Question

A foot path uniform width runs around the inside of a rectangular field 38m long and 32 m wide . If the path occupies 600 sq m , find the width .

Answer 1

Given :-

A foot path uniform width runs around the inside of a rectangular field 38 m long and 32 m wide .

The path occupies 600 sq.m.

To find :-

The measure of the width.

Solution :-

Given that

Length of a rectangular field (l) = 38 m

Breadth of the rectangular field

(b) = 32 m

We know that

Area of a rectangle = length×breadth sq.units

Area of the rectangular field = 38×32 sq.m

Therefore, Area of the field = 1216 sq.m

Let the width of the foot path be

(w) = X m

Given that

The width of the foot path runs around inside of the field .

Length of the inner field = l-2w

= 38-2(X) = 38-2X m

Breadth of the inner field = b-2w

= 32-2(X) = 32-2X m

Area of the inner field = (38-2X)×(32-2X) m²

=> Area = 38(32-2X)-2X(32-2X)

=> Area = 1216-76X-64X+4X²

=> Area = 1216-140X+4X²

Therefore,

Area of the inner field =

4X²-140X+1216 m²

Now Area of the path = Area of the outer rectangle - Area of the inner rectangle

=> 1216-(4X²-140X+1216)

=> 1216-4X²+140X-1216

=> -4X²+140X m²

Therefore, Area of the path

= -4X²+140X m²

According to the given problem

Area of the path = 600 m²

Therefore, -4X²+140X = 600

=> 600+4X²-140X = 0

=> 4X²-140X +600 = 0

=> 4(X²-35X+150) = 0

=> X²-35X+150 = 0/4

=> X²-35X+150 = 0

=> X²-5X-30X+150 = 0

=> X(X-5)-30(X-5) = 0

=> (X-5)(X-30) = 0

=> X-5 = 0 or X-30 = 0

=> X = 5 or X = 30

Therefore, width = 5 m and 30 m

Alternative Method:-

We have,

Length of the rectangular field (l) = 38 m

Breadth of the rectangular field (b)

= 32 m

Let the width of the path (w) = X m

We know that

Area of a path runs around inside the filed = (l+b-2w)2w sq.units

= (38+32-2X)2X m²

= (70-2X)(2X) m²

= 140X-4X² m²

According to the given problem

Area of the path = 600 m²

= 140X-4X² = 600

=> 600+4X²-140X = 0

=> 4X²-140X +600 = 0

=> 4(X²-35X+150) = 0

=> X²-35X+150 = 0/4

=> X²-35X+150 = 0

=> X²-5X-30X+150 = 0

=> X(X-5)-30(X-5) = 0

=> (X-5)(X-30) = 0

=> X-5 = 0 or X-30 = 0

=> X = 5 or X = 30

Therefore, width = 5 m and 30 m

Answer :-

The measure of the width of the foot path = 5 m and 30 m

Check :-

If w = 5 m then Area of the path = -4X²+140X becomes

= -4(5)²+140(5)

= -4(25)+140(5)

= -100+700

= 600 m²

and

If w = 30 m then Area of the path = -4X²+140X becomes

= -4(30)²+140(30)

= -4(900)+4200

= -3600+4200

= 600 m²

The path occupies 600 m² is true for the width of 5 m and 30 m

Verified the given relations in the given problem.

Used formulae:-

→ Area of a rectangle = length × breadth sq.units

→ Area of a path runs around inside the filed = (l+b-2w)2w sq.units

Answer 2

QUESTION :-

A foot path uniform width runs around the inside of a rectangular field 38m long and 32 m wide . If the path occupies 600 sq m , find the width .

ANSWER :-

width is 5 m and 30m

GIVEN :-

• A foot path uniform width runs around the inside of a rectangular field 38m long and 32 m wide

• If the path occupies 600 sq m

TO FIND :-

find the width = ?

SOLUTION :-

Let the width of path be x m .

Area of rectangular field = 38 × 32 = 1216 m²

Area of rectangular field without path

=(38 - 2x)(32 - 2x)

=1216 - 64x - 76x + 4x²

=1216 - 140x + 4x²

Area of the path =1216 - 1216 + 140x - 4x²

=140x - 4x²

⇒ 140x - 4x²=600

⇒ 4x²- 140x + 600 = 0

⇒ x²- 35x + 150 = 0

⇒ x²- 30x - 5x + 150 = 0

⇒ x (x-30) - 5 (x-30)=0

⇒x = 5 as x = 30

the width x = 5 as x = 30