Area of the Path
Here we will discuss about area of the path. It is observed that in square or rectangular gardens, parks, etc., some space in the form of path is left inside or outside or in between as cross paths. We will apply this concept for the areas of rectangle and square to determine the areas of different paths.
Workedout examples on Area of the Path:
1. A rectangular lawn of length 50 m and breadth 35 m is to be surrounded externally by a path which is 2 m wide. Find the cost of turfing the path at the rate of $3 per m^{2}.
Solution:
Length of the lawn = 50 m Breadth of the lawn = 35 m Area of the lawn = (50 × 35) m^{2} = 1750 m^{2} Length of lawn including the path = [50 + (2 + 2)] m = 54 cm Breadth of the lawn including the path = [35 + (2 + 2)] m = 39 m Area of the lawn including the path = 54 × 39 m^{2} = 2106 m^{2} Therefore, area of the path = (2106  1750) m^{2} = 356 m^{2} For 1 m^{2}, the cost of turfing the path = $ 3 For 356 m^{2}, the cost of turfing the path = $3 × 356 = $1068
2. A painting is painted on a cardboard 19 cm and 14 cm wide, such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.
Solution:
Length of the cardboard = 19 cm Breadth of the cardboard = 14 cm Area of the cardboard = 19 × 14 cm^{2} = 266 cm^{2} Length of the painting excluding the margin = [19  (1.5 + 1.5)] cm = 16 cm Breadth of the painting excluding the margin = 14  (1.5 + 1.5) = 11 cm Area of the painting excluding the margin = (16 × 11) cm^{2} = 176 cm^{2} Therefore, area of the margin = (266  176) cm^{2} = 90 cm^{2}
3. A square flowerbed is surrounded by a path 10 cm wide around it. If the area of the path is 2000 cm^{2}, find the area of the square flowerbed.
Solution:
In the adjoining figure,
ABCD is the square flowerbed. EFGH is the outer boundary of the path. Let each side of the flowerbed = x cm Then, the area of the square flowerbed ABCD (x × x) cm^{2} = x^{2} cm^{2} Now, the side of the square EFGH = (x + 10 + 10) cm = (x + 20) cm So, the area of square EFGH = (x + 20) (x + 20) cm^{2} = (x + 20)^{2} cm^{2} Therefore, area of the path = Area of EFGH  Area of ABCD = [(x + 20)^{2}  x^{2}] cm^{2} = [x^{2} + 400 + 40x  x^{2}] cm^{2} = (40x + 400) cm^{2} But the area of path given = 2000 cm^{2} Therefore, 40x + 400 = 2000 40x = 2000  400 40x = 1600 x = 1600/40 = 40 Therefore, side of square flowerbed =40 cm Therefore, the area of the square flowerbed = 40 × 40 cm^{2} = 1600 cm^{2}
● Mensuration
Area and Perimeter Perimeter and Area of Rectangle Perimeter and Area of Square Area of the Path Area and Perimeter of the Triangle Area and Perimeter of the Parallelogram Area and Perimeter of Rhombus Area of Trapezium Circumference and Area of Circle Units of Area Conversion Practice Test on Area and Perimeter of Rectangle Practice Test on Area and Perimeter of Square
● Mensuration  WorksheetsWorksheet on Area and Perimeter of Rectangles Worksheet on Area and Perimeter of Squares Worksheet on Area of the Path Worksheet on Circumference and Area of Circle Worksheet on Area and Perimeter of Triangle
7th Grade Math Problems
8th Grade Math Practice
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