Question
What is the area of a rhombus, whose sides are 25 cm and one of the diagonals is 14 cm?
This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
$336 \text{ cm}^2$
Rhombus Area Calculation
To find the area of the rhombus, we need both its diagonals. We are given the side length and one diagonal.
Given:
- Side length, $s = 25$ cm
- One diagonal, $d_1 = 14$ cm
Finding the Second Diagonal
The diagonals of a rhombus bisect each other at right angles. This divides the rhombus into four congruent right-angled triangles. Each triangle has:
- Hypotenuse = side of the rhombus = $25$ cm
- One leg = half of the known diagonal = $\frac{14 \text{ cm}}{2} = 7$ cm
- Other leg = half of the unknown diagonal ($d_2/2$)
Using the Pythagorean theorem ($a^2 + b^2 = c^2$):
$ (7 \text{ cm})^2 + \left(\frac{d_2}{2}\right)^2 = (25 \text{ cm})^2 $ $ 49 \text{ cm}^2 + \left(\frac{d_2}{2}\right)^2 = 625 \text{ cm}^2 $ $ \left(\frac{d_2}{2}\right)^2 = 625 \text{ cm}^2 - 49 \text{ cm}^2 $ $ \left(\frac{d_2}{2}\right)^2 = 576 \text{ cm}^2 $ $ \frac{d_2}{2} = \sqrt{576 \text{ cm}^2} $ $ \frac{d_2}{2} = 24 \text{ cm} $Therefore, the second diagonal is:
$ d_2 = 2 \times 24 \text{ cm} = 48 \text{ cm} $Calculating Rhombus Area
The area of a rhombus is calculated using the formula:
$ \text{Area} = \frac{1}{2} \times d_1 \times d_2 $Substituting the values of the diagonals:
$ \text{Area} = \frac{1}{2} \times 14 \text{ cm} \times 48 \text{ cm} $ $ \text{Area} = 7 \text{ cm} \times 48 \text{ cm} $ $ \text{Area} = 336 \text{ cm}^2 $The area of the rhombus is $336 \text{ cm}^2$. This matches option C.
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