Question 1209518
Here's how to solve this problem:

**1. Area of Rhombus:**

Let the side length of the rhombus be 's'.  The area of a rhombus can be calculated using its diagonals, but since we're dealing with midpoints and ratios, a simpler approach is to consider it as four congruent triangles.  We'll denote the area of rhombus PQRS as [PQRS].

**2. Area of Triangles:**

* **Triangle PQT:** Since W and T are midpoints, PW = WQ = s/2 and PT = TS = s/2.  Triangle PQT has half the base and half the height of triangle PQS (which is half of the rhombus). Therefore, [PQT] = (1/2) * (s/2) * (s/2) * sin(∠P) = (1/8) * s² * sin(∠P) = (1/4) * [PQS] = (1/8) * [PQRS].

* **Triangles PSW and QRS:** By the same logic, [PSW] = (1/8) * [PQRS].

**3. Area of Triangle PVW:**

We know that the medians of a triangle divide it into six equal areas. In triangle PQS, QT and SW are medians. Therefore, the area of triangle PVW is 1/6 the area of triangle PQS.
[PVW] = (1/6) * [PQS] = (1/6) * (1/2) [PQRS] = (1/12) * [PQRS]

**4. Area of Quadrilateral PWVT:**

The area of quadrilateral PWVT can be found by subtracting the areas of triangles PVW, PWT, and QWV from the area of triangle PQT.

*Note that PWT and QWV are also 1/6 of the area of the relevant larger triangle, which is 1/4 of the area of the rhombus.*
[PWT] = [QWV] = (1/6) * [PQT] = (1/6) * (1/8) [PQRS] = (1/48) [PQRS]

[PWVT] = [PQT] - [PVW] - [PWT]
[PWVT] = (1/8)[PQRS] - (1/12)[PQRS] - (1/48)[PQRS]
[PWVT] = ((6-4-1)/48) * [PQRS]
[PWVT] = (1/48) * [PQRS]

**5. Fraction of Rhombus:**

The area of quadrilateral PWVT is (1/48) of the area of rhombus PQRS.

Therefore, the fraction of rhombus PQRS that is quadrilateral PWVT is **1/48**.