Question

a footpath of a uniform width runs around the inside of a rectangular field 38m long and 32 wide .if the path occupies 600 square meter,find the width​

Answer 1

Given:

✭ Length of field = 38 m

✭ Breadth of field = 32 m

✭ Area of the path = 600 m²

To Find:

◈ Width of the path

Solution:

Let width of the path be x m

Hence,

➝ Length of the inner rectangle = 38 - 2x m

➝ Breadth of the inner rectangle = 32 - 2 x m

Now we have to find the area of the field,

»» Area of the field = 38 × 32

»» Area of the field = 1216 m²

Now,

➢ Area of inner rectangle = (38 - 2x) × (32 - 2x)

➢ Area of inner rectangle = 1216 - 76x - 64x + 4x²

➢ Area of inner rectangle = 1216 - 140x + 4x²

Now,

➳ Area of the path = Area of field - Area of inner rectangle

➳ 600 = 1216 -(1216 - 140x +4x²)

➳ 600 = 1216 - 1216 + 140x - 4x²

➳ 4x² - 140x + 600 = 0

[Dividing by 4]

➳ x² - 35x + 150 = 0

➳ x²- 30x - 5x + 150 = 0

➳ x (x - 30) - 5 (x - 30) = 0

➳ x = 30, x = 5

x = 30 is not possible here since then area would become negative

Hence

➳ x = 5

Therefore the width of the path is 5 m

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Answer 2

Given:

✫Length of footpath = 38m

✫Breadth of footpath = 32m

✫Area of footpath = 600m²

Find:

✫Width of footpath

Solution:

Let, the given Rectangular field be ABCD with

⍣Length AB = 38m

⍣Width BC = 32m

Let, the width of the path be x m

➜Length of rectangle excluding path is EF = (38 - 2x)m

➜Width of rectangle excluding path is FG = (32 - 2x)m

Area of rect(ABCD) - Area of rect(EFGH) = Area of Path

$\sf \Longrightarrow 38 \times 32 - (38 - 2x)(32 - 2x) = 600$

$\sf \Longrightarrow 38 \times 32 - 38 (32 - 2x) + 2x(32 - 2x) = 600$

$\sf \Longrightarrow 38 \times 32 - 1216 + 76x + 64x - 4 {x}^{2} = 600$

$\sf \Longrightarrow 1216- 1216 + 76x + 64x - 4 {x}^{2} = 600$

$\sf \Longrightarrow 0+ 140x + 4 {x}^{2} = 600$

$\sf \Longrightarrow 4 {x}^{2} - 140x + 600 = 0$

$\sf \Longrightarrow 4( {x}^{2} - 35x + 150) = 0$

$\sf \Longrightarrow {x}^{2} - 35x + 150= \dfrac{0}{4}$

$\sf \Longrightarrow {x}^{2} - 35x + 150= 0$

Solve it by middle split term

$\sf \Longrightarrow {x}^{2} - 30x - 5x+ 150= 0$

$\sf \Longrightarrow x(x - 30) - 5(x - 30) = 0$

$\sf \Longrightarrow (x - 30)(x - 5) = 0$

$\begin{gathered} \to\sf x - 30 = 0 \\ \sf x =30m \end{gathered} \qquad \qquad\begin{gathered} \sf x - 5 = 0 \\ \sf x =5m \end{gathered}$

Rejecting x = 30m

(because it doesn't satisfy the calculation of the area of the path i.e., 600m²)

we, get x = 5m

_______________

Hence, the width of the path = 5m