Question

A club swimming pool is 23 ft wide and 33 ft long. the club members want an exposed aggregate border in a strip of uniform width around the pool. they have enough material for 660 ft squaredft2. how wide can the strip​ be?

Answer 1

Let x = the width of the border

2x+23) by (2x + 33) are the overall dimensions of the pool and the border. Use the area to find x!

Overall area - pool area = 660 sq/ft (aggregate area)

(2x+23)*(2x+33) - (23*33) = 660

Solve.

4x^2+112x = 660

4x^2+112x−660=0 → subtract 660 on both sides

4(x−5)(x+33)=0 → factor

x−5=0 or x+33=0 → set both factors equal to 0

x = 5 or x = -33

In this case, we would use the positive solution, so the strip can be 5 ft. wide!

I hope this helps :)