Area question

Question

Length of rectanglular pool's 40 feet. Width's 24feet. A walkway of uniform width surrounds the pool. The combined area of the pool and walkway's 2240 square feet. Find the width of the walkway.

Answer 1

Hey Titaree -- pool takes up 40x24 or 960 sq.ft ... "stair-step" up to the 2240 ... 50x34 is 1700 ... 56x40 is 2240 ==> looks like a 16 ft-wide walkway ... Regards :)

Answer 2

From the question we can construct two equations:

  A) Area of Pool and Walkway: P+W=2240

            -given

  B) Area of Pool and Walkway: P+W=(40+2x)(24+2x)  

           -we are told that the walkway is of a uniform width, x.  The rectangle formed by the walkway is therefore x higher, x lower, x longer on the right, and x longer on the left than the rectangle formed by the pool.  Sketching this out makes it easier to see.

From equations A and B we get (40+2x)(24+2x)=P+W=2240

Solving this:

   (40+2x)(24+2x)=2240

   4x^2+80x+48x+960=2240

   4x^2+128x+960=2240

   4x^2+128x-1280=0

   divide the equation by four: x^2+32x-320=0

   factor: (x+40)(x-8)=0

   x=-40,8

If x=-40, the walkway would be inside the pool, which does not make sense.  So the answer is 8 ft.

Answer 3

Alexandra H. has the correct answer. Updating to reflect this.

Answer 4

A walkway that is x feet wide adds 2x feet to the combined length and 2x feet to the combined width. The combined length is 40+2x, the combined width is 24+2x. The combined ares is then combined LENGTH x WIDTH.

(2x + 40) (2x + 24) = 2240
4x^2 + 128x - 1280 = 0

Now x can be solved using the Quadratic Equation
ax^2 + bx + c = 0
Where x = { -b +/- sqrt(b^2 - 4ac) } / 2a
In this case a=4, b=128, and c=-1280

x = { -128 +/- sqrt(128^2 - 4(4)(-1280) } / 2(4)

x has a positive and negative solution, in this case the negative solution is ignored.

x = 8 feet = width of walkway.