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.
1 answer:
Answer:
(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.
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