Question

The three small rectangles have the same dimensions

Column A	Column B
\(\frac{PS}{RS}\)	\(\frac{1}{2}\)

Hint:

In geometry problems with composite shapes, assign variables to the dimensions of the basic components. Then, use the properties of the larger shape (e.g., opposite sides of a rectangle are equal) to form equations and solve for the relationships between the variables.

Answer 1

Step 1: Understanding the Concept:
The problem involves determining the ratio of the side lengths of a large rectangle that is composed of three identical smaller rectangles. The key is to express the sides of the large rectangle in terms of the dimensions of the smaller ones.
Step 2: Key Formula or Approach:
Let the dimensions of the small identical rectangles be length \(l\) and width \(w\), with \(l>w\). We will use the geometric constraints of the figure to find a relationship between \(l\) and \(w\).
Step 3: Detailed Explanation:
Let's analyze the figure. The large rectangle QRSP is divided into one rectangle on top and two side-by-side rectangles on the bottom. Since all three small rectangles are identical, they all have dimensions \(l \times w\).
The two bottom rectangles are standing vertically. Their width is \(w\) and their height is \(l\).
The bottom side of the large rectangle, PS, is the sum of the widths of these two rectangles.
\[ PS = w + w = 2w \] The height of the large rectangle, RS, is the same as the height of the vertically oriented bottom-right rectangle.
\[ RS = l \] The top rectangle is lying horizontally. Its width must match the total width of the two rectangles below it. Therefore, its width is \(2w\). Its height is \(w\).
Since the dimensions of this top rectangle must also be \(l\) and \(w\), we can see that its longer side is \(l\) and its shorter side is \(w\). From the figure, the width of the top rectangle is its longer side.
\[ l = 2w \] This establishes the relationship between the length and width of the small rectangles.
Now, we can calculate the ratio in Column A.
\[ \frac{PS}{RS} = \frac{2w}{l} \] Substitute the relationship \(l = 2w\) into the expression:
\[ \frac{PS}{RS} = \frac{2w}{2w} = 1 \] The value of Column A is 1.
Step 4: Final Answer:
We are comparing Column A and Column B.
Column A = 1
Column B = \(\frac{1}{2}\)
Since \(1>\frac{1}{2}\), the quantity in Column A is greater.