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Question

Find the area of a rhombus whose perimeter is 164 cm and one diagonal is of length 80 cm.

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
$720 \text{ cm}^2$

To find the area of a rhombus, we can use the formula:

\(A = \frac{1}{2} \times d_1 \times d_2\)

where \(d_1\) and \(d_2\) are the lengths of the diagonals.

Given:

  • Perimeter of the rhombus = 164 cm
  • One diagonal \((d_1) = 80 \text{ cm}\)

Since all sides of a rhombus are equal, the side \((s)\) is:

\(s = \frac{\text{Perimeter}}{4} = \frac{164}{4} = 41 \text{ cm}\)

Next, using the Pythagorean theorem in one of the right triangles formed by the diagonals:

\(\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = s^2\)

Substituting the known values:

\(\left(\frac{80}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = 41^2\)

Simplifying:

\(1600 + \left(\frac{d_2}{2}\right)^2 = 1681\)

\(\left(\frac{d_2}{2}\right)^2 = 81\)

\(\frac{d_2}{2} = 9\)

\(d_2 = 18 \text{ cm}\)

Now compute the area:

\(A = \frac{1}{2} \times 80 \times 18 = 720 \text{ cm}^2\)

Therefore, the area of the rhombus is 720 cm².

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