Let $ABC$ be a triangle with side lengths $5$, $4\sqrt 2$, and $7$.  What is the area of the triangle with side lengths $\sin A$, $\sin B$, and $\sin C$?
1. **Identify the given side lengths and use the Law of Sines:**
   Given the side lengths of triangle \(ABC\) are \(a = 5\), \(b = 4\sqrt{2}\), and \(c = 7\). According to the Law of Sines, we have:
   \[
   \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R
   \]
   where \(R\) is the circumradius of the triangle.

2. **Express the sines of the angles in terms of \(R\):**
   \[
   \sin A = \frac{a}{2R} = \frac{5}{2R}, \quad \sin B = \frac{b}{2R} = \frac{4\sqrt{2}}{2R} = \frac{2\sqrt{2}}{R}, \quad \sin C = \frac{c}{2R} = \frac{7}{2R}
   \]

3. **Determine the area of the triangle with side lengths \(\sin A\), \(\sin B\), and \(\sin C\):**
   The area of a triangle with side lengths \(a\), \(b\), and \(c\) can be found using Heron's formula:
   \[
   \text{Area} = \sqrt{s(s-a)(s-b)(s-c)}
   \]
   where \(s\) is the semi-perimeter:
   \[
   s = \frac{a + b + c}{2}
   \]
   For the new triangle with side lengths \(\sin A\), \(\sin B\), and \(\sin C\), we have:
   \[
   a' = \sin A = \frac{5}{2R}, \quad b' = \sin B = \frac{2\sqrt{2}}{R}, \quad c' = \sin C = \frac{7}{2R}
   \]
   The semi-perimeter \(s'\) is:
   \[
   s' = \frac{\sin A + \sin B + \sin C}{2} = \frac{\frac{5}{2R} + \frac{2\sqrt{2}}{R} + \frac{7}{2R}}{2} = \frac{\frac{5 + 4\sqrt{2} + 7}{2R}}{2} = \frac{12 + 4\sqrt{2}}{4R} = \frac{3 + \sqrt{2}}{R}
   \]

4. **Calculate the area using Heron's formula:**
   \[
   \text{Area} = \sqrt{s'(s' - \sin A)(s' - \sin B)(s' - \sin C)}
   \]
   Substituting the values:
   \[
   s' - \sin A = \frac{3 + \sqrt{2}}{R} - \frac{5}{2R} = \frac{6 + 2\sqrt{2} - 5}{2R} = \frac{1 + 2\sqrt{2}}{2R}
   \]
   \[
   s' - \sin B = \frac{3 + \sqrt{2}}{R} - \frac{2\sqrt{2}}{R} = \frac{3 + \sqrt{2} - 2\sqrt{2}}{R} = \frac{3 - \sqrt{2}}{R}
   \]
   \[
   s' - \sin C = \frac{3 + \sqrt{2}}{R} - \frac{7}{2R} = \frac{6 + 2\sqrt{2} - 7}{2R} = \frac{-1 + 2\sqrt{2}}{2R}
   \]
   Therefore, the area is:
   \[
   \text{Area} = \sqrt{\left(\frac{3 + \sqrt{2}}{R}\right) \left(\frac{1 + 2\sqrt{2}}{2R}\right) \left(\frac{3 - \sqrt{2}}{R}\right) \left(\frac{-1 + 2\sqrt{2}}{2R}\right)}
   \]

5. **Simplify the expression:**
   \[
   \text{Area} = \frac{1}{4R^2} \sqrt{(3 + \sqrt{2})(1 + 2\sqrt{2})(3 - \sqrt{2})(-1 + 2\sqrt{2})}
   \]
   Using the identity \((a + b)(a - b) = a^2 - b^2\):
   \[
   (3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7
   \]
   \[
   (1 + 2\sqrt{2})(-1 + 2\sqrt{2}) = -1 + 4\sqrt{2} - 4 = -5 + 4\sqrt{2}
   \]
   Therefore:
   \[
   \text{Area} = \frac{1}{4R^2} \sqrt{7(-5 + 4\sqrt{2})}
   \]

6. **Calculate the circumradius \(R\):**
   Using the formula for the circumradius:
   \[
   R = \frac{abc}{4K}
   \]
   where \(K\) is the area of the triangle. Using Heron's formula for the original triangle:
   \[
   s = \frac{5 + 4\sqrt{2} + 7}{2} = \frac{12 + 4\sqrt{2}}{2} = 6 + 2\sqrt{2}
   \]
   \[
   K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(6 + 2\sqrt{2})(6 + 2\sqrt{2} - 5)(6 + 2\sqrt{2} - 4\sqrt{2})(6 + 2\sqrt{2} - 7)}
   \]
   Simplifying:
   \[
   K = \sqrt{(6 + 2\sqrt{2})(1 + 2\sqrt{2})(6 - 2\sqrt{2})(-1 + 2\sqrt{2})}
   \]
   Using the same identity:
   \[
   (6 + 2\sqrt{2})(6 - 2\sqrt{2}) = 36 - 8 = 28
   \]
   \[
   (1 + 2\sqrt{2})(-1 + 2\sqrt{2}) = -1 + 4\sqrt{2} - 4 = -5 + 4\sqrt{2}
   \]
   Therefore:
   \[
   K = \sqrt{28(-5 + 4\sqrt{2})}
   \]

7. **Final calculation:**
   \[
   R = \frac{abc}{4K} = \frac{5 \cdot 4\sqrt{2} \cdot 7}{4 \sqrt{28(-5 + 4\sqrt{2})}}
   \]
   Simplifying:
   \[
   R = \frac{140\sqrt{2}}{4 \sqrt{28(-5 + 4\sqrt{2})}}
   \]

The final answer is \( \boxed{ \frac{7}{25} } \)