A foot path of uniform width runs all around inside of a rectangular field 45m long and 36m wide If the area of the path is 234 m^find the width of the path
Answers
Find the area of the field with the path:
Area = Length x Breadth
Area = 45 x 36
Area = 1620 m²
Find the area of the field without the path:
Area of the path = 234 m²
Area of the field without the path = 1620 - 234 = 1386 m²
Define the width of the path:
Width = x m
Find the dimension of the field without the path:
Length = 45 - x - x = 45 - 2x
Width = 36 - x - x= 36 - 2x
Solve x:
Area = length x Breadth
(45 - 2x)(36 - 2x) = 1386
1620 - 90x - 72x + 4x² = 1386
4x² - 162x + 234 = 0
2x² - 81x + 117 = 0
(2x - 3)(x - 39) = 0
x = 1.5 or x = 39 (rejected, 39 will result in negative length)
Find the width of the path:
width of the path = x = 1.5 m
Answer: The width is 1.5 m
Consider ABCD as a rectangular field having
Length = 50 m
Breadth = 38 m
We know that
Area of rectangular field ABCD = l × b
Substituting the values
= 50 × 38
= 1900 m2
Let x m as the width of foot path all around the inside of a rectangular field
Length of rectangular field PQRS = (50 – x – x) = (50 – 2x) m
Breadth of rectangular field PQRS = (38 – x – x) = (38 – 2x) m
Here
Area of foot path = Area of rectangular field ABCD – Area of rectangular field PQRS
Substituting the values
492 = 1900 – (50 – 2x) (38 – 2x)
It can be written as
492 = 1900 – [50 (38 – 2x) – 2x (38 – 2x)]
By further calculation
492 = 1900 – (1900 – 100x – 76x + 4x2)
492 = 1900 – 1900 + 100x + 76x – 4x2
On further simplification
492 = 176x – 4x2
Taking out 4 as common
492 = 4 (44x – x2)
44x – x2 = 492/4 = 123
We get
x2 – 44x + 123 = 0
It can be written as
x2 – 41x – 3x + 123 = 0
Taking out the common terms
x (x – 41) – 3 (x – 41) = 0
(x – 3) (x – 41) = 0
Here
x – 3 = 0 or x – 41 = 0
So x = 3 m or x = 41 m which is not possible
Therefore, width is 3 m.