Question

a foot path of uniform with runs All Around The Inside of a rectangular field of 50 M long and 38 M wide if the area of the path is 492 m square find its weight

Answer 1

A rectangular field , which Length = 50m and width = 38 m .
Let width of path = P
length of small rectangular field = 50 -2P
width of small rectangular field = 38 -2P
then,

area of path = area of big rectangular field - area of small rectangular field

Here, given,
area of path = 492 m²
area of rectangle = length × width

492 = 50×38 - (50-2P )(38 -2P)

492 = 1900 - 1900 +176P -4P²

492 = 176P -4P²

4P² - 176P + 492 = 0

P² - 44P + 123 = 0

P² -41P - 3P + 123 = 0

P(P -41) -3(P -41) = 0

(P -3)(P -41) = 0

P = 3 and 41

but P ≠ 41 because P can never greater then width of rectangular field
so , P = 3

hence, width of path = 3m

Answer 2

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.

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