Consider parallelogram 𝐴 𝐵 𝐶 𝐷 ABCD. Point 𝑋 X is on side 𝐶 𝐷 CD and 𝐴 𝑋 AX intersects 𝐵 𝐷 BD at point 𝑌 Y. Given that the area of △ 𝐴 𝐵 𝑌 △ABY is 45 and the area of △ 𝑋 𝐷 𝑌 △XDY is 20, find the area of the unshaded portion of the parallelogram.

To solve this problem, we need to analyze the given geometric configuration in the parallelogram \( ABCD \). 

### Step 1: Understand the problem
- We are given a parallelogram \( ABCD \).
- Point \( X \) is on side \( CD \), and line segment \( AX \) intersects diagonal \( BD \) at point \( Y \).
- The area of triangle \( ABY \) is 45, and the area of triangle \( XDY \) is 20.

We need to find the area of the unshaded portion of the parallelogram.

### Step 2: Establish relationships using area properties
In a parallelogram, the diagonals bisect each other, so if we consider the triangles formed by these diagonals, they will share common bases and heights.

#### Consider the triangles \( ABY \) and \( XDY \):
- Triangle \( ABY \) and triangle \( XDY \) share the point \( Y \), which lies on diagonal \( BD \).
- Let \( S \) represent the total area of the parallelogram \( ABCD \).

### Step 3: Use the area information
1. The area of \( \triangle ABY \) is given as 45.
2. The area of \( \triangle XDY \) is given as 20.

### Step 4: Calculate the area of the parallelogram
Notice that the diagonal \( BD \) of the parallelogram divides it into two congruent triangles, \( \triangle ABD \) and \( \triangle BCD \). These two triangles have equal areas, which sum up to half of the total area of the parallelogram. 

The area of \( \triangle ABD \) is given as the area of \( \triangle ABY \) plus the area of \( \triangle ADY \). Therefore:

\[
\text{Area of } \triangle ABD = \text{Area of } \triangle ABY + \text{Area of } \triangle ADY
\]

But \( \text{Area of } \triangle ADY = \text{Area of } \triangle XDY \) because \( Y \) is the point where the diagonals intersect, dividing the parallelogram into triangles of equal height, and the base \( DY = XY \). Hence:

\[
\text{Area of } \triangle ABD = 45 + 20 = 65
\]

Since the area of \( \triangle ABD \) is half the total area of the parallelogram:

\[
S = 2 \times 65 = 130
\]

### Step 5: Find the unshaded area
To find the unshaded portion, we need to subtract the areas of \( \triangle ABY \) and \( \triangle XDY \) from the total area of the parallelogram:

\[
\text{Unshaded Area} = S - (\text{Area of } \triangle ABY + \text{Area of } \triangle XDY)
\]
\[
\text{Unshaded Area} = 130 - (45 + 20) = 130 - 65 = 65
\]

### Final Answer:
The area of the unshaded portion of the parallelogram is **65** square units.

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Would you like more details on any part of this solution?

Here are five follow-up questions related to this problem:
1. How would the solution change if the point \( X \) was moved along the side \( CD \)?
2. What would be the area of triangle \( ABC \) if we were given its dimensions?
3. How does the property of diagonals in a parallelogram help in solving area problems?
4. Can we solve this problem using vectors or coordinates? How would that approach look?
5. What are the implications of the areas of triangles sharing the same vertex in a parallelogram?

**Tip:** In parallelograms, diagonals split the figure into two equal-area triangles, which is crucial in area-related problems.

Learn how to find the area of the unshaded portion in a parallelogram problem involving intersections and diagonal areas. This problem explores the properties of parallelograms, including how diagonals divide the figure into triangles with equal areas. Suitable for high school students studying geometry and area calculations.

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