If the sum of the lengths of the diagonals of a rhombus of side 4 cm is 10 cm, what is its area ?

Options

A. 8 `cm^(2)`
B. 9 `cm^(2)`
C. 10 `cm^(2)`
D. 12 `cm^(2)`

Answer

To find the area of a rhombus when the sum of the lengths of its diagonals is given, we can follow these steps:

### Step-by-Step Solution

1. **Understand the Problem**:
We have a rhombus with each side measuring 4 cm, and the sum of the lengths of the diagonals (d1 + d2) is 10 cm. We need to find the area of the rhombus.

**Hint**: Remember that the diagonals of a rhombus bisect each other at right angles.

2. **Set Up the Variables**:
Let one diagonal be \( d_1 = x \) cm. Therefore, the other diagonal will be \( d_2 = 10 - x \) cm.

**Hint**: The sum of the diagonals is given as 10 cm, so if one diagonal is \( x \), the other must be \( 10 - x \).

3. **Use the Properties of the Rhombus**:
The diagonals bisect each other at right angles. Thus, we can form right triangles using half of each diagonal. The lengths of the halves of the diagonals are \( \frac{x}{2} \) and \( \frac{10 - x}{2} \).

**Hint**: Use the Pythagorean theorem since the diagonals intersect at right angles.

4. **Apply the Pythagorean Theorem**:
In triangle formed by half diagonals and the side of the rhombus:
\[ \left(\frac{x}{2}\right)^2 + \left(\frac{10 - x}{2}\right)^2 = 4^2 \]
Simplifying this gives:
\[ \frac{x^2}{4} + \frac{(10 - x)^2}{4} = 16 \]
Multiplying through by 4 to eliminate the fraction:
\[ x^2 + (10 - x)^2 = 64 \]

5. **Expand and Simplify**:
Expanding \( (10 - x)^2 \):
\[ x^2 + (100 - 20x + x^2) = 64 \]
Combine like terms:
\[ 2x^2 - 20x + 100 = 64 \]
Rearranging gives:
\[ 2x^2 - 20x + 36 = 0 \]
Dividing the entire equation by 2:
\[ x^2 - 10x + 18 = 0 \]

6. **Solve the Quadratic Equation**:
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -10, c = 18 \):
\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 18}}{2 \cdot 1} \]
\[ x = \frac{10 \pm \sqrt{100 - 72}}{2} \]
\[ x = \frac{10 \pm \sqrt{28}}{2} \]
\[ x = \frac{10 \pm 2\sqrt{7}}{2} \]
\[ x = 5 \pm \sqrt{7} \]

7. **Find the Lengths of the Diagonals**:
If \( d_1 = 5 + \sqrt{7} \) then \( d_2 = 10 - (5 + \sqrt{7}) = 5 - \sqrt{7} \).

**Hint**: Check both cases for \( d_1 \) and \( d_2 \).

8. **Calculate the Area of the Rhombus**:
The area \( A \) of a rhombus is given by:
\[ A = \frac{1}{2} \times d_1 \times d_2 \]
Substituting the values:
\[ A = \frac{1}{2} \times (5 + \sqrt{7}) \times (5 - \sqrt{7}) \]
Using the difference of squares:
\[ A = \frac{1}{2} \times (25 - 7) = \frac{1}{2} \times 18 = 9 \text{ cm}^2 \]

### Final Answer:
The area of the rhombus is **9 cm²**.