Let's denote the lengths of the rectangle sides as \(AB = a\) and \(AD = b\). Also, let \(AE = x\) and \(DF = y\).

The area of a triangle can be calculated using the formula \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\).

Given that the areas of triangles \(ABE\), \(ADF\), and \(CEF\) are \(5\), \(5\), and \(10\) respectively, we can write the following equations:

\[ \frac{1}{2} \cdot AE \cdot AB = 5 \Rightarrow \frac{1}{2} \cdot x \cdot a = 5 \Rightarrow xa = 10 \quad \text{(1)} \]

\[ \frac{1}{2} \cdot DF \cdot AD = 5 \Rightarrow \frac{1}{2} \cdot y \cdot b = 5 \Rightarrow yb = 10 \quad \text{(2)} \]

\[ \frac{1}{2} \cdot CE \cdot BC = 10 \Rightarrow \frac{1}{2} \cdot (a - x) \cdot a = 10 \Rightarrow \frac{a^2}{2} - ax = 20 \quad \text{(3)} \]

Adding equations \(\text{(1)}\) and \(\text{(2)}\):

\[ xa + yb = 10 + 10 \Rightarrow xa + yb = 20 \quad \text{(4)} \]

Substitute \(yb\) from equation \(\text{(2)}\) into equation \(\text{(4)}\):

\[ xa + 10 = 20 \Rightarrow xa = 10 \]

Substitute this value of \(xa\) into equation \(\text{(3)}\):

\[ \frac{a^2}{2} - 10 = 20 \Rightarrow \frac{a^2}{2} = 30 \Rightarrow a^2 = 60 \]

Since \(a\) is a length of the rectangle, it must be positive, so we take the positive square root:

\[ a = \sqrt{60} = 2 \sqrt{15} \]

Now, substitute this value of \(a\) into equation \(\text{(1)}\):

\[ x \cdot 2 \sqrt{15} = 10 \Rightarrow x = \frac{10}{2 \sqrt{15}} = \frac{\sqrt{15}}{3} \]

Similarly, substitute \(xa\) from equation \(\text{(1)}\) into equation \(\text{(4)}\):

\[ yb + 10 = 20 \Rightarrow yb = 10 \]

Substitute this value of \(yb\) into equation \(\text{(2)}\):

\[ y \cdot b = 10 \Rightarrow y = \frac{10}{b} \]

Now we can use equation \(\text{(3)}\) to find the value of \(b\):

\[ \frac{a^2}{2} - ax = 20 \Rightarrow \frac{(2 \sqrt{15})^2}{2} - \frac{\sqrt{15}}{3} \cdot 2 \sqrt{15} = 20 \]

Simplify:

\[ 30 - 10 = 20 \Rightarrow 20 = 20 \]

This is always true, so any value of \(b\) will work. Let's take \(b = 1\) for simplicity.

Substitute \(b = 1\) into equation \(\text{(2)}\):

\[ y \cdot 1 = 10 \Rightarrow y = 10 \]

Now, the sides of the rectangle are \(a = 2 \sqrt{15}\) and \(b = 1\), so the area of the rectangle \(ABCD\) is:

\[ \text{Area} = a \cdot b = 2 \sqrt{15} \cdot 1 = 2 \sqrt{15} \]

So, the area of the rectangle \(ABCD\) is \( \boxed{2 \sqrt{15}} \).