Collins Maths Solutions Class 7 Chapter 20 Perimeter and Area
Welcome to NCTB Solution. Here with this post we are going to help 7th class students for the Solutions of Collins Maths Class 7 Mathematics, Chapter 20, Perimeter and Area. Here students can easily find chapter 7 solutions with exercise wise explanation. Students will find proper solutions for Exercise 20.1, 20.2, 20.3, 20.4 and 20.5 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily. Here in this post all the solutions are based on ICSE latest Syllabus.
Perimeter and Area Exercise 20.1 Solution :
Question no – (1)
Solution :
The area of rectangular park = 240 m²
Its breadth = 15 m
∴ length = 240/15 m
= 16 m
As we know that,
Perimeter of rectangle = 2 (length + breadth)
∴ Perimeter,
= 2 (16 + 15)
= 2 × 31 m
= 62 m
Therefore, the perimeter of the park will be 62 m,
Question no – (2)
Solution :
Square shaped garden side = 13 ft.
Cost of fencing = Rs 100 per foot.
∴ 1st perimeter of garden,
= (13 × 4) …..(Perimeter of square = (4 × side)
= 52 ft.
∴ Now total cost of fencing,
= (52 × 100)
= 5200 Rs.
Therefore, the cost of fencing the garden will be 5200 Rs.
Question no – (3)
Solution :
Area of shaded region,
= (12)² – (6)²
= 144 – 36
= 108 cm²
Therefore, the area of shaded region will be 108 cm².
Question no – (4)
Solution :
The area of the rectangle is = 22.5 cm²
Length of the rectangle is = 7.5 cm
Breadth of the rectangle is,
= 22.5/7.5 cm
= 3 cm.
∴ Perimeter of the figure,
= 2 (7.5 + 3) …..2 (length + breadth)
= 2 × 10.5
= 21 cm
Therefore, the perimeter of the given figure will be 21 cm.
Question no – (5)
Solution :
Area of field = 1200 cm²
∴ 1/2 × x × 3x/2 = 1200
= 3x²/4 = 1200
= 3x² = 4800
= x² = 4800/3
= x = 1600
= x = 40 cm
∴ Base = 40 cm
Perimeter and Area Exercise 20.2 Solution :
Question no – (1)
Solution :
(a) Perimeter, = 2 (length + Breadth)
= 2 (7 + 4)
= (2 × 11)
= 22 m
Therefore, the perimeter of the given figure will be 22 m.
(b) Perimeter, = (4 × sides)
= (4 × 9)
= 36 cm.
Hence, the perimeter of the given figure will be 36 cm.
(c) Perimeter, = 2 (length + breadth)
= 2 (14 + 4)
= 2 × 21
= 42 cm
Thus, the perimeter of the given figure will be 42 cm.
(d) Perimeter, = (4 × sides)
= (4 × 11) m
= 44 m
Therefore, the perimeter of the given figure will be 44 m.
Question no – (2)
Solution :
As we know that,
Perimeter of triangle = Side + Side + Side (Sum of 3 sides)
(a) Perimeter,
= (5 + 6 4 cm
= 15 cm.
So, the perimeter of the given figure will be 15 cm.
(b) Perimeter,
= (3 × 5) cm ∵ (three sides are same)
= 15 cm.
Thus, the perimeter of the given figure will be 15 cm.
(c) Perimeter,
= (2 × 30 + 60) ∵ (two sides are same)
= (60 + 60)
= 120 mm
Hence, the perimeter of the given figure will be 120 mm.
Question no – (3)
Solution :
(a) Parallelogram,
Length = 15 cm
Breadth = 6 cm
∴ Perimeter,
= (2 × 15 + 2 × 6)
= (30 + 12) cm
= 42 cm.
Therefore, the perimeter will be 42 cm.
(b) Rhombus,
Length = 7 cm.
∴ Perimeter,
= (4 × 7)
= 28 cm
Therefore, the perimeter will be 28 cm.
Question no – (4)
Solution :
(a) Perimeter = 85 cm
Side length = 18 cm and 36 cm
∴ Missing side length,
= 85 – (36 +18) cm
= 85 – 54 cm
= 31 cm.
Therefore, the length of the missing side will be 31 cm.
(b) Perimeter = 76 cm
One side length = 46 cm
∴ Missing both side length,
= (76 – 46) / 2
= 30/2
= 15 cm
Therefore, the length of the both missing side will be 15 cm.
Question no – (5)
Solution :
Kennel base dimensions = 10 m by 4 m
∴ Perimeter,
= 2( 10 + 4) m
= (2 × 14) m
= 28 m
Therefore, Julie will required 28 metre fencing.
Question no – (6)
Solution :
1st we need to find the perimeter of the park,
(Perimeter of square = 4 × side)
Perimeter of the square park,
= (4 × 600)
= 2400 m
In 2 round distance cover,
= (2 × 2400) m
= 4800 m
= 4.8 km.
Therefore, Vivek will cover 4.8 km.
Question no – (7)
Solution :
Given, Perimeter is = 17.28 m
Let, side = a
∴ 3a = 17.28
= 17.28/3
= a = 5.76 m
Therefore, length of the each side will be 5.76 m.
Question no – (8)
Solution :
Given, Parallelogram,
Length = 14 m
Breadth = 7 m
∴ Perimeter of parallelogram,
= (2 × length + 2 × breadth)
= (2 × 14 + 2 × 7)
= (28 + 14) m
= 42 m
Therefore, the perimeter of the parallelogram will be 42 m.
Perimeter and Area Exercise 20.3 Solution :
Question no – (1)
Solution :
(a) Given, Base = 25ft
Height = 16.25 ft.
∴ Area of the parallelogram,
= (25 × 16.25) ft²
= 406.25 ft²
Therefore, the area will be 406.25 ft²
(b) Given, Base = 20 cm
Height = 0.75 m
∴ Area of the parallelogram,
= (20 × 75) cm²
= 1500 cm²
Therefore, the are will be 1500 cm².
Question no – (2)
Solution :
Given, Perimeter of a rectangular garden = 326 sq. ft
Its length is = 101 ft.
∴ Breadth of the garden,
= 2 (length + breadth) = 326
= length + breadth = 326/2
= 163
Breadth = 163 – l
= (163 – 101) ft
= 62 ft.
Therefore, the breadth of the garden will be 62 ft.
Question no – (3)
Solution :
1st, we need to find perimeter,
Perimeter = 252/4.50
= 56 m
∴ 4 × side = 56
= Side = 56/4
= 14 m
Now, area,
= (14)² m²
= 196 m²
Therefore, the length will be 14 m and area will be 196m².
Question no – (4)
Solution :
Given, Cost of flooring = 3500
The length of the hall = 40 m
Rate of the flooring is = 3.50 Rs.
Now, Area,
= 3500/3.50
= 1000 sq. m
∴ (length × breadth) = 1000
= (40 × b) = 1000
= 40 × b = 1000
= b = 1000 ÷ 40
= 25 m
Therefore, the breadth of the hall will be 25m and area will be 1000 sq. m.
Question no – (5)
Solution :
Square area = 625 sq. m
∴ a² = 625
= a = √625
= a = 25 m
∴ Side of square = 25 m.
∴ Perimeter of square = 4 × 25
= 100 m
Now, breadth of square,
= 2 (32 + b) = 100
= 32 + b = 50
= b = 50 – 32
= b = 18 m
Therefore, the area of the square will be 25 m, and breadth will be 18 m.
Question no – (6)
Solution :
As we know that,
Area of triangle = (1/2 × b × h) × 3
∴ Area of triangle,
= (1/2 × 3 × 4) × 3
= 18 cm²
Now, Area of rectangle,
= (18 × 7) cm² ∵ (Area of rectangle = Length × Breadth)
= 126 cm²
Now, Area of shaded region,
= (126 – 18) cm²
= 108 cm²
Question no – (7)
Solution :
Area of triangle ABC = 1/2 × 3 × 7.5
= 52.5 = 7.5 b/2
= b = 52.5 × 2/7.5
= b = 14 cm
∴ h = 7.5 cm
Now, New, triangle of DBC
= 1/2 × (14 × 4.5)
= 31.5 cm²
Therefore, the area of triangle DBC will be 31.5 cm².
Question no – (9)
Solution :
Let, one side = a
And another = 2.5 a
∴ Perimeter = 70
= 2 (2.5 a + 1a) = 70
= 7a = 70
= a = 70/7
= a = 10 m
Another side = (2.5 × 10) m²
= 25 m²
Now, Area,
= (25 × 10) m²
= 250 m²
Therefore, its area will be 250 m² and measure of all its sides will be 25 m².
Question no – (10)
Solution :
The area of right angled triangle is = 84 sq. m
Its base is = 24 cm
And, its hypotenuse is = 25 cm.
∴ 1/2 × length × 24 = 84
= 24 length = 84 × 2
= length = 84 × 2/24
= 7 cm
Now, the perimeter,
= (7 + 24 + 25) cm
= 56 cm.
Therefore, the perimeter of the right angled triangle will be 56 cm.
Perimeter and Area Exercise 20.4 Solution :
Question no – (1)
Solution :
(a) Given radii is = 12 m
As we know that,
Circumference = 2πr
∴ 2 × 22/7 × 12
= 75.42 m
Therefore, the circumference will be 75.42 m.
(b) Given radii is = 19.7 cm
∴ Circumference,
= 2 × 22/7 19.7
= 123.82 cm
Therefore, circumference will be 123.82 cm.
(c) Given radii is = 22.5
∴ Circumference,
= (2 × 22/7 × 22.5) m
= 141.42 m
Therefore, circumference will be 141.42 m.
(d) Given radii is = 46.6 cm
∴ Circumference,
= (2 × 22/7 × 46.6)
= 252.91 cm
Therefore, circumference will be 252.91 cm.
Question no – (2)
Solution :
(a) Given diameter is 36 cm.
∴ r = 36/2
= 18
∴ Circumference,
= 2 × 22/7 × 18
= 113.14 cm.
Therefore, the circumference will be 113.14 cm.
(b) Given diameter is = 50 mm
∴ r = 50/2 mm
= 25 mm
∴ Circumference,
= 2 × 22/7 × 25
= 157.14 mm
Therefore, the circumference will be 157.14 mm.
(c) Given diameter is = 102 cm.
∴ r = 102/2 cm
= 51 cm
∴ Circumference,
= 2 × 22/7 × 51
= 320.57 cm.
(d) Given diameter is = 16 m
∴ r = 16 /2
= 8 m
∴ Circumference,
= 2 × 22/7 × 8
= 50.28 m
Therefore, the circumference will be 50.28 m.
Question no – (3)
Solution :
(a) Here, 2πr = 35.8 m
= r = 35.8/2 × 3.14
= 5.70 mm
∴ Diameter = (2 × 5.70) m
= 11.40 m
Therefore, the diameter will be 11.40 m.
(b) Here, 2πr = 101.7
= r = 101.7/2 × 3.14
= 16.19 cm
∴ Diameter,
= (2 × 16.19) cm
= 32.38 cm.
Therefore, the diameter will be 32.38 cm.
(c) Here, 2πr = 125.6 mm
= r = 125.6/2 × 3.14 m
= 125.6/6.28
= 20 mm
∴ Diameter,
= (2 × 20) mm
= 40 mm.
Therefore, the diameter will be 40 mm.
(d) Here, 2πr = 37.7
= r = 37.7/2 × 3.14
= 37.7/6.28
= 6 m
∴ Diameter,
= (2 × 6) m
= 12 m
Therefore, the diameter will be 12 m.
Question no – (4)
Solution :
Diameter = 4 m
∴ r = 2 m
Now, distance cover in 10 rotation,
= (10 × 2 × 22/7 × 2)
= 88/7
= 125.71 m
Therefore, the road roller will cover 125.71 metre in 10 rotation.
Question no – (5)
Solution :
Ratio of circumference,
= 2πr : 2πr²
= 2π × 5 : 2π × 8
= 5 : 8
Therefore, the ratio of their circumference will be 5 : 8.
Question no – (6)
Solution :
Area of rectangular field,
= (15 × 11) cm²
= 165 cm²
Now, circular part,
D = 8 cm
= r = 4 cm
Circumference,
= π × 4
= 4 × 22/7
= 12.57
Perimeter of rectangular field,
= 2 ( 15 + 11)
= 2 × 26
= 52 cm.
Therefore, the new perimeter of the field 52 cm.
Question no – (7)
Solution :
(a) Given radii is = 21 m
= Area = π × (21)²
= 22/7 × 21 × 21
= 1386 m²
Therefore, the area will be 1386 m²
(b) Given radii is = 32 mm²
Circumference = 2 × 22/7 × 32
= 2.1 14 mm²
Area = π × 32 × 32
= 22/7 × 32 × 32
= 3218.28 mm²
Therefore, the area will be 3218.28 mm²
(c) Given radii is = 49 cm
Circumference,
= (2 × 22/7 × 49)
= 308 cm²
Area = 22/7 × 49 × 49
= 7546 cm²
Therefore, the area will be 7546 cm²
(d) Given radii is = 63 m
Area = 22/7 × 63 × 63
= 12474 m²
Therefore, the area will be 12474 m²
Question no – (8)
Solution :
(a) Diameter = 28 m
Radii = 14 m
∴ Area = 22/7 × 14 × 14
= 616 m²
Therefore, the area will be 616 m².
(b) Diameter = 40 cm
Radii = 20 cm
Area = 22/7 × 20 × 20
= 1257.17 cm²
Therefore, the area will be 1257.17 cm².
(c) Diameter = 5 mm
Radii = 28 m
Area = 22/7 × 28 × 28
= 2464 mm²
Therefore, the area will be 2464 mm²
Question no – (9)
Solution :
(a) Area = 706.5 m²
= πr² = 706.5
= r² = 706.5/3.14
= 225
∴ r = 15 m
Radii = 15 m
∴ Diameter = (2 × 15) m
= 30 m.
(b) Area = 254.34 cm²
= πr² = 254.34
= r² = 254.34/3.14
= 81
= 9 cm
Now, diameter,
= (2 × 9) cm
= 18 cm
Therefore, the diameter will be 18 cm.
(c) Area 379.94 mm²
∴ πr²= 379.94 mm²
= r² = 379.94/3.14
= 121
∴ r = 11 mm
Now, Diameter,
= (2 × 11) mm
= 22 mm
Therefore, the diameter will be 22 mm.
Question no – (10)
Solution :
Given radius = 14 m
∴ Area = (22/7 × 14 × 14) m²
= 616 m²
1/2 sq. m cover with 1 kg fertilizer.
∴ 1 sq. m cover with 1 × 1/2
∴ 616 sq. m cover with,
= 616 × 2
= 1232 kg.
Therefore, 1232 kg fertilizer will required.
Question no – (11)
Solution :
Area of rectangle,
= (7 × 15) cm²
= 105 cm²
Small semicircle,
= 22/7 × 7 × 7/2
= 154/2
= 77 cm2
Large semicircle,
= 22/7 × 15/2 × 15/2
= 176.78
Now, total,
= (105 + 77 + 176.78) cm²
= 358.78 cm²
Therefore, the total area of the given figure will be 358.78 cm
Question no – (12)
Solution :
Given area = 39.47 cm²
= πr²/4 = 39.47
= r² = 39.47 × 4/3.14
= 157.88/3.14
= 50.28
∴ radius = 7.08 cm
Therefore, the radius of the circle will be 7.08 cm.
Perimeter and Area Exercise 20.5 Solution :
Question no – (1)
Solution :
(a) Lateral surface area,
= 4a²
= 4 × 8²
= 256 sq.cm
∴ Total surface area = 6a²
= 6 × 8²
= 384 sq. cm²
Therefore, lateral surface area will be 256 sq. cm and total surface area will be 384 sq. cm
(b) Lateral surface area,
= 4 × (12.5)² cm²
= 625 cm²
Total surface area,
= 6 × (12.5)²
= 937.5 cm²
Therefore, lateral surface area will be 625 cm² and total surface area will be 937.5 cm².
(c) Lateral surface area,
= 4 × (25)²
= 2500 m²
Total surface area,
= 6 × (25)²
= 3750 m²
Therefore, lateral surface area will be 2500 m² and total surface area will be 3750 m².
(d) Lateral surface area,
= 4 × (16.5)²
= 1089 m²
Total surface area,
= 6 × (16.5)²
= 1633.5 m²
Therefore, lateral surface area will be 1089 m² and total surface area will be 1633.5 m².
Question no – (2)
Solution :
(a) Lateral surface area of cuboid,
= 2 (l + b) × h
= 2 (6 + 3) × 2
= 2 × 9 × 2
= 36 cm²
Total surface area,
= 2 (lb + bh + hl) sq. unit
= 2 (6 × 3 + 3 × 2 + 2 × 6)
= 2 (18 + 6 + 12) cm²
= 2 × 36
= 72 cm²
(b) Lateral surface area of cuboid,
= 2 (l + b) × h
= 2 (13.5 + 10) × 2.5
= 2 × 23.5 × 2.5
= 117.5 m²
Total surface area,
= 2 (lb + bh + hl)
= 2 (13.5 ×10 + 10 × 2.5 × 13.5)
= 2 (13.5 + 25 +33.75) m²
= 2 × 193.75
= 387.5m²
(c) Lateral surface area of cuboid,
= 2 (l + b) × h
= 2 (4.5 +3.5) × 1.5
= 2 × 8 × 1.5
= 24 cm²
Total surface area,
= 2 (lb + bh + hl)
= 2 (4.5 × 3.5 + 3.5 × 1.5 × 4.5) cm²
= 2 × (15.75 + 5.25 + 6.75) cm²
= 55.5 cm²
Question no – (3)
Solution :
We know,
Total surface = Lateral surface area + Area of top + Area of base
∴ 1750 = L.S.A + 250 + 250
= L.S.A = 1750 – 500
= L.S.A = 1250 Sq. m
Therefore, the lateral surface area of the box will be 1250 sq. m.
Question no – (4)
Solution :
Material required,
= 6 × (13.5)²
= 1093.5 m²
Therefore, 1093.5 m² me7tarial will be required.
Question no – (5)
Solution :
Total surface area,
= 2 (lb + bh + hl) sq. unit
= 2 (15 × 12 + 12 × 7 + 15 ×7)
= 2 (180 + 84 + 105 cm²
= 2 × 369
= 738 m²
∴ Total cost,
= (738 × 12.50)
= 9225 Rs.
Therefore, it will cost 9225 Rs.
Question no – (6)
Solution :
Total surface area,
= 2 (lb + bh + hl) sq. unit
= 2 (50 × 35 + 35 × 16 + × 16 × 50)
= 2 (1750 + 560 + 800)
= 2 (3110) cm²
= 6220 cm²
∴ Total cost,
= (6220 × 0.10)
= 622 Rs
Therefore, the cost of the carton will be 622 Rs.
Question no – (7)
Solution :
L.S.A = 2 (15 + 6) × 3
= 2 × 21 × 3
= 126 sq. m
∴ Total cost,
= (126 × 50)
= 6300 Rs.
Therefore, the cost of the painting will be 6300 Rs.
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