NCERT Solutions Class 7 Mathematics Chapter 11 Perimeter and Area

NCERT Solutions Class 7 Mathematics Chapter 11 Perimeter and Area have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 7 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 7 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 7 Mathematics are an important part of exams for Class 7 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 7 Mathematics and also download more latest study material for all subjects. Chapter 11 Perimeter and Area is an important topic in Class 7, please refer to answers provided below to help you score better in exams

Chapter 11 Perimeter and Area Class 7 Mathematics NCERT Solutions

Class 7 Mathematics students should refer to the following NCERT questions with answers for Chapter 11 Perimeter and Area in Class 7. These NCERT Solutions with answers for Class 7 Mathematics will come in exams and help you to score good marks

Chapter 11 Perimeter and Area NCERT Solutions Class 7 Mathematics

 

Exercise 11.1

Q.1) The length and breadth of a rectangular piece of land are 500 π‘š and 300 π‘š respectively.
Find:
(i) Its area.
(ii) The cost of the land, if 1π‘š2 of the land costs 𝑅𝑠. 10,000.
Sol.1) Given: Length of a rectangular piece of land = 500 π‘š and Breadth of a rectangular piece of land = 300 π‘š
(i) Area of a rectangular piece of land = πΏπ‘’π‘›π‘”π‘‘β„Ž Γ— π΅π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž
= 500 Γ— 300 = 1,50,000 π‘š2
(ii) Since, the cost of 1π‘š2 land = 𝑅𝑠. 10,000
Therefore, the cost of 1,50,000 π‘š2 land = 10,000 Γ— 1,50,000
= 𝑅𝑠. 1,50,00,00,000

Q.2) Find the area of a square park whose perimeter is 320 π‘š
Sol.2) Given: Perimeter of square park = 320 π‘š
β‡’ 4 Γ— 𝑠𝑖𝑑𝑒 = 320
β‡’ side = 320 4 = 80 π‘š
Now, Area of square park = 𝑠𝑖𝑑𝑒 Γ— 𝑠𝑖𝑑𝑒
= 80 Γ— 80 = 6400 π‘š2
Thus, the area of square park is 6400 π‘š2.

Q.3) Find the breadth of a rectangular plot of land, if its area is 440 π‘š2 and the length is 22 π‘š. Also find its perimeter.
Sol.3) Area of rectangular park = 440 π‘š2
β‡’ length x breadth = 440 π‘š2
β‡’ 22 Γ— π‘π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž = 440
β‡’ breadth = 440/22 = 20 π‘š
Now, Perimeter of rectangular park = 2 (π‘™π‘’π‘›π‘”π‘‘β„Ž + π‘π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž)
= 2 (22 + 20) = 2 Γ— 42 = 84 π‘š
Thus, the perimeter of rectangular park is 84 π‘š.

Q.4) The perimeter of a rectangular sheet is 100 π‘π‘š. If the length is 35 π‘π‘š, find its breadth. Also find the area.
Sol.4) Perimeter of the rectangular sheet = 100π‘π‘š
β‡’ 2 (π‘™π‘’π‘›π‘”π‘‘β„Ž + π‘π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž) = 100 π‘π‘š
β‡’ 2 (35 + π‘π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž) = 100
β‡’ 35 + π‘π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž = 100/2
β‡’ 35 + π‘π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž = 50
β‡’ breadth = 50 – 35
β‡’ breadth = 15 π‘π‘š
Now, Area of rectangular sheet = π‘™π‘’π‘›π‘”π‘‘β„Ž Γ— π‘π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž
= 35 Γ— 15 = 525 cπ‘š2
Thus, breadth and area of rectangular sheet are 15 π‘π‘š and 525 π‘π‘š2 respectively.

Q.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 π‘π‘š, find the breadth of the rectangular park.
Sol.5) Given: The side of the square park = 60 π‘š
The length of the rectangular park = 90 π‘š
According to the question,
Area of square park = Area of rectangular park
β‡’ 𝑠𝑖𝑑𝑒 Γ— 𝑠𝑖𝑑𝑒 = π‘™π‘’π‘›π‘”π‘‘β„Ž Γ— π‘π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž
β‡’ 60 Γ— 60 = 90 Γ— π‘π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž
β‡’ breadth = 60Γ—60/90 = 40 π‘š
Thus, the breadth of the rectangular park is 40 π‘š

Q.6) A wire is in the shape of a rectangle. Its length is 40 π‘π‘š and breadth is 22 π‘π‘š. 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?
Sol.6) According to the question, Perimeter of square = Perimeter of rectangle
β‡’ 4 x side = 2 (π‘™π‘’π‘›π‘”π‘‘β„Ž + π‘π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž)
β‡’ 4 x side = 2 (40 + 22)
β‡’ 4 x side = 2 Γ— 62
β‡’ side = 2Γ—62/4 = 31 π‘π‘š
Thus, the side of the square is 31 cm.
Now, Area of rectangle = π‘™π‘’π‘›π‘”π‘‘β„Ž π‘₯ Γ— π‘π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž
= 40 Γ— 22 = 880 π‘π‘š2
And Area of square = 𝑠𝑖𝑑𝑒 Γ— 𝑠𝑖𝑑𝑒
= 31 Γ— 31 = 961 π‘π‘š2
Therefore, on comparing, the area of square is greater than that of rectangle.

Q.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.
Sol.7) Perimeter of rectangle = 130 π‘π‘š
β‡’ 2 (length + breadth) = 130 π‘π‘š
β‡’ 2 (length + 30) = 130
β‡’ length + 30 = 130/2
β‡’ length + 30 = 65
β‡’ length = 65 – 30 = 35 π‘π‘š
Now area of rectangle = length x breadth = 35 Γ— 30 = 1050 π‘π‘š2
Thus, the area of rectangle is 1050 π‘π‘š2.

Q.8) A door of length 2 π‘š and breadth 1 π‘š is fitted in a wall. The length of the wall is 4.5 π‘š and the breadth is 3.6 π‘š. Find the cost of white washing the wall, if the rate of white washing the wall is 𝑅𝑠. 20 π‘π‘’π‘Ÿ π‘š2
Sol.8) Area of rectangular door = length x breadth
= 2 π‘š Γ— 1 π‘š = 2 π‘š2
Area of wall including door = length x breadth
= 4.5 π‘š Γ— 3.6 π‘š = 16.2 π‘š2

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area

Now,
Area of wall excluding door = Area of wall including door – Area of door
= 16.2 – 2 = 14.2 π‘š2
Since, The rate of white washing of 1 π‘š2 the wall = 𝑅𝑠. 20
Therefore, the rate of white washing of 14.2 π‘š2 the wall
= 20 Γ— 14.2 = 𝑅𝑠. 284
Thus, the cost of white washing the wall excluding the door is 𝑅𝑠. 284.

Exercise 11.2

Q.1) Find the area of each of the following parallelogram:

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-1

Sol.1) We know that the area of parallelogram = base x height
(a) Here base = 7 cm and height = 4 cm
∴ Area of parallelogram = 7 x 4 = 28 cm2
(b) Here base = 5 cm and height = 3 cm
∴ Area of parallelogram = 5 x 3 = 15 cm2
(c) Here base = 2.5 cm and height = 3.5 cm
∴ Area of parallelogram = 2.5 x 3.5 = 8.75 cm2
(d) Here base = 5 cm and height = 4.8 cm
∴ Area of parallelogram = 5 x 4.8 = 24 cm2
(e) Here base = 2 cm and height = 4.4 cm
∴ Area of parallelogram = 2 x 4.4 = 8.8 cm2

Q.2) Find the area of each of the following triangles :
Sol.2) We know that the area of triangle = 1/2 Γ— π‘π‘Žπ‘ π‘’ Γ— β„Žπ‘’π‘–π‘”β„Žπ‘‘
(a) Here, base = 4 cm and height = 3 cm
∴ Area of triangle = 1/2 Γ— 4 Γ— 6 = 6π‘π‘š2
(b) Here, base = 5 cm and height = 3.2 cm
∴ Area of triangle = 1/2 Γ— 5 Γ— 3.2 = 8π‘π‘š2
(c) Here, base = 3 cm and height = 4 cm
∴ Area of triangle = 1/2 Γ— 3 Γ— 4 = 6π‘π‘š2
(d) Here, base = 3 cm and height = 2 cm
∴ Area of triangle = 1/2 Γ— 3 Γ— 2 = 3π‘π‘š2

Q.3) Find the missing values:

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-2

Sol.3) We know that the area of parallelogram = base x height
(a) Here, base = 20 cm and area = 246π‘π‘š2
∴ Area of parallelogram = base x height
β‡’ 246 = 20 Γ— β„Žπ‘’π‘–π‘”β„Žπ‘‘
β‡’ height = 246/20 = 12.3π‘π‘š
(b) Here, height = 15 cm and area = 154.5 π‘π‘š2
∴ Area of parallelogram = base x height
β‡’ 154.5 = base x 15
β‡’ base = 154.5/15 = 10.3 π‘π‘š
(c) Here, height = 8.4 cm and area = 48.72 π‘π‘š2
∴ Area of parallelogram = base x height
β‡’ 48.72 = π‘π‘Žπ‘ π‘’ Γ— 8.4
β‡’ base = 48.72/8.4 = 5.8 π‘π‘š
(d) Here, base = 15.6 cm and area = 16.38 π‘π‘š2
∴ Area of parallelogram = base x height
β‡’ 16.38 = 15.6 x height
β‡’ height = 16.38/15.6 = 1.05π‘π‘š

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-3

Sol.4) We know that the area of triangle = 1/2 Γ— π‘π‘Žπ‘ π‘’ Γ— β„Žπ‘’π‘–π‘”β„Žπ‘‘
In first row, base = 15 cm and area = 87π‘π‘š2

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-4

Q.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. Find:
(a) the area of the parallelogram PRS (b) QN, if PS = 8 cm
Sol.5) Given: 𝑆𝑅 = 12 π‘π‘š, 𝑄𝑀 = 7.6 π‘π‘š, 𝑃𝑆 = 8 π‘π‘š.
(a) Area of parallelogram = base x height
= 12 Γ— 7.6 = 91.2 π‘π‘š2
(b) Area of parallelogram = base x height
β‡’ 91.2 = 8 Γ— 𝑄𝑁
β‡’ 𝑄𝑁 = 91.2/8 = 11.4 π‘π‘š

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-5

Q.6) 𝐷𝐿 and 𝐡𝑀 are the heights on sides 𝐴𝐡 and 𝐴𝐷 respectively of parallelogram ABCD (Fig 11.24). If the area of the parallelogram is 1470π‘π‘š2, 𝐴𝐡 = 35 π‘π‘š and 𝐴𝐷 = 49 π‘π‘š, find the length of 𝐡𝑀 and 𝐷𝐿.
Sol.6) Given: Area of parallelogram = 1470π‘π‘š2 Base (AB) = 35 cm
and π‘π‘Žπ‘ π‘’ (𝐴𝐷) = 49 π‘π‘š
Since Area of parallelogram = base x height
β‡’ 1470 = 35 x DL
β‡’ DL = 1470/35
β‡’ 𝐷𝐿 = 42 π‘π‘š
Again, Area of parallelogram = base x height
β‡’ 1470 = 49 x BM
β‡’ 𝐡𝑀 = 1470/49
β‡’ 𝐡𝑀 = 30 π‘π‘š

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-6
Thus, the lengths of DL and BM are 42 cm and 30 cm respectively.

Q.7) Ξ” 𝐴𝐡𝐢 is right angled at 𝐴 (Fig 11.25). AD is perpendicular to BC. If 𝐴𝐡 = 5 π‘π‘š, 𝐡𝐢 = 13 π‘π‘š and 𝐴𝐢 = 12 π‘π‘š, find the area of Ξ” 𝐴𝐡𝐢. Also, find the length of 𝐴𝐷.
Sol.7) In right angles triangle BAC,
AB = 5 cm and AC = 12 π‘π‘š
Area of triangle = 1/2 Γ— π‘π‘Žπ‘ π‘’ Γ— β„Žπ‘’π‘–π‘”β„Žπ‘‘ 
= 1/2 Γ— 𝐴𝐡 Γ— 𝐴𝐢 = 1/2 Γ— 5 Γ— 12 = 30π‘π‘š2
Now, in Ξ” 𝐴𝐡𝐢,

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-7

Area of triangle ABC = 1/2 Γ— 𝐡𝐢 Γ— 𝐴𝐷
β‡’ 30 = 1/2 Γ— 13 Γ— 𝐴𝐷
β‡’ 𝐴𝐷 = 30 Γ— 2/13
= 60/13 π‘π‘š

Q.8) Ξ” 𝐴𝐡𝐢 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 Ξ” 𝐴𝐡𝐢. What will be the height from C to AB i.e., CE?
Sol.8) In Ξ” ABC, AD = 6 cm and BC = 9 cm
Area of triangle = 1/2 Γ— π‘π‘Žπ‘ π‘’ Γ— β„Žπ‘’π‘–π‘”β„Žπ‘‘
= 1 2 Γ— 𝐡𝐢 Γ— 𝐴𝐷 = 1/2 Γ— 9 Γ— 6 = 27 π‘π‘š2
Again, Area of triangle = 1/2 Γ— π‘π‘Žπ‘ π‘’ Γ— β„Žπ‘’π‘–π‘”β„Žπ‘‘ = 1/2 Γ— 𝐴𝐡 Γ— 𝐢𝐸
β‡’ 27 = 1/2 Γ— 7.5 Γ— 𝐢𝐸
β‡’ 𝐢𝐸 = 27 Γ— 2/7.5
β‡’ 𝐢𝐸 = 7.2 π‘π‘š
Thus, height from 𝐢 to 𝐴𝐡 i.e., 𝐢𝐸 is 7.2 π‘π‘š.

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-21

Exercise 11.3

Q.1) Find the circumference of the circles with the following radius: (π‘‘π‘Žπ‘˜π‘’ πœ‹ = 22/7)
(a) 14 cm (b) 28 mm (c) 21 cm
Sol.1) (a) Circumference of the circle = 2Ο€π‘Ÿ = 2 Γ— 22/7 Γ— 14 = 88 π‘π‘š
(b) Circumference of the circle = 2Ο€π‘Ÿ = 2 Γ— 22/7 Γ— 28 = 176 π‘šπ‘š
(c) Circumference of the circle = 2Ο€π‘Ÿ = 2 Γ— 22/7 Γ— 21 = 132 π‘π‘š

Q.2) Find the area of the following circles, given that:
(a) radius = 14 mm (b) diameter = 49 m (c) radius 5 cm
Sol.2) (a) Area of circle = Ο€π‘Ÿ2 = 22/7 Γ— 14 Γ— 14
= 22 Γ— 2 Γ— 14 = 616 π‘šπ‘š2
(b) Diameter = 49 π‘š 
∴ radius = 49/2 = 24.5 π‘š
∴ Area of circle = Ο€π‘Ÿ2 = 22/7 Γ— 24.5 Γ— 24.5
= 22 Γ— 3.5 Γ— 24.5 = 1886.5 π‘š2
(c) Area of circle = Ο€π‘Ÿ2 = 22/7 Γ— 5 Γ— 5 = 550/7 π‘π‘š2

Q.3) If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet.
Sol.3) Circumference of the circular sheet = 154 m
β‡’ 2Ο€π‘Ÿ = 154 π‘š
β‡’ π‘Ÿ = 154/2Ο€
β‡’ π‘Ÿ = 154Γ—7/2Γ—22 = 24.5π‘š
Now Area of circular sheet = Ο€π‘Ÿ2 = 22/7 Γ— 24.5 Γ— 24.5
= 22 Γ— 3.5 Γ— 24.5 = 1886.5 π‘š2
Thus, the radius and area of circular sheet are 24.5 π‘š and 1886.5 π‘š2 respectively.

Q.4) A gardener wants to fence a circular garden of diameter 21 π‘š. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also, find the costs of the rope, if it cost Rs.4 per meter.
Sol.4) Diameter of the circular garden = 21 π‘š
∴ Radius of the circular garden = 21/2 π‘š
Now Circumference of circular garden = 2Ο€π‘Ÿ = 2 Γ— 22/7 Γ— 21/2 = 22 Γ— 3 = 66 π‘š
The gardener makes 2 rounds of fence so the total length of the rope of fencing = 2 Γ—
2Ο€π‘Ÿ = 2 Γ— 66 = 132 π‘š
Since, the cost of 1-meter rope = 𝑅𝑠. 4
Therefore, cost of 132-meter rope = 4 Γ— 132 = 𝑅𝑠. 528

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-22

Q.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 =3.14 Ο€ )
Sol.5) Radius of circular sheet (R) = 4 cm and radius of removed circle (r) = 3 cm
Area of remaining sheet = Area of circular sheet – Area of removed circle
= π𝑅2 βˆ’ Ο€ π‘…2 = Ο€( 𝑅2 - π‘…2)
= Ο€(42 βˆ’ 32) = Ο€ (16 βˆ’ 9)
= 3.14 Γ— 7 = 21.98 π‘π‘š2
Thus, the area of remaining sheet is21.98 π‘π‘š2

Q.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.
(π‘‡π‘Žπ‘˜π‘’ = 3.14 Ο€ )
Sol.6) Diameter of the circular table cover = 1.5 m
∴ Radius of the circular table cover = 1.5 2 m
Circumference of circular table cover = 2Ο€π‘Ÿ
= 2 Γ— 3.14 Γ— 1.5/2
Therefore, the length of required lace is 4.71 π‘š.
Now the cost of 1 π‘š π‘™π‘Žπ‘π‘’ = 𝑅𝑠. 15
Then the cost of 4.71 m lace = 15 Γ— 4.71 = 𝑅𝑠. 70.65
Hence, the cost of 4.71 π‘š lace is 𝑅𝑠. 70.65.

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

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-23

Sol.7) Diameter = 10 π‘π‘š
∴ Radius = 10/2 = 5π‘π‘š
According to question,
Perimeter of figure = Circumference of semi-circle + diameter = Ο€π‘Ÿ + 𝐷
= 22/7 Γ— 5 + 10 = 110/7 + 10
= 110+70/7 = 180/7 = 25.71π‘π‘š
Thus, the perimeter of the given figure is 25.71 π‘π‘š.

Q.8) Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is π‘…𝑠. 15/π‘š2. (π‘‡π‘Žπ‘˜π‘’ = 3.14 Ο€ )
Sol.8) Diameter of the circular table top = 1.6 m
∴ Radius of the circular table top = 1.6/2 = 0.8 π‘š
Area of circular table top = Ο€π‘Ÿ2
= 3.14 Γ— 0.8 Γ— 0.8 = 2.0096 π‘š2
Now cost of 1π‘š2 polishing = Rs.15
Then cost of 2.0096 π‘š2 polishing = 15 Γ— 2.0096
= 𝑅𝑠. 30.14 (π‘Žπ‘π‘π‘Ÿπ‘œπ‘₯. )
Thus, the cost of polishing a circular table top is 𝑅𝑠. 30.14 (π‘Žπ‘π‘π‘Ÿπ‘œπ‘₯.)

Q.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, the circle or the square?
Sol.9) Total length of the wire = 44 cm
∴ the circumference of the circle = 2Ο€π‘Ÿ = 44 π‘π‘š
β‡’ 2 Γ— 22/7 Γ— π‘Ÿ = 44
β‡’ π‘Ÿ = 44 Γ— 7/2 Γ— 22 = 7π‘π‘š
Now Area of the circle = Ο€ π‘Ÿ2 = 22/7 Γ— 7 Γ— 7 = 154 π‘π‘š2
Now the wire is converted into square.
Then perimeter of square = 44 cm
β‡’ 4 x side = 44
β‡’ side = 44/4 = 11π‘π‘š
Now area of square = 𝑠𝑖𝑑𝑒 Γ— 𝑠𝑖𝑑𝑒 = 11 Γ— 11 = 121 π‘π‘š2
Therefore, on comparing,
the area of circle is greater than that of square, so the circle enclosed more area.

Q.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 1 cm are removed. (as shown in the adjoining figure).
Find the area of the remaining sheet.

Sol.10) Radius of circular sheet (R) = 14 cm and
Radius of smaller circle (r) = 3.5 cm
Length of rectangle (l) = 3 cm and
breadth of rectangle (b) = 1 cm
According to question,
Area of remaining sheet=Area of circular sheet– (Area of two smaller circle + Area of rectangle)
= πœ‹π‘…2 βˆ’ [2(πœ‹π‘Ÿ2) + (𝑙 Γ— 𝑏)]
= 22/7 Γ— 14 Γ— 14 βˆ’ [(2 Γ— 22/7 Γ— 3.5 Γ— 3.5) βˆ’ (3 Γ— 1)]
= 22 Γ— 14 Γ— 2 βˆ’ [44 Γ— 0.5 Γ— 3.5 + 3]
= 616 βˆ’ 80
= 536 π‘π‘š2
Therefore the area of remaining sheet is 536π‘π‘š2.

Q.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 = 3.14 Ο€ )
Sol.11) Radius of circle = 2 cm and
side of aluminium square sheet = 6 π‘π‘š
According to question,
Area of aluminium sheet left = Total area of aluminium sheet – Area of circle
= 𝑠𝑖𝑑𝑒 Γ— 𝑠𝑖𝑑𝑒 βˆ’ Ο€π‘Ÿ2
= 6 Γ— 6 – 22/7 Γ— 2 Γ— 2
= 36 – 12.56 = 23.44 π‘π‘š2
Therefore, the area of aluminium sheet left is 23.44 π‘π‘š2

Q.12) The circumference of a circle is 31.4 cm. Find the radius and the area of the circle. (Take =3.14 Ο€ )
Sol.12) The circumference of the circle = 31.4 cm
β‡’ 2Ο€π‘Ÿ = 31.4
β‡’ 2 Γ— 3.14 Γ— π‘Ÿ = 31.4
β‡’ π‘Ÿ = 31.4/2Γ—3.14 = 5 π‘π‘š
Then area of the circle = Ο€π‘Ÿ2 = 3.14 Γ— 5 Γ— 5 = 78.5 π‘π‘š2
Therefore, the radius and the area of the circle are 5 π‘π‘š and 78.5 π‘π‘š2 respectively.

Q.13) A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66π‘š. What is the area of this path? (Take = 3.14 Ο€ )

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-19

Sol.13) Diameter of the circular flower bed = 66 π‘š
∴ Radius of circular flower bed (π‘Ÿ) = 66/2 = 33π‘š
∴ Radius of circular flower bed with 4 m wide path (𝑅) = 33 + 4 = 37π‘š
According to the question,
Area of path = Area of bigger circle – Area of smaller circle
= π𝑅2 βˆ’ Ο€π‘Ÿ2 = Ο€(𝑅2 - 𝑅2)
= Ο€[((37)2 βˆ’ (33)2)
= 3.14[(37 + 33)(37 βˆ’ 33)]
= 3.14 Γ— 70 Γ— 4 = 879.20 π‘š2
Therefore, the area of the path is 879.20 π‘š2

Q.14) A circular flower garden has an area of 314π‘š2. A sprinkler at the centre of the garden can cover an area that has a radius of 12 π‘š. Will the sprinkler water the entire garden? (Take = 3.14 Ο€ )
Sol.14) Circular area by the sprinkler = Ο€π‘Ÿ2
= 3.14 Γ— 12 Γ— 12
= 3.14 Γ— 144
= 452.16 π‘š2
Area of the circular flower garden = 314 π‘š2
Since Area of circular flower garden is smaller than area by sprinkler.
Therefore, the sprinkler will water the entire garden.

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

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-17

Sol.15) Radius of outer circle (π‘Ÿ) = 19 π‘š
∴ Circumference of outer circle = 2Ο€π‘Ÿ = 2 Γ— 3.14 Γ— 19 = 119.32 π‘š
Now radius of inner circle (π‘Ÿβ€²) = 19 – 10 = 9 π‘š
∴ Circumference of inner circle = 2 β€² Ο€π‘Ÿ = 2 Γ— 3.14 Γ— 9 = 56.52 π‘š
Therefore, the circumferences of inner and outer circles are 56.52 π‘š and 119.32 π‘š
respectively.

Q.16) How many times a wheel of radius 28 cm must rotate to go 352 m?
Sol.16) Let wheel must be rotate 𝑛 times of its circumference.
Radius of wheel = 28 π‘π‘š and Total distance = 352 π‘š = 35200 π‘π‘š
∴ Distance covered by wheel = n x circumference of wheel
β‡’ 35200 = 𝑛 Γ— 2Ο€π‘Ÿ
β‡’ 35200 = 𝑛 Γ— 2 Γ— 22/7 Γ— 28
β‡’ π‘› = 35200 Γ— 7/2 Γ— 22Γ—28
β‡’ 𝑛 = 200 revolutions
Thus, wheel must rotate 200 times to go 352 π‘š

Q.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? (π‘‡π‘Žπ‘˜π‘’ = 3.14 Ο€)
Sol.17) In 1 hour, minute hand completes one round means makes a circle.
Radius of the circle (π‘Ÿ) = 15 π‘π‘š
Circumference of circular clock = 2Ο€π‘Ÿ
= 2 Γ— 3.14 Γ— 15 = 94.2 π‘π‘š
Therefore, the tip of the minute hand moves 94.2 π‘π‘š in 1 β„Žπ‘œπ‘’π‘Ÿ.

Exercise 11.4

Q.1) A garden is 90 π‘š long and 75 π‘š broad. A path 5 π‘š wide is to be built outside and around it.
Find the area of the path. Also find the area of the garden in hectares.
Sol.1) Length of rectangular garden = 90 π‘š
and breadth of rectangular garden = 75 π‘š

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-18

Outer length of rectangular garden with path = 90 + 5 + 5 = 100 π‘š
Outer breadth of rectangular garden with path = 75 + 5 + 5 = 85 π‘š
Outer area of rectangular garden with path = length x breadth
= 100 Γ— 85 = 8,500 π‘š2
Inner area of garden without path = length x breadth
= 90 Γ— 75 = 6,750 π‘š2
Now, Area of path = Area of garden with path – Area of garden without path
= 8,500 – 6,750 = 1,750π‘š2
Since, 1π‘š2 = (1/10000)β„Žπ‘’π‘π‘‘π‘Žπ‘Ÿπ‘’π‘ 
Therefore, 6,750 π‘š2 = 6750/10000 = 0.675 β„Žπ‘’π‘π‘‘π‘Žπ‘Ÿπ‘’π‘ .

Q.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.

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-16

Sol.2) Length of rectangular park = 125 π‘š,
Breadth of rectangular park = 65 π‘š
and Width of the path = 3 π‘š
Length of rectangular park with path = 125 + 3 + 3 = 131 π‘š
Breadth of rectangular park with path = 65 + 3 + 3 = 71 π‘š
∴ Area of path = Area of park with path – Area of park without path
= (𝐴𝐡 Γ— 𝐴𝐷) – (𝐸𝐹 Γ— 𝐸𝐻)
= (131 Γ— 71) – (125 Γ— 65)
= 9301 – 8125 = 1,176 π‘š2
Thus, area of path around the park is 1,176 π‘š2.

Q.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.

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-15

Sol.3) Length of painted cardboard = 8 cm and breadth of painted card = 5 cm
Since, there is a margin of 1.5 cm long from each of its side.
Therefore reduced length = 8 – (1.5 + 1.5) = 8 – 3 = 5 π‘π‘š
And reduced breadth = 5 – (1.5 + 1.5) = 5 – 3 = 2 π‘π‘š
∴ Area of margin = Area of cardboard (ABCD) – Area of cardboard (EFGH)
= (𝐴𝐡 Γ— 𝐴𝐷) – (𝐸𝐹 Γ— 𝐸𝐻)
= (8 Γ— 5) – (5 Γ— 2)
= 40 – 10 = 30 π‘π‘š2
Thus, the total area of margin is 30 π‘π‘š2.

Q.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.
(ii) the cost of cementing the floor of the verandah at the rate of 𝑅𝑠. 200 π‘π‘’π‘Ÿ π‘š2

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-14

Sol.4) (i) The length of room = 5.5 m and width of the room = 4 m
The length of room with verandah = 5.5 + 2.25 + 2.25 = 10 π‘š
The width of room with verandah = 4 + 2.25 + 2.25 = 8.5 π‘š
Area of verandah = Area of room with verandah – Area of room without verandah = Area of
ABCD – Area of EFGH
= (𝐴𝐡 Γ— 𝐴𝐷) – (𝐸𝐹 Γ— 𝐸𝐻)
= (10 Γ— 8.5) – (5.5 Γ— 4)
= 85 – 22 = 63 π‘š2
(ii) The cost of cementing 1π‘š2 the floor of verandah = 𝑅𝑠. 200
The cost of cementing 63π‘š2 the floor of verandah = 200 Γ— 63 = 𝑅𝑠. 12,600

Q.5) A path 1 π‘š wide is built along the border and inside a square garden of side 30 π‘š. Find:
(i) the area of the path.
(ii) the cost of planting grass in the remaining portion of the garden at the rate of π‘…𝑠. 40 π‘π‘’π‘Ÿ π‘š2

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-13

Sol.5) (i) Side of the square garden = 30 m
and Width of the path along the border = 1 π‘š
Side of square garden without path = 30 – (1 + 1) = 30 – 2 = 28 π‘š
Now Area of path = Area of ABCD – Area of EFGH
= (𝐴𝐡 Γ— 𝐴𝐷) – (𝐸𝐹 Γ— 𝐸𝐻)
= (30 Γ— 30) – (28 Γ— 28)
= 900 – 784 = 116 π‘š2
(ii) Area of remaining portion = 28 Γ— 28 = 784 π‘š2
The cost of planting grass in 1π‘š2 of the garden = 𝑅𝑠. 40
The cost of planting grass in 784 π‘š2 of the garden = 𝑅𝑠. 40 Γ— 784 = 𝑅𝑠. 31,360

Q.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.
Sol.6) Here, 𝑃𝑄 = 10 π‘š and 𝑃𝑆 = 300 π‘š, 𝐸𝐻 = 10 π‘š and
𝐸𝐹 = 700 π‘š and 𝐾𝐿 = 10 π‘š and 𝐾𝑁 = 10 π‘š
Area of roads = Area of PQRS + Area of EFGH – Area of KLMN
[ KLMN is taken twice, which is to be subtracted]
= 𝑃𝑆 Γ— 𝑃𝑄 + 𝐸𝐹 Γ— 𝐸𝐻 – 𝐾𝐿 Γ— 𝐾𝑁
= (300 Γ— 10) + (700 Γ— 10)– (10 Γ— 10)
= 3000 + 7000 – 100 = 9,900 π‘š2
Area of road in hectares = 1π‘š2 = (1/10000)β„Žπ‘’π‘π‘‘π‘Žπ‘Ÿπ‘’π‘ 
∴ 9,900 π‘š2 = 9900/10000 = 0.99 β„Žπ‘’π‘π‘‘π‘Žπ‘Ÿπ‘’π‘ 

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-12

(ii) Now, Area of park excluding cross roads = Area of park – Area of road
= (𝐴𝐡 Γ— 𝐴𝐷) – 9,900
= (700 Γ— 300) – 9,900
= 2,10,000 – 9,900 = 2,00,100 π‘š2
= (200100/10000)β„Žπ‘’π‘π‘‘π‘Žπ‘Ÿπ‘’π‘  = 20.01 β„Žπ‘’π‘π‘‘π‘Žπ‘Ÿπ‘’π‘ 

Q.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 π‘π‘’π‘Ÿ π‘š2.
Sol.7) Here, 𝑃𝑄 = 3 π‘š and 𝑃𝑆 = 60 π‘š, 𝐸𝐻 = 3 π‘š and
𝐸𝐹 = 9 and 𝐾𝐿 = 3 π‘š and 𝐾𝑁 = 3 π‘š
Area of roads = Area of PQRS + Area of EFGH – Area of KLMN
[ KLMN is taken twice, which is to be subtracted]
= 𝑃𝑆 Γ— 𝑃𝑄 + 𝐸𝐹 Γ— 𝐸𝐻 – 𝐾𝐿 Γ— 𝐾𝑁
= (60 Γ— 3) + (90 Γ— 3)– (3 Γ— 3)
= 180 + 270 βˆ’ 9 = 441 π‘š2

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-11

(ii) The cost of 1π‘š2 constructing the roads = 𝑅𝑠. 110
The cost of 441 π‘š2 constructing the roads = 𝑅𝑠. 110 Γ— 441
= 𝑅𝑠. 48,510
Therefore, the cost of constructing the roads = 𝑅𝑠. 48,510

Q.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? (Take = 3.14 Ο€)

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-10

Sol.8) Radius of pipe = 4 π‘π‘š
Wrapping cord around circular pipe = 2Ο€r
= 2 Γ— 3.14 Γ— 4 = 25.12 π‘π‘š
Again, wrapping cord around a square = 4 x side
= 4 x 4 = 16 cm
Remaining cord = Cord wrapped on pipe – Cord wrapped on square
= 25.12 – 16 = 9.12 cm
Thus, she has left 9.12 cm cord.

Q.9) 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.

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-9

Sol.9) Length of rectangular lawn = 10 m, breadth of the rectangular lawn = 5 m and radius of the circular flower bed = 2 π‘š
(i) Area of the whole land = length x breadth = 10 Γ— 5 = 50π‘š2
(ii) Area of flower bed = Ο€r2 = 3.14 Γ— 2 Γ— 2 = 12.56 π‘š2
(iii) Area of lawn excluding the area of the flower bed = area of lawn – area of flower bed
= 50 – 12.56 = 37.44 π‘š2
(iv) The circumference of the flower bed = 2Ο€r
= 2 Γ— 3.14 Γ— 2 = 12.56 π‘š

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

""NCERT-Solutions-Class-7-Mathematics-Perimeter-and-Area-8

Sol.10) (i) Here, 𝐴𝐡 = 18 π‘π‘š, 𝐡𝐢 = 10 π‘π‘š, 𝐴𝐹 = 6 π‘π‘š, 𝐴𝐸 = 10 π‘π‘š and 𝐡𝐸 = 8 π‘π‘š
Area of shaded portion = Area of rectangle ABCD – (Area of Ξ” 𝐹𝐴𝐸 + area of Ξ” 𝐸𝐡𝐢)
= (𝐴𝐡 Γ— 𝐡𝐢) – (1/2 Γ— 𝐴𝐸 Γ— 𝐴𝐹 + 1/2 Γ— 𝐡𝐸 Γ— 𝐡𝐢)
= (18 Γ— 10) – (1/2 Γ— 10 Γ— 6 + 1/2 Γ— 8 Γ— 10).
= 180 – (30 + 40)
= 180 – 70 = 110 π‘π‘š2
(ii) Here, 𝑆𝑅 = π‘†π‘ˆ + π‘ˆπ‘… = 10 + 10 = 20 π‘π‘š, 𝑄𝑅 = 20 π‘π‘š
𝑃𝑄 = 𝑆𝑅 = 20 π‘π‘š, 𝑃𝑇 = 𝑃𝑆 – 𝑇𝑆 = 20 – 10 π‘π‘š, 𝑇𝑆 = 10 π‘π‘š, π‘†π‘ˆ = 10 π‘π‘š

𝑄𝑅 = 20 π‘π‘š and π‘ˆπ‘… = 10 π‘π‘š
Area of shaded region = Area of square PQRS – Area of Ξ” 𝑄𝑃𝑇 – Area of Ξ” π‘‡π‘†π‘ˆ – Area of Ξ” π‘ˆπ‘„π‘…
= (𝑆𝑅 Γ— 𝑄𝑅) βˆ’ 1/2 Γ— 𝑃𝑄 Γ— 𝑃𝑇 – 1/2 Γ— 𝑆𝑇 Γ— π‘†π‘ˆ – 1/2
= 20 Γ— 20 – 1 2 Γ— 20 Γ— 10 – 1/2 Γ— 10 Γ— 10 – 1/2 Γ— 20 Γ— 10
= 400 – 100 – 50 – 100 = 150 π‘π‘š2

Q.11) Find the area of the equilateral 𝐴𝐡𝐢𝐷. Here, 𝐴𝐢 = 22 π‘π‘š, 𝐡𝑀 = 3 π‘π‘š, 𝐷𝑁 = 3 π‘π‘š and π΅π‘€ βŠ₯ 𝐴𝐢, 𝐷𝑁 βŠ₯ 𝐴𝐢.
Sol.11) Here, 𝐴𝐢 = 22 π‘π‘š, 𝐡𝑀 = 3 π‘π‘š, 𝐷𝑁 = 3 π‘π‘š
Area of quadrilateral 𝐴𝐡𝐢𝐷𝐹 = Area of Ξ” 𝐴𝐡𝐢 + Area of Ξ” 𝐴𝐷𝐢
= 1/2 Γ— 𝐴𝐢 Γ— 𝐡𝑀 + 1/2 Γ— 𝐴𝐢 Γ— 𝐷𝑁
= 1/2 Γ— 22 Γ— 3 + 1/2 Γ— 22 Γ— 3
= 3 Γ— 11 + 3 Γ— 11
= 33 + 33 = 66 π‘π‘š2
Thus, the area of quadrilateral ABCD is π‘π‘š2

Chapter 01 Integers
NCERT Solutions Class 7 Mathematics Chapter 1 Integers
Chapter 02 Fractions and Decimals
NCERT Solutions Class 7 Mathematics Chapter 2 Fractions and Decimals
Chapter 03 Data Handling
NCERT Solutions Class 7 Mathematics Chapter 3 Data Handling
Chapter 04 Simple Equations
NCERT Solutions Class 7 Mathematics Chapter 4 Simple Equations
Chapter 05 Lines and Angles
NCERT Solutions Class 7 Mathematics Chapter 5 Lines and angles
Chapter 06 The Triangle and its Properties
NCERT Solutions Class 7 Mathematics Chapter 6 Triangle and its properties
Chapter 07 Congruence of Triangles
NCERT Solutions Class 7 Mathematics Chapter 7 Congruence of Triangle
Chapter 08 Comparing Quantities
NCERT Solutions Class 7 Mathematics Chapter 8 Comparing Quantities
Chapter 09 Rational Numbers
NCERT Solutions Class 7 Mathematics Chapter 9 Rational Numbers
Chapter 10 Practical Geometry
NCERT Solutions Class 7 Mathematics Chapter 10 Practical Geometry
Chapter 11 Perimeter and Area
NCERT Solutions Class 7 Mathematics Chapter 11 Perimeter and Area
Chapter 12 Algebraic Expressions
NCERT Solutions Class 7 Mathematics Chapter 12 Algebraic Expressions
Chapter 13 Exponents and Powers
NCERT Solutions Class 7 Mathematics Chapter 13 Exponents and Power
Chapter 14 Symmetry
NCERT Solutions Class 7 Mathematics Chapter 14 Symmetry
Chapter 15 Visualising Solid Shapes
NCERT Solutions Class 7 Mathematics Chapter 15 Visualizing Solid Shapes
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NCERT Solutions Class 7 Mathematics Chapter 11 Perimeter and Area

The above provided NCERT Solutions Class 7 Mathematics Chapter 11 Perimeter and Area is available on our website www.studiestoday.com for free download in Pdf. You can read the solutions to all questions given in your Class 7 Mathematics textbook online or you can easily download them in pdf.Β The answers to each question in Chapter 11 Perimeter and Area of Mathematics Class 7 has been designed based on the latest syllabus released for the current year. We have also provided detailed explanations for all difficult topics in Chapter 11 Perimeter and Area Class 7 chapter of Mathematics so that it can be easier for students to understand all answers.Β These solutions of Chapter 11 Perimeter and Area NCERT Questions given in your textbook for Class 7 Mathematics have been designed to help students understand the difficult topics of Mathematics in an easy manner. These will also help to build a strong foundation in the Mathematics. There is a combination of theoretical and practical questions relating to all chapters in Mathematics to check the overall learning of the students of Class 7.

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