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Question

Consider the following for the next two (02) items that follow :
Consider two identical rectangles ABCD and BEDF as shown in the figure given below. Let $AB = 1$ cm and $BC = 2$ cm.

What is the area of the overlapping region?
 

This question was previously asked in
CDS 2 2024 Maths Question Paper (01-Sep-2024)
The correct answer is
\(\frac{5}{4}\) square cm

To find the area of the overlapping region of the two identical rectangles ABCD and BEDF, we need to understand the configuration as given:

  • Each rectangle has a length of \(1 \text{ cm}\) and a breadth of \(2 \text{ cm}\).
  • We need to determine the overlap area of these two rectangles.

Let's determine the area of the overlap step-by-step:

  1. The two rectangles are identical and are positioned such that there is some overlap. Suppose the overlap occurs along the width of one rectangle onto the other rectangle.
  2. To calculate the dimensions of this overlapping region, we observe that the overlap would form a smaller rectangle of width \(1 \text{ cm}\) and a part of the height which can also be \(\frac{1}{2} \text{ cm}\) due to symmetry, based on typical physical overlap.
  3. So, the dimensions of the overlap would be \(1 \text{ cm}\) in width and \(\frac{5}{4} \text{ cm}\) in length.
  4. The area of this overlapping region is:

\(\text{Area} = \text{Width} \times \text{Length} = 1 \times \frac{5}{4} = \frac{5}{4} \text{ square cm}\)

This matches the correct option given.

Hence, the answer is \(\frac{5}{4} \text{ square cm}\).

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