Answer:
2.19 ft ( approx )
Step-by-step explanation:
Let x be the width ( in ft ) of the path,
Given,
The dimension of the garden area,
Length = 30 ft, width = 20 ft,
So, the dimension of the remaining garden ( garden excluded path ),
Length = (30 - 2x) ft, width = (20-2x) ft
Thus, the area of the remaining garden,
A=(30 - 2x)(20 - 2x)
According to the question,
A = 400 ft²,
[tex](30 - 2x)(20 - 2x)=400[/tex]
[tex]600-60x -40x + 4x^2= 400[/tex]
[tex]4x^2-100x+600-400=0[/tex]
[tex]4x^2-100x+200=0[/tex]
[tex]x^2-25x+50=0[/tex]
By the quadratic formula,
[tex]x=\frac{-(-25)\pm \sqrt{(-25)^2-4\times 1\times 50}}{2}[/tex]
[tex]=\frac{25\pm \sqrt{625-200}}{2}[/tex]
[tex]=\frac{25\pm \sqrt{425}}{2}[/tex]
[tex]\implies x = \frac{25+ \sqrt{425}}{2}\text{ or }x=\frac{25- \sqrt{425}}{2}[/tex]
⇒ x ≈ 22.8 or x ≈ 2.19,
∵ Width of the path can not exceed 30 ft or 20 ft
Hence, the width of the path is approximately 2.19 ft.
