The pool has area [tex]160\,\mathrm{ft}^2[/tex].
Let [tex]x[/tex] be the width of the sidewalk. Then the combined area of the pool and sidewalk is [tex](10+x)(16+x)=160+26x+x^2[/tex], so that the area of the sidewalk alone is [tex]26x+x^2[/tex].
We're told this area is [tex]155\,\mathrm{ft}^2[/tex], so
[tex]26x+x^2=155\implies x^2+26x-155=(x-5)(x+31)=0\implies x=5[/tex]
Answer:
The width of the sidewalk is 5.0 ft.
Step-by-step explanation:
Given,
The dimension of the rectangular swimming pool is 16 ft × 10 ft,
So, the area of the pool = 16 × 10 = 160 ft²,
Let x be the uniform width of the cement sidewalk,
So, the dimension of the area covered by both swimming pool and sidewalk = (16+x) ft × (10+x) ft,
Thus, the combined area of the swimming pool and sidewalk = (16+x)(10+x) ft²
Also, the area of the sidewalk = The combined area - Area of the pool,
= (16+x)(10+x) - 160
According to the question,
[tex](16+x)(10+x)-160 = 155[/tex]
[tex](16+x)(10+x)=315[/tex]
[tex]160+16x+10x+x^2=315[/tex]
[tex]x^2+26x-155=0[/tex]
By the quadratic formula,
[tex]x=\frac{-26\pm \sqrt{676+620}}{2}[/tex]
[tex]x=\frac{-26\pm 36}{2}[/tex]
[tex]\implies x=5\text{ or } x = -31[/tex]
Side can not be negative,
Hence, the width of the sidewalk is 5.0 ft.
