SOLUTION: Find the ratio of the area of the red region to the area of the yellow region. Enter your answer as a fraction.

Question 1210480: Find the ratio of the area of the red region to the area of the yellow region. Enter your answer as a fraction.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(2136)   (Show Source): You can put this solution on YOUR website!
This is a trick question! The description is inconsistent.
* If the **red region is a triangle** and the **yellow region is the same triangle**, their areas must be **equal**.
* If the **yellow region is the same triangle, half as big**, the yellow region cannot be the *same* triangle. It must be a **similar triangle** whose dimensions are scaled down.
Let's proceed with the most logical interpretation for a geometry/ratio problem: the yellow triangle is **similar** to the red triangle, with its linear dimensions scaled by a factor of $1/2$.
## 📐 Ratio of Areas for Similar Figures
When two geometric figures are **similar** (the same shape but different sizes), the ratio of their areas is equal to the square of the ratio of their corresponding linear dimensions (e.g., sides, heights).
Let:
* $A_R$ be the area of the **red triangle**.
* $A_Y$ be the area of the **yellow triangle**.
* $s_R$ be a linear dimension (side length) of the red triangle.
* $s_Y$ be the corresponding linear dimension of the yellow triangle.
---
### Step 1: Determine the Ratio of Linear Dimensions
The problem states the yellow triangle is "half as big." In geometry, "half as big" usually means the linear dimensions are halved:
$$\frac{s_Y}{s_R} = \frac{1}{2}$$
### Step 2: Determine the Ratio of Areas
The ratio of the areas of two similar figures is the square of the ratio of their corresponding linear dimensions:
$$\frac{A_Y}{A_R} = \left(\frac{s_Y}{s_R}\right)^2$$
$$\frac{A_Y}{A_R} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$
### Step 3: Find the Required Ratio
The question asks for the ratio of the area of the **red region** to the area of the **yellow region** ($\frac{A_R}{A_Y}$):
$$\frac{A_R}{A_Y} = \frac{1}{\frac{A_Y}{A_R}} = \frac{1}{\frac{1}{4}} = \mathbf{4}$$
The ratio of the area of the red region to the area of the yellow region is **$\frac{4}{1}$**.
***
If you strictly follow the first part of the statement ("The red region is a triangle, and the yellow region is the **same triangle**"):
* $A_R = A_Y$
* Ratio $\frac{A_R}{A_Y} = \frac{1}{1}$
However, this makes the second part ("half as big") irrelevant or contradictory. The answer $\frac{4}{1}$ is the correct solution based on the intended mathematical meaning of a scaled figure.
Do you have another geometry problem you'd like to solve?
Answer by ikleyn(53266)   (Show Source): You can put this solution on YOUR website!
.
Find the ratio of the area of the red region to the area of the yellow region. Enter your answer as a fraction.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        I am strongly disagree with the solution given by @CPhill in his post.

First, his interpretation is far-fetched. Nothing in the problem points
to triangle and its midline and so on.

But even if to accept his interpretation, his answer 1/4 for the area ratios
is .

Yes, the ratio 1/4 is valid as the ratio of the area of smaller triangle to the area of the whole triangle.

But the question is about the ratio of the areas of differently colored parts.

The ratio of the area of smaller triangle to the area of the remaining part
of the whole triangle is then 1/3,

and the correct should be then .

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