In triangle PQR , PX and QY are medians which intersect at O and also XZ is parallel to QY such that Z lies on PR. Find the ratio of area of triangle QOX to that of quadrilateral ΟΥΖΧ?

Given the problem in triangle \( PQR \):

1. \( PX \) and \( QY \) are medians that intersect at \( O \), which means \( O \) is the centroid of the triangle.
2. \( XZ \) is parallel to \( QY \) and \( Z \) lies on \( PR \).

### To solve for the ratio of the area of \( \triangle QOX \) to that of quadrilateral \( OYZX \):

#### 1. **Properties of Medians and Centroid:**
   - The centroid \( O \) divides each median into a 2:1 ratio, with the longer part being on the vertex side.

#### 2. **Areas of Triangles in the Context of Medians:**
   - The area of \( \triangle QOX \) can be related to the area of \( \triangle PQR \) since \( O \) divides \( \triangle PQR \) into smaller triangles of equal area.
   - Since \( O \) is the centroid, each of the smaller triangles \( \triangle POX \), \( \triangle QOY \), and \( \triangle ROZ \) has an area that is \( \frac{1}{3} \) of the area of \( \triangle PQR \).

#### 3. **Parallel Line Properties:**
   - Since \( XZ \parallel QY \), quadrilateral \( OYZX \) is a trapezoid. The area of this quadrilateral can be calculated using the area formula for a trapezoid or by subtracting the area of triangle \( QOX \) from \( \triangle POX \).

#### 4. **Area Calculations:**
   - The area of \( \triangle QOX \) is \( \frac{1}{6} \) of the area of \( \triangle PQR \) because \( O \) divides \( \triangle PQX \) into two smaller triangles of equal area.
   - The quadrilateral \( OYZX \) consists of \( \triangle POX \) minus \( \triangle QOX \). Since the area of \( \triangle POX \) is \( \frac{1}{3} \) of \( \triangle PQR \), and \( \triangle QOX \) is \( \frac{1}{6} \) of \( \triangle PQR \), the area of quadrilateral \( OYZX \) is:
     \[
     \text{Area of } OYZX = \frac{1}{3} - \frac{1}{6} = \frac{1}{6} \text{ of the area of } \triangle PQR
     \]

#### 5. **Ratio Calculation:**
   - Since both \( \triangle QOX \) and quadrilateral \( OYZX \) have the same area \( \frac{1}{6} \) of the area of \( \triangle PQR \), the ratio of the areas is:
     \[
     \text{Ratio} = \frac{\text{Area of } \triangle QOX}{\text{Area of } OYZX} = \frac{\frac{1}{6}}{\frac{1}{6}} = 1:1
     \]

### Final Answer:
The ratio of the area of \( \triangle QOX \) to that of quadrilateral \( OYZX \) is **1:1**.

Would you like any further details on this problem or any other questions related to this topic? Here are some related questions:

1. What are the properties of a centroid in a triangle?
2. How does the centroid divide the medians in a triangle?
3. What is the significance of parallel lines in triangle geometry?
4. How can the area of a trapezoid be calculated in different scenarios?
5. How do we find the area of quadrilaterals formed by medians and parallel lines?

**Tip:** In problems involving centroids and medians, remember that the centroid divides the medians into segments with a 2:1 ratio, which is crucial in calculating areas and understanding the geometry.

Explore the solution to finding the ratio of the area of triangle QOX to that of quadrilateral OYZX in triangle PQR. Understand the properties of medians, centroids, and parallel lines in geometry.

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