Question 1210466
This is a classic geometry problem that can be solved by breaking down the areas within the rectangle.

## 📏 Setting Up the Area Equations

Let $[X]$ denote the area of polygon $X$.
The rectangle is $ABCD$. Let $A_{rect}$ be its total area.
The problem uses a diagram that is not visible, but based on the context (the ratio $\frac{[AEP]}{[DFP]} = 5$ and the area fractions), we must assume **$P$ is a point on the side $CD$** and **$E$ and $F$ are points on the sides $AB$ and $BC$ respectively**.

However, the question contains **contradictory information** about the red area:
* "The red area is **1/2** of the rectangle"
* "The red area is **1/3** of the rectangle"

Assuming the most common configuration for problems involving areas and a point on a side of a rectangle, let's first analyze the areas of the triangles defined by the point $P$.

Let $h$ be the height $AD$ and $w$ be the width $AB$. $A_{rect} = w \cdot h$.

**Assumption based on standard geometry problems (The Point P is on the side CD):**
* The **red region** is typically the area of the **trapezoid $AECD$** or the area of the **triangle $ADP$ and $BCP$ combined**.
* The **yellow region** is typically the area of the **triangle $ABP$**.

### Case 1: The Red Region is $\frac{1}{2}$ of the Rectangle

The only region inside a rectangle that is guaranteed to be **half** the area is a triangle with its base on one side and its opposite vertex on the opposite side.
This suggests the **yellow region** (let's call it $A_{yellow}$) is the **triangle $\triangle ABP$**, and $P$ is on $CD$.
If $P$ is on $CD$, then:
$$[ABP] = \frac{1}{2} \cdot AB \cdot h = \frac{1}{2} A_{rect}$$
This means the **Yellow Area** is $\frac{1}{2} A_{rect}$.
Consequently, the **Red Area** ($A_{red}$) must be the remaining area:
$$A_{red} = A_{rect} - [ABP] = A_{rect} - \frac{1}{2} A_{rect} = \frac{1}{2} A_{rect}$$
This matches the first piece of information: **"The red area is 1/2 of the rectangle."**

Under this valid geometric setup:
* $A_{red} = \frac{1}{2} A_{rect}$
* $A_{yellow} = \frac{1}{2} A_{rect}$

### Finding the Ratio

The ratio of the area of the red region to the area of the yellow region is:
$$\frac{A_{red}}{A_{yellow}} = \frac{\frac{1}{2} A_{rect}}{\frac{1}{2} A_{rect}} = \mathbf{\frac{1}{1}}$$

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## 🚫 Addressing the Contradictory Information

The information $\frac{[AEP]}{[DFP]} = 5$ and the second statement **"the red area is 1/3 of the rectangle"** cannot be reconciled with the standard geometric setup where $P$ is on $CD$.

* If $A_{red} = \frac{1}{3} A_{rect}$, then $A_{yellow} = 1 - \frac{1}{3} = \frac{2}{3} A_{rect}$.
* The ratio would be $\frac{A_{red}}{A_{yellow}} = \frac{1/3}{2/3} = \frac{1}{2}$.

However, the information that the red area is $1/2$ of the rectangle is the **only one consistent with the yellow area being $\triangle ABP$ with $P$ on $CD$**, a setup that is very common and would make the second ratio $\frac{[AEP]}{[DFP]} = 5$ necessary for finding the lengths of $AE$ and $DF$, which are irrelevant if the red and yellow areas are simply the two halves of the rectangle.

Since the problem is likely testing the fundamental area property of a rectangle, we **must prioritize the information that makes the yellow area a known fraction of the rectangle's area.**

**The most consistent and standard interpretation is that:**
1.  **$P$ lies on the side $CD$.**
2.  The **yellow region** is the triangle $\triangle ABP$.
3.  The **red region** is the remaining area ($A_{red} = [ADP] + [BCP]$).
4.  Therefore, $A_{yellow} = \frac{1}{2} A_{rect}$ and $A_{red} = \frac{1}{2} A_{rect}$.

The ratio $\frac{[AEP]}{[DFP]} = 5$ and the second statement "the red area is 1/3 of the rectangle" are likely extraneous or errors in the problem statement.

The required ratio is:
$$\frac{\text{Area of Red Region}}{\text{Area of Yellow Region}} = \frac{1/2}{1/2} = \frac{1}{1}$$