fish tank problem

Let the width of the strip of yard around the pool be $w$ metres. Then the whole yard is a rectangle that is $11 + 2w$ metres long and $8 + 2w$ metres wide - you can see this because there's a strip of width $w$ on the left of the pool, and a strip of equal size on the right, and similarly for the top and bottom.

So what's the area of the yard + pool? How can you relate that to stuff you know?


The information that the strip of yard surrounding the pool is of uniform width is important. The following calculatins are done in meters.

See the picture below ($w$ denotes the width and $m$ is meters; please note that some of the $w$'s are rotated!):

enter image description here

Let a and b be the dimensions of the rectangular backyard. You know that $a\cdot b = 1120$.

As the strip of surrounding yard is of uniform width, you also know that

$a - 11 = b - 8$,

because $(a - 11 = 2w = b - 8)$.

Now you can isolate a:

$a = b + 3$,

thus you must solve $(b+3)\cdot b = 1120$. The positive solution to this equation is $b=32$. Thus the width of the strip is $\frac{32-8}{2}$=12. Here I have just isolated $w$ in the equation $8 + 2w = b$, see the figure.

Tags:

Area