To solve the problem, we follow these steps:
Step 1: Set up coordinates
Place square (ABCD) with (A(0,0)), (B(1,0)), (C(1,1)), (D(0,1)).
- (E) on (AB): (E(x,0))
- (F) on (AD): (F(0,x))
Step 2: Calculate the area of the bounded figure
The area (y) of the figure bounded by (EF), (EC), (FC) is the area of the square minus the areas of three triangles ((\triangle AEF), (\triangle EBC), (\triangle FDC)):
- Area of (\triangle AEF) (right triangle with legs (x)): (\frac{1}{2}x^2)
- Area of (\triangle EBC) (right triangle with base (1-x), height (1)): (\frac{1}{2}(1-x))
- Area of (\triangle FDC) (same as (\triangle EBC)): (\frac{1}{2}(1-x))
Total area of the three triangles:
[ \frac{1}{2}x^2 + \frac{1}{2}(1-x) + \frac{1}{2}(1-x) = \frac{1}{2}x^2 - x + 1 ]
Thus:
[ y = 1 - \left(\frac{1}{2}x^2 - x + 1\right) = -\frac{1}{2}x^2 + x ]
Step 3: Determine the domain
(x) must satisfy (0 \leq x \leq 1) (since (E) and (F) lie on the sides of the square).
Final Answer:
The functional relationship is (\boxed{y = -\frac{1}{2}x^2 + x}) with domain (x \in [0,1]).
(\boxed{y = -\frac{1}{2}x^2 + x}) (for (0 \leq x \leq 1))
But since the question asks for the functional relationship, the main answer is:
(\boxed{y = -\dfrac{1}{2}x^2 + x})
(\boxed{y = -\dfrac{1}{2}x^2 + x})
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