The figure shows a square floor plan with a smaller square area that will accommodate a combination fountain and pool. The floor with the fountain pool area removed has an area of 33 Square meters and a perimeter of 36 meters. Find the dimensions of the floor and the dimensions of the square that will accommodate the fountain and pool.

  (x, y) = (7, 4) meters

Step-by-step explanation:

The area of the floor without the removal is x^2, so with the smaller square removed, it is x^2 -y^2.

The perimeter of the floor is the sum of all side lengths, so is 4x +2y.

The given dimensions tell us ...

  x^2 -y^2 = 33

  4x +2y = 36

From the latter equation, we can write an expression for y:

  y = 18 -2x

Substituting this into the first equation gives ...

  x^2 -(18 -2x)^2 = 33

  x^2 -(324 -72x +4x^2) = 33

  3x^2 -72x + 357 = 0 . . . . write in standard form

  3(x -7)(x -17) = 0 . . . . . factor

Solutions to this equation are x=7 and x=17. However, for y > 0, we must have x < 9.

  y = 18 -2(7) = 4

The floor dimension x is 7 meters; the inset dimension y is 4 meters.

m∠5 + m∠3 = m∠1   is true.

m∠5 + m∠6 = 180 is true

m∠2 +m∠3 = m∠6   is true

m∠2 +m∠3 +m∠5 = 180   is true