To find the value of \( x \), we can use the fact that the area of a triangle scales by the square of the scale factor when transformed by a linear transformation.
Let the area of the original triangle be \( A_1 = 12.5 , \text{cm}^2 \) and the area of the transformed triangle be \( A_2 = 25 , \text{cm}^2 \).
The scale factor \( k \) can be found using the relationship between the areas as follows:
\[ \frac{A_2}{A_1} = k^2 \]
Plugging in the values:
\[ \frac{25}{12.5} = k^2 \]
\[ 2 = k^2 \]
Taking the square root gives us:
\[ k = \sqrt{2} \]
Now, we need to analyze the transformation matrix \( T \), which is given in the form:
\[ T = \begin{pmatrix} x + 4 & 5 \ 6 & x \end{pmatrix} \]
The scaling factor from the transformation matrix can be calculated using the determinant of the matrix. The determinant \( \text{det}(T) \) can be expressed as:
\[ \text{det}(T) = (x + 4)(x) - (6)(5) \]
Calculating this gives:
\[ \text{det}(T) = x^2 + 4x - 30 \]
The absolute value of the determinant affects the scaling of the area, so we want the absolute value of \( \text{det}(T) \) to equal the square of the scaling factor:
\[ |\text{det}(T)| = k^2 \]
Since \( k^2 = 2 \), we have:
\[ |x^2 + 4x - 30| = 2 \]
This gives us two cases to solve:
- \( x^2 + 4x - 30 = 2 \)
- \( x^2 + 4x - 30 = -2 \)
Case 1:
\[ x^2 + 4x - 30 - 2 = 0 \] \[ x^2 + 4x - 32 = 0 \]
Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-32)}}{2 \cdot 1} \] \[ x = \frac{-4 \pm \sqrt{16 + 128}}{2} \] \[ x = \frac{-4 \pm \sqrt{144}}{2} \] \[ x = \frac{-4 \pm 12}{2} \]
This gives us:
- \( x = \frac{8}{2} = 4 \)
- \( x = \frac{-16}{2} = -8 \)
Case 2:
\[ x^2 + 4x - 30 + 2 = 0 \] \[ x^2 + 4x - 28 = 0 \]
Using the quadratic formula again: \[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-28)}}{2 \cdot 1} \] \[ x = \frac{-4 \pm \sqrt{16 + 112}}{2} \] \[ x = \frac{-4 \pm \sqrt{128}}{2} \] \[ x = \frac{-4 \pm 8\sqrt{2}}{2} \] \[ x = -2 \pm 4\sqrt{2} \]
Thus, from both cases, we have the solutions for \( x \):
- \( x = 4 \)
- \( x = -8 \)
- \( x = -2 + 4\sqrt{2} \)
- \( x = -2 - 4\sqrt{2} \)
You can choose the value of \( x \) based on any additional context or restrictions provided in the problem. All four values satisfy the transformations regarding areas.