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,
![(30 - 2x)(20 - 2x)=400](https://tex.z-dn.net/?f=%2830%20-%202x%29%2820%20-%202x%29%3D400)
![600-60x -40x + 4x^2= 400](https://tex.z-dn.net/?f=600-60x%20-40x%20%2B%204x%5E2%3D%20400)
![4x^2-100x+600-400=0](https://tex.z-dn.net/?f=4x%5E2-100x%2B600-400%3D0)
![4x^2-100x+200=0](https://tex.z-dn.net/?f=4x%5E2-100x%2B200%3D0)
![x^2-25x+50=0](https://tex.z-dn.net/?f=x%5E2-25x%2B50%3D0)
By the quadratic formula,
![x=\frac{-(-25)\pm \sqrt{(-25)^2-4\times 1\times 50}}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-%28-25%29%5Cpm%20%5Csqrt%7B%28-25%29%5E2-4%5Ctimes%201%5Ctimes%2050%7D%7D%7B2%7D)
![=\frac{25\pm \sqrt{625-200}}{2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B25%5Cpm%20%5Csqrt%7B625-200%7D%7D%7B2%7D)
![=\frac{25\pm \sqrt{425}}{2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B25%5Cpm%20%5Csqrt%7B425%7D%7D%7B2%7D)
![\implies x = \frac{25+ \sqrt{425}}{2}\text{ or }x=\frac{25- \sqrt{425}}{2}](https://tex.z-dn.net/?f=%5Cimplies%20x%20%3D%20%5Cfrac%7B25%2B%20%5Csqrt%7B425%7D%7D%7B2%7D%5Ctext%7B%20or%20%7Dx%3D%5Cfrac%7B25-%20%5Csqrt%7B425%7D%7D%7B2%7D)
⇒ 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.