The figure to the right 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 27 square meters and a perimeter of 30 meters. Find the dimensions of the floor and the dimensions of the square that will accommodate the fountain and pool.
Each length of the large square floor is and each length of the small square is
(Simplify your answer. Type integers or decimals.)

Step 1 ：The problem is asking for the dimensions of the large square floor and the smaller square. We know that the area of the large square floor minus the area of the smaller square is 27 square meters. We also know that the perimeter of the large square floor is 30 meters.

Step 2 ：We can use these two pieces of information to set up two equations. The area of a square is given by the formula \(A = s^2\), where \(s\) is the side length of the square. The perimeter of a square is given by the formula \(P = 4s\).

Step 3 ：Let's denote the side length of the large square as \(x\) and the side length of the smaller square as \(y\). Then we have the following two equations: \(x^2 - y^2 = 27\) and \(4x = 30\).

Step 4 ：We can solve these two equations to find the values of \(x\) and \(y\). The solution to the system of equations gives two possible values for \(y\), \(-3\sqrt{13}/2\) and \(3\sqrt{13}/2\). However, since \(y\) represents a length, it cannot be negative. Therefore, the side length of the smaller square is \(3\sqrt{13}/2\). The side length of the large square is \(15/2\).

Step 5 ：Final Answer: The dimensions of the large square floor are \(\boxed{\frac{15}{2}}\) meters and the dimensions of the smaller square are \(\boxed{\frac{3\sqrt{13}}{2}}\) meters.