In ∆EFG, XY ∥ FG, area of the quadrilateral XFGY = 44 m2. If EX : XF = 2 : 3, then find the area of ∆EXY (in m2).

Given that in \(\Delta EFG\), line \(XY \parallel FG\), the area of quadrilateral \(XFGY\) is 44 m², and the ratio \(EX : XF = 2 : 3\), we need to find the area of \(\Delta EXY\).

### Step 1: Understand the ratio
The ratio \(EX : XF = 2 : 3\) implies that \(EX = \frac{2}{5}EF\) and \(XF = \frac{3}{5}EF\), where \(EF\) is the entire length from \(E\) to \(F\).

### Step 2: Use the property of similar triangles
Since \(XY \parallel FG\), triangles \(\Delta EXY\) and \(\Delta EFG\) are similar.

### Step 3: Determine the area ratio
The area of similar triangles is proportional to the square of the corresponding sides. Here, the corresponding side ratio between \(\Delta EXY\) and \(\Delta EFG\) is:

\[
\text{Side ratio} = \frac{EX}{EF} = \frac{\frac{2}{5}EF}{EF} = \frac{2}{5}
\]

Thus, the area ratio is:

\[
\text{Area ratio} = \left(\frac{2}{5}\right)^2 = \frac{4}{25}
\]

### Step 4: Calculate the area of \(\Delta EXY\)
Let the area of \(\Delta EFG\) be \(A\). The area of \(\Delta EXY\) is then \(\frac{4}{25}A\).

The area of quadrilateral \(XFGY\) is the difference between the area of \(\Delta EFG\) and \(\Delta EXY\), given by:

\[
\text{Area of } XFGY = A - \frac{4}{25}A = \frac{21}{25}A
\]

We know that \(\text{Area of } XFGY = 44\) m², so:

\[
\frac{21}{25}A = 44
\]

Solve for \(A\):

\[
A = 44 \times \frac{25}{21} = \frac{1100}{21} \text{ m}^2
\]

### Step 5: Find the area of \(\Delta EXY\)
Now, the area of \(\Delta EXY\) is:

\[
\text{Area of } \Delta EXY = \frac{4}{25}A = \frac{4}{25} \times \frac{1100}{21} = \frac{4400}{525} = \frac{88}{10.5} \approx 8.38 \text{ m}^2
\]

The area of \(\Delta EXY\) is approximately \(8.38\) m².

Would you like to know more about how the properties of similar triangles were applied here? Also, here are some related questions to deepen your understanding:

1. How does the parallel condition affect the similarity of triangles?
2. What are the consequences of the side ratio on the area ratio in similar triangles?
3. How can you find the area of other sections if a different segment ratio was given?
4. Can the same method be applied if \(XY\) was not parallel to \(FG\)?
5. What would happen to the areas if the ratio \(EX:XF\) was different?

**Tip:** When dealing with similar triangles, remember that their corresponding sides are proportional, and their areas are proportional to the square of the corresponding side ratio.

Learn how to find the area of triangle EXY in a geometry problem involving similar triangles and parallel lines. Given the area of quadrilateral XFGY, and the ratio EX : XF = 2 : 3, discover step-by-step calculations and the application of properties of similar triangles. Suitable for high school geometry students.

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