If the sum of the lengths of the diagonals of a rhombus is 10 m and if its area is `9m^(2)`, then what is the sum of the square of the diagonals ?

Options

A. 36 `m^(2)`
B. 64 `m^(2)`
C. 80 `m^(2)`
D. 100 `m^(2)`

Answer

To solve the problem step by step, we will use the properties of a rhombus and the given information about the diagonals and area.

### Step-by-Step Solution:

1. **Identify the variables**:
Let the lengths of the diagonals be \( d_1 \) and \( d_2 \).
According to the problem, we have:
\[ d_1 + d_2 = 10 \quad \text{(1)} \]
and the area of the rhombus is given by:
\[ \text{Area} = \frac{1}{2} d_1 d_2 = 9 \quad \text{(2)} \]

2. **From equation (2), express \( d_1 d_2 \)**:
Rearranging equation (2):
\[ d_1 d_2 = 9 \times 2 = 18 \quad \text{(3)} \]

3. **Use equations (1) and (3) to form a quadratic equation**:
From equation (1), we can express \( d_2 \) in terms of \( d_1 \):
\[ d_2 = 10 - d_1 \]
Substituting this into equation (3):
\[ d_1(10 - d_1) = 18 \]
Expanding this gives:
\[ 10d_1 - d_1^2 = 18 \]
Rearranging this into standard quadratic form:
\[ d_1^2 - 10d_1 + 18 = 0 \quad \text{(4)} \]

4. **Solve the quadratic equation (4)**:
We can use the quadratic formula:
\[ d_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -10, c = 18 \):
\[ d_1 = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 18}}{2 \cdot 1} \]
\[ d_1 = \frac{10 \pm \sqrt{100 - 72}}{2} \]
\[ d_1 = \frac{10 \pm \sqrt{28}}{2} \]
\[ d_1 = \frac{10 \pm 2\sqrt{7}}{2} \]
\[ d_1 = 5 \pm \sqrt{7} \]
Thus, we have two possible values for \( d_1 \):
\[ d_1 = 5 + \sqrt{7} \quad \text{and} \quad d_2 = 5 - \sqrt{7} \]

5. **Find the sum of the squares of the diagonals**:
We need to find \( d_1^2 + d_2^2 \). Using the identity:
\[ d_1^2 + d_2^2 = (d_1 + d_2)^2 - 2d_1d_2 \]
Substituting the known values from equations (1) and (3):
\[ d_1^2 + d_2^2 = (10)^2 - 2(18) \]
\[ d_1^2 + d_2^2 = 100 - 36 \]
\[ d_1^2 + d_2^2 = 64 \]

### Final Answer:
The sum of the squares of the diagonals is \( 64 \, m^2 \).