The smaller diagonal of a rhombus is equal to length of its sides. If length of each side is 6 cm, what is the area (in `cm^(2)` ) of an equilateral triangle whose side is equal to the bigger diagonal of the rhombus ?

Options

A. `18 sqrt(3)`
B. `27 sqrt(3)`
C. `32 sqrt(3)`
D. `36 sqrt(3)`

Answer

To solve the problem, we will follow these steps:

1. **Identify the properties of the rhombus**: In a rhombus, the diagonals bisect each other at right angles. Let the smaller diagonal be denoted as \(d_1\) and the larger diagonal as \(d_2\).

2. **Given values**: We know that the smaller diagonal \(d_1\) is equal to the length of the sides of the rhombus. Since each side of the rhombus is given as 6 cm, we have:
\[ d_1 = 6 \text{ cm} \]

3. **Use the relationship between the diagonals and the sides**: The relationship between the diagonals and the sides of a rhombus can be expressed using the formula:
\[ \text{side}^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 \]
Substituting the known values:
\[ 6^2 = \left(\frac{6}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 \]
Simplifying this gives:
\[ 36 = 3^2 + \left(\frac{d_2}{2}\right)^2 \]
\[ 36 = 9 + \left(\frac{d_2}{2}\right)^2 \]
\[ \left(\frac{d_2}{2}\right)^2 = 36 - 9 = 27 \]
\[ \frac{d_2}{2} = \sqrt{27} = 3\sqrt{3} \]
Therefore, the larger diagonal \(d_2\) is:
\[ d_2 = 2 \times 3\sqrt{3} = 6\sqrt{3} \text{ cm} \]

4. **Calculate the area of the equilateral triangle**: The area \(A\) of an equilateral triangle with side length \(s\) is given by the formula:
\[ A = \frac{\sqrt{3}}{4} s^2 \]
Here, the side length \(s\) is equal to the larger diagonal \(d_2\):
\[ A = \frac{\sqrt{3}}{4} (6\sqrt{3})^2 \]
\[ A = \frac{\sqrt{3}}{4} \times 36 \times 3 \]
\[ A = \frac{\sqrt{3}}{4} \times 108 \]
\[ A = 27\sqrt{3} \text{ cm}^2 \]

Thus, the area of the equilateral triangle is \(27\sqrt{3} \text{ cm}^2\).