We can solve this problem by utilizing the properties of medians and similar triangles in a trapezoid.

Median Property: A median of a trapezoid divides the bases into segments with equal lengths. In this case, since MN is a median, we have:

PQ=MQ+QP

RS=SR+RN

Similar Triangles: Since MN is a median, it intersects diagonals PR and QS at their midpoints (X and Y respectively).

This creates two pairs of similar triangles:

Triangle MPX is similar to Triangle NQY (due to AA similarity - corresponding angles at P and Q are congruent since they are alternate interior angles of parallel lines PQ and RS, and shared angle at M/N)

Triangle MRX is similar to Triangle NSX (due to AA similarity - corresponding angles at R and S are congruent since they are alternate interior angles, and shared angle at X)

Leveraging Similarity: Since triangles MPX and NQY are similar, we have the following proportion:

NQMP=NYMX

Similarly, from triangles MRX and NSX:

NSMR=NXMX

Relating Segments: We are given that XY = 5, which translates to MX + NY = 5. Since X is the midpoint of PR, we know MX = XR and similarly, NY = YN. Substituting these into the first equation from step 3:

NQMP=NYMX=YNXR

Similarly, since X is the midpoint of QS, we know MX = XS and similarly, NX = NY. Substituting these into the second equation from step 3:

NSMR=NXMX=NYXS

Connecting to Bases: We know that RS = 28, which can be further divided using the median property:

RS=SR+RN=MR+NS+RN=MR+NS

Combining Information: We now have two equations relating the ratios of segments on the bases (MP/NQ and MR/NS) to the segments along the diagonals (XR/YN and XS/NY). We also have an expression for RS in terms of segments along the median (MR + NS).

Since we are looking for PQ, we can rewrite it using the median property:

PQ=MQ+QP=NQ+MR

Solving for PQ: To solve for PQ, we can try to eliminate terms involving segments along the diagonals (XR, YN, XS) from the equation for PQ.

One way to achieve this is to notice that the ratios XR/YN and XS/NY appear in both equations we derived in step 5. If we can express one ratio in terms of the other, we can eliminate it.

From the equation for RS:

RS=MR+NS

Dividing both sides by NQ, we get:

NQRS=NQMR+NQNS

Since PQ = NQ + MR, we can substitute:

PQ−MRRS=1 (because MR/NQ and NS/NQ come from the ratios relating segments on the bases to those on the diagonals)

Substituting RS with its value (28) and rearranging the equation:

PQ = MR + PQ−MR28

Simplifying and Solving: This equation might seem complex at first glance, but it can be simplified. We can multiply both sides by PQ - MR to get:

PQ^2 - MRP = 28

We can rewrite MRP as (PQ - RS) * RS using the expression for RS from step 6. This gives us:

PQ^2 - (PQ - 28) * 28 = 28

Expanding the equation and rearranging:

PQ^2 - 28PQ + 784 = 28

PQ^2 - 28PQ - 756 = 0

Factoring the equation:

(PQ - 42)(PQ + 18) = 0

Therefore, the possible values for PQ are 42 and -18. Since PQ represents the length of a base of a trapezoid, it cannot be negative. So, the only valid solution for PQ is:

PQ = 42