Question

A rectangular swimming pool that is 10 ft wide by 16 ft long is surrounded by a cement sidewalk of uniform width. If the area of the sidewalk is 155 ft2, what is its width? A = lw 2.5 ft 5.0 ft 6.2 ft 15.5 ft

Answer 1

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,

$(16+x)(10+x)-160 = 155$

$(16+x)(10+x)=315$

$160+16x+10x+x^2=315$

$x^2+26x-155=0$

By the quadratic formula,

$x=\frac{-26\pm \sqrt{676+620}}{2}$

$x=\frac{-26\pm 36}{2}$

$\implies x=5\text{ or } x = -31$

Side can not be negative,

Hence, the width of the sidewalk is 5.0 ft.

Answer 2

The pool has area $160\,\mathrm{ft}^2$.

Let $x$ be the width of the sidewalk. Then the combined area of the pool and sidewalk is $(10+x)(16+x)=160+26x+x^2$, so that the area of the sidewalk alone is $26x+x^2$.

We're told this area is $155\,\mathrm{ft}^2$, so

$26x+x^2=155\implies x^2+26x-155=(x-5)(x+31)=0\implies x=5$