5. Given is a rectangle $A B C D$ and a point $E$ on side $\overline{B C}$. The line segment $\overline{D E}$ divides the rectangle $A B C D$ into two parts such that the areas of these parts are in the ratio $6: 1$. What is the ratio of the lengths of segments $\overline{C E}$ and $\overline{B E}$?

## Problems worth 10 points:
First method:

Sketch:

![](https://cdn.mathpix.com/cropped/2024_05_30_e1c3024da477c84dc560g-5.jpg?height=480&width=611&top_left_y=1876&top_left_x=288)

Let \( |AB| = |DC| = a \), \( |BE| = x \), \( |EC| = y \). Then \( |AD| = |BC| = x + y \).

This divides the given rectangle \( ABCD \) into a right triangle \( ECD \) and a right trapezoid \( ABED \) with \( \overline{AD} \) and \( \overline{BE} \) as the bases, and \( \overline{AB} \) as the height.

\[ P_{\text{ABED}} = \frac{(x + y) + x}{2} \cdot a \]
\[ P_{ECD} = \frac{y}{2} \cdot a \]

1 POINT

The area of the right triangle is less than half the area of the rectangle, so \( P_{ABED} : P_{ECD} = 6 : 1 \). Applying the formulas for the areas of these shapes, we get:

```
\(\frac{(2x + y) \cdot a}{2} : \frac{y \cdot a}{2} = 6 : 1\)
\((2x + y) : y = 6 : 1\)
\(2x + y = 6y\)
\(2x = 5y\)
\(x : y = 5 : 2\),
or \( |BE| : |CE| = 5 : 2 \) or \( |CE| : |BE| = 2 : 5 \).
```

1 POINT

1 POINT

TOTAL 6 POINTS

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