Question

(i) A footpath of uniform width runs all around the inside of a rectangular field 54 m long and 35 m wide. If the area of the path is 420 m²., find the width of the path.
(ii) A carpet is laid on the floor of a room 8 m by 5 m. There is a border of constant width all around the carpet. If the area of the border is 12 m², find its width.

Answer 1

(i) Area of the rectangular field = $54\times 35=1890{\mathrm{m}}^2$

Let the width of the path be x m. The path is shown in the following diagram:


Length of the park excluding the path = (54 $-$ 2x) m
Breadth of the park excluding the path = (35 $-$ 2x ) m

Thus, we have:
Area of the path = 420 m²
$$\begin{gathered} \Rightarrow 420=54\times 35-(54-2x)(35-2x) \\ \Rightarrow 420=1890-\left(1890-70x-108x+4x^2\right) \\ \Rightarrow 420=-4x^2+178x \\ \Rightarrow 4x^2-178x+420=0 \\ \Rightarrow 2x^2-89x+210=0 \\ \Rightarrow 2{\mathrm{x}}^2-84\mathrm{x}-5\mathrm{x}+210=0 \\ \Rightarrow 2\mathrm{x}(\mathrm{x}-42)-5(\mathrm{x}-42)=0 \\ \Rightarrow (\mathrm{x}-42)(2x-5)=0 \\ \Rightarrow x-42=0\mathrm{or}2x-5=0 \\ \Rightarrow x=42\mathrm{or}\mathrm{x}=2.5 \end{gathered}$$

The width of the path cannot be more than the breadth of the rectangular field.
∴ x = 2.5 m

​Thus, the path is 2.5 m wide.

(ii) Let the width of the border be x m.
The length and breadth of the carpet are 8 m and 5 m, respectively.
Area of the carpet = $8\times 5=40{\mathrm{m}}^2$
Length of the carpet without border = $(8-2x)$
Breadth of carpet without border = $(5-2x)$
Area of the border = 12 m²
Area of the carpet without border = $(8-2x)(5-2x)$
$$\begin{gathered} \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ 12=40-\left[(8-2x)(5-2x)\right] \\ \Rightarrow 12=40-(40-26x+4x^2) \\ \Rightarrow 12=26x-4x^2 \end{gathered}$$

$$\begin{gathered} \Rightarrow 26x-4x^2=12 \\ \Rightarrow 4x^2-26x+12=0 \\ \Rightarrow 2x^2-13x+6=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow (2x-1)(x-6)=0 \\ \Rightarrow 2x-1=0\mathrm{and}x-6=0 \\ \Rightarrow x=\frac{1}{2}\mathrm{and}x=6 \end{gathered}$$

Because the border cannot be wider than the entire carpet, the width of the carpet is $\frac{1}{2}$ m, i.e., 50 cm.