While doing her homework for a Momentum Learning class, Valencia draws two intersecting segments $AB = 10$ and $CD = 7$ on a plane. Across all possible configurations of  those two segments, determine the maximum possible area of quadrilateral $ACBD$.

To determine the maximum possible area of quadrilateral \(ACBD\) formed by two intersecting segments \(AB = 10\) and \(CD = 7\), we need to consider two cases: when the quadrilateral is convex and when it is non-convex.

1. **Convex Case:**
   - Let segment \(AB\) intersect segment \(CD\) at point \(O\).
   - Draw lines parallel to \(AB\) passing through points \(C\) and \(D\).
   - Draw lines parallel to \(CD\) passing through points \(A\) and \(B\).
   - Denote \(KLMN\) as the parallelogram defined by the intersection of these pairs of parallel lines.
   - The area of quadrilateral \(ACBD\) is half the area of parallelogram \(KLMN\):
     \[
     [ACBD] = \frac{[KLMN]}{2}
     \]
   - The area of parallelogram \(KLMN\) is given by:
     \[
     [KLMN] = KL \cdot KN \cdot \sin \angle LKN
     \]
   - Since \(KL = AB\) and \(KN = CD\), we have:
     \[
     [KLMN] = AB \cdot CD \cdot \sin w
     \]
   - Therefore, the area of quadrilateral \(ACBD\) is:
     \[
     [ACBD] = \frac{1}{2} \cdot AB \cdot CD \cdot \sin w
     \]
   - The maximum area occurs when \(\sin w = 1\), which happens when \(AB \perp CD\):
     \[
     [ACBD] = \frac{1}{2} \cdot AB \cdot CD
     \]

2. **Non-Convex Case:**
   - Assume \(AB\) intersects \(CD\) such that segment \(AB\) does not intersect segment \(CD\), and let \(A\) be the farthest point from line \(CD\).
   - Let \(M\) and \(N\) be the projections of \(A\) and \(B\) on line \(CD\), respectively.
   - Let \(O\) be the projection of \(B\) on segment \(AM\).
   - The area of quadrilateral \(ACBD\) can be expressed as the difference of two triangles:
     \[
     [ACBD] = [BCD] - [ACD]
     \]
   - The area of triangle \(BCD\) is:
     \[
     [BCD] = \frac{1}{2} \cdot CD \cdot AO
     \]
   - The area of triangle \(ACD\) is:
     \[
     [ACD] = \frac{1}{2} \cdot CD \cdot BN
     \]
   - Therefore, the area of quadrilateral \(ACBD\) is:
     \[
     [ACBD] = \frac{1}{2} \cdot CD \cdot (AO - BN)
     \]
   - Since \(AO = AB \cdot \sin \angle OAB\) and \(BN = 0\) (as \(B\) lies on \(CD\)), we have:
     \[
     [ACBD] = \frac{1}{2} \cdot CD \cdot AB \cdot \sin w
     \]
   - The maximum area occurs when \(\sin w = 1\), which happens when \(AB \perp CD\):
     \[
     [ACBD] = \frac{1}{2} \cdot AB \cdot CD
     \]

In both cases, the maximum area of quadrilateral \(ACBD\) is:
\[
\frac{1}{2} \cdot AB \cdot CD = \frac{1}{2} \cdot 10 \cdot 7 = 35
\]

The final answer is \(\boxed{35}\).