height

= 2.5 × 3.5

= 8.75 cm2

(d) From the figure,

Height of the parallelogram = 4.8 cm

Base of the parallelogram = 5 cm

Then,

Area of the parallelogram = Base × Height

= 5 × 4.8

= 24 cm2

(e) From the figure,

Height of the parallelogram = 4.4 cm

Base of the parallelogram = 2 cm

Then,

Area of the parallelogram = Base × Height

= 2 × 4.4

= 8.8 cm2

Question 6 :

Find the area of each of the following triangles.

(a)

(b)

(c)

(d)

(a) From the figure,

Base of the triangle = 4 cm

Height of the height = 3 cm

Then,

Area of the triangle = ½ × Base × Height

= ½ × 4 × 3

= 1 × 2 × 3

= 6 cm2

(b) From the figure,

Base of the triangle = 3.2 cm

Height of the height = 5 cm

Then,

Area of the triangle = ½ × Base × Height

= ½ × 3.2 × 5

= 1 × 1.6 × 5

= 8 cm2

(c) From the figure,

Base of the triangle = 3 cm

Height of the height = 4 cm

Then,

Area of the triangle = ½ × Base × Height

= ½ × 3 × 4

= 1 × 3 × 2

= 6 cm2

(d) From the figure,

Base of the triangle = 3 cm

Height of the height = 2 cm

Then,

Area of the triangle = ½ × Base × Height

= ½ × 3 × 2

= 1 × 3 × 1

= 3 cm2

Question 7 :

Find the missing values.

 S.No. Base Height Area of the Parallelogram a. 20 cm 246 cm2 b. 15 cm 154.5 cm2 c. 8.4 cm 48.72 cm2 d. 15.6 cm 16.38 cm2

(a) From the table,

Base of the parallelogram = 20 cm

Height of the parallelogram =?

Area of the parallelogram = 246 cm2

Then,

Area of the parallelogram = Base × Height

246 = 20 × height

Height = 246/20

Height = 12.3 cm

∴ Height of the parallelogram is 12.3 cm.

(b) From the table,

Base of the parallelogram =?

Height of the parallelogram =15 cm

Area of the parallelogram = 154.5 cm2

Then,

Area of the parallelogram = Base × Height

154.5 = Base × 15

Base = 154.5/15

Base = 10.3 cm

∴ Base of the parallelogram is 10.3 cm.

(c) From the table,

Base of the parallelogram =?

Height of the parallelogram =8.4 cm

Area of the parallelogram = 48.72 cm2

Then,

Area of the parallelogram = Base × Height

48.72 = Base × 8.4

Base = 48.72/8.4

Base = 5.8 cm

∴ Base of the parallelogram is 5.8 cm.

(d) From the table,

Base of the parallelogram = 15.6 cm

Height of the parallelogram =?

Area of the parallelogram = 16.38 cm2

Then,

Area of the parallelogram = Base × Height

16.38 = 15.6 × Height

Height = 16.38/15.6

Height = 1.05 cm

∴ Height of the parallelogram is 1.05 cm.

 S.No. Base Height Area of the Parallelogram a. 20 cm 12.3 cm 246 cm2 b. 10.3 cm 15 cm 154.5 cm2 c. 5.8 cm 8.4 cm 48.72 cm2 d. 15.6 cm 1.05 16.38 cm2

Question 8 :

Find the missing values.

 Base Height Area of Triangle 15 cm 87 cm2 31.4 mm 1256 mm2 22 cm 170.5 cm2

(a) From the table,

Height of the triangle =?

Base of the triangle = 15 cm

Area of the triangle = 16.38 cm2

Then,

Area of the triangle = ½ × Base × Height

87 = ½ × 15 × Height

Height = (87 × 2)/15

Height = 174/15

Height = 11.6 cm

∴ Height of the triangle is 11.6 cm.

(b) From the table,

Height of the triangle =31.4 mm

Base of the triangle =?

Area of the triangle = 1256 mm2

Then,

Area of the triangle = ½ × Base × Height

1256 = ½ × Base × 31.4

Base = (1256 × 2)/31.4

Base = 2512/31.4

Base = 80 mm = 8 cm

∴ Base of the triangle is 80 mm or 8 cm.

(c) From the table,

Height of the triangle =?

Base of the triangle = 22 cm

Area of the triangle = 170.5 cm2

Then,

Area of the triangle = ½ × Base × Height

170.5 = ½ × 22 × Height

170.5 = 1 × 11 × Height

Height = 170.5/11

Height = 15.5 cm

∴ Height of the triangle is 15.5 cm.

Exercise 11.3

Question 1 :

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)

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 = 2πr

44 = 2 × (22/7) × r

44 = 44/7 × r

(44 × 7)/44 = r

r = 7 cm

Area of the circle = πr2

= 22/7 × 72

= 22/7 × 7 ×7

= 22 × 7

= 154 cm2

Now,

If the wire is bent into a square,

Length of the wire = perimeter of the square

44 = 4 x side

44 = 4s

s = 44/4

s = 11cm

Area of square = (side)2 =  112

= 121 cm2

Therefore, the circle has more area than the square.

Question 2 :

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)

From the question, it is given that,

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

The breadth of the rectangle = 1 cm

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

Now,

Area of the circular card sheet = πr2

= 22/7 × 142

= 22/7 × 14 × 14

= 22 × 2 × 14

= 616 cm2

Area of the 2 small circles = 2 × πr2

= 2 × (22/7 × 3.52)

= 2 × (22/7 × 3.5 × 3.5)

= 2 × ((22/7) × 12.25)

= 2 × 38.5

= 77 cm2

Area of the rectangle = Length × Breadth

= 3 × 1

= 3 cm2

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 cm2

Hence, the area of the remaining sheet is 536 cm2.

Question 3 :

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)

From the question, it is given that,

Diameter of the flower bed = 66 m

Then,

The radius of the flower bed = d/2

= 66/2

= 33 m

Area of flower bed = πr2

= 3.14 × 332

= 3.14 × 1089

= 3419.46 m

Now we have to find the area of the flower bed and path together

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

Area of the flower bed and path together = πr2

= 3.14 × 372

= 3.14 × 1369

= 4298.66 m

Finally,

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

= 4298.66 – 3419.46

= 879.2 m2

Hence, the area of the path is 879.2 m2.

Question 4 :

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)

From the question, it is given that,

Area of the circular flower garden = 314 m2

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

Area of the circular flower garden = πr2

314 = 3.14 × r2

314/3.14 = r2

r2 = 100

r = √100

r = 10 m

∴ The radius of the circular flower garden is 10 m.

Since the sprinkler can cover an area of a radius of 12 m, the sprinkler will water the whole garden.

Question 5 :

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

From the figure,

The radius of the inner circle = outer circle radius – 10

= 19 – 10

= 9 m

Circumference of the inner circle = 2πr

= 2 × 3.14 × 9

= 56.52 m

Then,

The radius of the outer circle = 19 m

Circumference of the outer circle = 2πr

= 2 × 3.14 × 19

= 119.32 m

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

Question 6 :

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

From the question, it is given that,

The radius of the wheel = 28 cm

Total distance = 352 m = 35200 cm

Circumference of the wheel = 2πr

= 2 × 22/7 × 28

= 2 × 22 × 4

= 176 cm

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

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 to go 352 m.

Question 7 :

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)

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

= 2πr

= 2 × 3.14 × 15

= 94.2 cm

Therefore, the minute hand moves 94.2 cm in 1 hour.

Question 8 :

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

(a) 14 cm

(b) 28 mm

(c) 21 cm

(a) Given, the radius of the circle = 14 cm

Circumference of the circle = 2πr

= 2 × (22/7) × 14

= 2 × 22 × 2

= 88 cm

(b) Given, the radius of the circle = 28 mm

Circumference of the circle = 2πr

= 2 × (22/7) × 28

= 2 × 22 × 4

= 176 mm

(c) Given, the radius of the circle = 21 cm

Circumference of the circle = 2πr

= 2 × (22/7) × 21

= 2 × 22 × 3

= 132 cm

Question 9 :

Find the area of the following circles, given that:

(a) Radius = 14 mm (Take π = 22/7)

(b) Diameter = 49 m

(a) Given, the radius of the circle = 14 mm

Then,

Area of the circle = πr2

= 22/7 × 142

= 22/7 × 196

= 22 × 28

= 616 mm2

(b) Given, the diameter of the circle (d) = 49 m

We know that radius (r) = d/2

= 49/2

= 24.5 m

Then,

Area of the circle = πr2

= 22/7 × (24.5)2

= 22/7 × 600.25

= 22 × 85.75

= 1886.5 m2

(c) Given, the radius of the circle = 5 cm

Then,

Area of the circle = πr2

= 22/7 × 52

= 22/7 × 25

= 550/7

= 78.57 cm2

Question 10 :

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)

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 = πr2

= 3.14 × 0.82

= 3.14 × 0.8 ×0.8

= 2.0096 m2

Cost of polishing 1 m2 area = ₹ 15 [given]

Cost of polishing 2.0096 m2 area = ₹ 15 × 2.0096

= ₹ 30.144

Hence, the cost of polishing a 2.0096 m2 area is ₹ 30.144.

Question 11 :

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

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 = πr + 2r

= 3.14(5) + 2(5)

= 5 [3.14+ 2]

= 5 [5.14]

Therefore, the perimeter of the semicircle = 25.7 cm

Question 12 :

Saima wants to put a lace on the edge of a circular table cover with a diameter of 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)

From the question, it is given that,

Diameter of the circular table cover = 1.5 m

We know that radius (r) = d/2

= 1.5/2

= 0.75 m

Then,

Circumference of the circular table cover = 2πr

= 2 × 3.14 × 0.75

= 4.71 m

So, the length of the lace required = 4.71 m.

Cost of 1 m lace = ₹ 15 [given]

Cost of 4.71 m lace = ₹ 15 × 4.71

= ₹ 70.65

So, the cost of the lace = ₹ 70.65.

Question 13 :

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)

From the question, it is given that,

The radius of circular sheet R = 4 cm

A circle of radius to be removed r = 3 cm

Then,

The area of the remaining sheet = πR2 – πr2

= π (R2 – r2)

= 3.14 (42 – 32)

= 3.14 (16 – 9)

= 3.14 × 7

= 21.98 cm2

So, the area of the remaining sheet is 21.98 cm2.

Question 14 :

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

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 = 2πr

= 2 × (22/7) × 10.5

= 462/7

= 66 m

So, the length of rope required = 2 × 66 = 132 m.

Cost of 1 m rope = ₹ 4 [given]

Cost of 132 m rope = ₹ 4 × 132

= ₹ 528

So, the cost of the rope = ₹ 528.

Question 15 :

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 leftover aluminium sheet? (Take π = 3.14)

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 the square aluminium sheet and circle to find out the remaining area.

Now,

Area of the square = side2

Hence, the area of the square aluminium sheet = 62 = 36 cm2

Area of the circle = πr2

= 3.14 × 22

= 3.14 × 2 × 2

= 3.14 × 4

= 12.56 cm2

Now,

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

= 36 – 12.56

= 23.44 cm2

Hence, the area of the aluminium sheet left is 23.44 cm2.

Question 16 :

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

From the question, it is given that,

Circumference of the circle = 154 m

Then,

We know that, circumference of the circle = 2πr

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 = πr2

= 22/7 × (24.5)2

= 22/7 × 600.25

= 22 × 85.75

= 1886.5 m2

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

Question 17 :

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

From the question, it is given that,

Circumference of a circle = 31.4 cm

We know that,

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 = πr2

= 3.14 × (5cm)2

= 3. 14 × 25 cm2

= 78.5 cm2

Therefore, the radius of the circle is 5 cm, and the area of the circle is 78.5 cm2.

Exercise 11.4

Question 1 :

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.

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

(i) From the question it is given that,

Length of the room (L) = 5.5 m

Breadth of the room (B) = 4 m

Then,

Area of the room = length × breadth

= 5.5 × 4

= 22 m2

From the figure,

The new length and breadth of the room when verandah is included is 10 m and 8.5 m, respectively.

New area of the room when verandah is included = 10 × 8.5

= 85 m2

The area of verandah = Area of the room when verandah is included – Area of the room

= 85 – 22

= 63 m2

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

Then the cost of cementing the 63 m2 area of floor of the verandah = 200 × 63

= ₹ 12600

Question 2 :

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

(i) the area of the 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.

(i) From the figure,

Length of rectangular lawn = 10 m

Breadth of rectangular lawn = 5 m

Area of the rectangular lawn = Length × Breadth

= 10 × 5

= 50 m2

(ii) From the figure,

Radius of the flower bed = 2 m

Area of the flower bed = πr2

= 3.14 × 22

= 3.14 × 4

= 12.56 m2

(iii) The area of the lawn excluding the area of the flower bed = Area of rectangular lawn –

Area of flower bed

= 50 – 12.56

= 37.44 m2

(iv) The circumference of the flower bed = 2πr

= 2 × 3.14 × 2

= 12.56 m

Question 3 :

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

(i)

(ii)

(i) To find the area of EFDC, first we have to find the area of ΔAEF, ΔEBC and rectangle ABCD

Area of ΔAEF = ½ × Base × Height

= ½ × 6 × 10

= 1 × 3 × 10

= 30 cm2

Area of ΔEBC = ½ × Base × Height

= ½ × 8 × 10

= 1 × 4 × 10

= 40 cm2

Area of rectangle ABCD = length × breadth

= 18 × 10

= 180 cm2

Then,

Area of EFDC = ABCD area – (ΔAEF + ΔEBC)

= 180 – (30 + 40)

= 180 – 70

= 110 cm2

(ii) To find the area of ΔQTU, first we have to find the area of ΔSTU, ΔTPQ, ΔQRU and square PQRS

Area of ΔSTU = ½ × Base × Height

= ½ × 10 × 10

= 1 × 5 × 10

= 50 cm2

Area of ΔTPQ = ½ × Base × Height

= ½ × 10 × 20

= 1 × 5 × 20

= 100 cm2

Area of ΔQRU = ½ × Base × Height

= ½ × 10 × 20

= 1 × 5 × 20

= 100 cm2

Area of square PQRS = Side2

= 20 × 20

= 400 cm2

Then,

Area of ΔQTU = PQRS area – (ΔSTU + ΔTPQ + ΔQRU)

= 400 – (50 + 100 + 100)

= 400 – 250

= 150 cm2

Question 4 :

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.

From the question it is given that,

Length of the garden (L) = 90 m

Breadth of the garden (B) = 75 m

Then,

Area of the garden = length × breadth

= 90 × 75

= 6750 m2

From the figure,

The new length and breadth of the garden when path is included is 100 m and 85 m, respectively.

New area of the garden = 100 × 85

= 8500 m2

The area of path = New area of the garden including path – Area of garden

= 8500 – 6750

= 1750 m2

For 1 hectare = 10000 m2

Hence, area of garden in hectare = 6750/10000

= 0.675 hectare

Question 5 :

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.

From the question it is given that,

Length of the park (L) = 125 m

Breadth of the park (B) = 65 m

Then,

Area of the park = length × breadth

= 125 × 65

= 8125 m2

From the figure,

The new length and breadth of the park when path is included is 131 m and 71 m, respectively.

New area of the park = 131 × 71

= 9301 m2

The area of path = New area of the park including path – Area of park

= 9301 – 8125

= 1176 m2

Question 6 :

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.

From the question it is given that,

Length of the cardboard (L) = 8 cm

Breadth of the cardboard (B) = 5 cm

Then,

Area of the cardboard = length × breadth

= 8 × 5

= 40 cm2

From the figure,

The new length and breadth of the cardboard when margin is not included is 5 cm and 2 cm, respectively.

New area of the cardboard = 5 × 2

= 10 cm2

The area of margin = Area of the cardboard when margin is including – Area of the

cardboard when margin is not including

= 40 – 10

= 30 cm2

Question 7 :

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 ₹ 40 per m2.

(i) From the question it is given that,

Side of square garden (s) = 30 m

Then,

Area of the square garden = S2

= 302

= 30 × 30

= 900 m2

From the figure,

The new side of the square garden when path is not included is 28 m.

New area of the room when verandah is included = 282

= 28 × 28

= 784 m2

The area of path = Area of the square garden when path is included – Area of the square

Garden when path is not included

= 900 – 784

= 116 m2

(ii) Given, the cost of planting the grass in the remaining portion of the garden at the rate of

= ₹ 40 per m2

Then the cost of planting the grass in 784 m2 area of the garden = 784 × 40

= ₹ 31360

Question 8 :

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.

From the question it is given that,

Length of the park (L) = 700 m

Breadth of the park (B) = 300 m

Then,

Area of the park = length × breadth

= 700 × 300

= 210000 m2

Let us assume that ABCD is the one cross road and EFGH is another cross road in the park.

The length of ABCD cross road = 700 m

The length of EFGH cross road = 300 m

Both cross road have the same width = 10 m

Then,

= 700 × 10

= 7000 m2

= 300 × 10

= 3000 m2

Area of the IJKL at center = length × breadth

= 10 × 10

= 100 m2

Area of the roads = Area of ABCD + Area of EFGH – Area of IJKL

= 7000 + 3000 – 100

= 10000 – 100

= 9900 m2

We know that, for 1 hectare = 10000 m2

Hence, area of roads in hectare = 9900/10000

= 0.99 hectare

Finally, Area of the park excluding roads = Area of park – Area of the roads

= 210000 – 9900

= 200100 m2

= 200100/10000

= 20.01 hectare

Question 9 :

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)

From the question it is given that,

Radius of a circular pipe = 4 cm

Side of a square = 4 cm

Then,

Perimeter of the circular pipe = 2πr

= 2 × 3.14 × 4

= 25.12 cm

Perimeter of the square = 4 × side of the square

= 4 × 4

= 16 cm

So, the length of cord left with Pragya = Perimeter of circular pipe – Perimeter of square

= 25.12 – 16

= 9.12 cm

Yes, 9.12 cm cord is left.

Question 10 :

Find the area of the quadrilateral ABCD.

Here, AC = 22 cm, BM = 3 cm,

DN = 3 cm, and BM ⊥ AC, DN ⊥ AC

From the it is given that,

AC = 22 cm, BM = 3 cm DN = 3 cm and BM ⊥ AC, DN ⊥ AC

To find the area of quadrilateral ABCD, first we have to find the area of ΔABC, and ΔADC Area of ΔABC = ½ × Base × Height

= ½ × 22 × 3

= 1 × 11 × 3

= 33 cm2

Area of ΔADC = ½ × Base × Height

= ½ × 22 × 3

= 1 × 11 × 3

= 33 cm2

Then,

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

= 33 + 33

= 66 cm2

Question 11 :

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 ₹ 110 per m2.

(i) From the question it is given that,

Length of the field (L) = 90 m

Breadth of the field (B) = 60 m

Then,

Area of the field = length × breadth

= 90 × 60

= 5400 m2

Let us assume that ABCD is the one cross road and EFGH is another cross road in the park.

The length of ABCD cross road = 90 m

The length of EFGH cross road = 60 m

Both cross road have the same width = 3 m

Then,

= 90 × 3

= 270 m2

= 60 × 3

= 180 m2

Area of the IJKL at center = length × breadth

= 3 × 3

= 9 m2

Area of the roads = Area of ABCD + Area of EFGH – Area of IJKL

= 270 + 180 – 9

= 450 – 9

= 441 m2

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

Then the cost of constructing the 441 m2 roads = 441 × 110

= ₹ 48510

The NCERT solution for Class 7 Chapter 11: Perimeter and Area is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.

Yes, the NCERT solution for Class 7 Chapter 11: Perimeter and Area is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. Perimeter and Areaally, they can solve the practice questions and exercises that allow them to get exam-ready in no time.

You can get all the NCERT solutions for Class 7 Maths Chapter 11 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.

Yes, students must practice all the questions provided in the NCERT solution for Class 7 Maths Chapter 11: Perimeter and Area as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.

Students can utilize the NCERT solution for Class 7 Maths Chapter 11 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.

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