The length of one side of a rhombus and one of the diagonal are 6 cm each. The area of the rhombus (in `cm^(2)`) is:

Options

A. `27sqrt(3)`
B. 18
C. `9sqrt(3)`
D. `18sqrt(3)`

Answer

To find the area of the rhombus given one side and one diagonal, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Length of one side of the rhombus (AD) = 6 cm
- Length of one diagonal (AC) = 6 cm

2. **Label the Rhombus**:
- Let the vertices of the rhombus be A, B, C, and D.
- The diagonals AC and BD intersect at point O, which is the midpoint of both diagonals.

3. **Determine the Lengths of the Halves of the Diagonals**:
- Since O is the midpoint, we have:
- OA = OC = 6 cm / 2 = 3 cm
- Let OD = OB = x cm (we need to find x).

4. **Use the Pythagorean Theorem**:
- In triangle OAD, we can apply the Pythagorean theorem:
\[ AD^2 = OA^2 + OD^2 \]
- Substituting the known values:
\[ 6^2 = 3^2 + x^2 \]
\[ 36 = 9 + x^2 \]
\[ x^2 = 36 - 9 = 27 \]
\[ x = \sqrt{27} = 3\sqrt{3} \text{ cm} \]

5. **Calculate the Length of the Other Diagonal (BD)**:
- Since OD = OB = 3√3 cm, we can find the full length of diagonal BD:
\[ BD = OD + OB = 3\sqrt{3} + 3\sqrt{3} = 6\sqrt{3} \text{ cm} \]

6. **Calculate the Area of the Rhombus**:
- The area of a rhombus can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times d_1 \times d_2 \]
- Here, \(d_1 = AC = 6 \text{ cm}\) and \(d_2 = BD = 6\sqrt{3} \text{ cm}\):
\[ \text{Area} = \frac{1}{2} \times 6 \times 6\sqrt{3} \]
\[ = \frac{1}{2} \times 36\sqrt{3} = 18\sqrt{3} \text{ cm}^2 \]

### Final Answer:
The area of the rhombus is \(18\sqrt{3} \text{ cm}^2\).

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