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We are given a rectangle \( ABCD \) and asked to find the area of triangle \( PBQ \). The areas of the smaller triangles are labeled as follows:

- Area of triangle \( APB = 36 \)
- Area of triangle \( BCQ = 20 \)
- Area of triangle \( PDQ = 16 \)

Now let's work step by step:

1. Since \( ABCD \) is a rectangle, triangles that share the same height and are between the same parallel sides must have proportional areas based on their base lengths.

2. The entire quadrilateral \( PBCQ \) is made up of triangles \( PBQ \), \( APB \), and \( BCQ \). Similarly, quadrilateral \( PDQ \) includes triangles \( PDQ \) and \( PBQ \).

3. Let's assume the height from point \( B \) to line \( AD \) is the same for all triangles. By comparing areas, we will find the base relationships and solve for the area of \( PBQ \).

I'll now perform the calculation to find the area of triangle \( PBQ \).

36, 20, and 16 are the areas of the smaller triangles indicated below. If ABCD is a rectangle, find the area of the triangle PBQ.

Solve for the area of triangle PBQ using given triangle areas in rectangle ABCD. The problem involves proportional triangle areas and basic geometry principles. Suitable for students in Grades 9-11.

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