The problem involves two triangles, ABC and DEF, which are mathematically similar. Here's the given information and the task you need to solve:

Given:

• Triangles ABC and DEF are mathematically similar.
• Side AB of triangle ABC has length $2x + 10$ cm.
• Side DE of triangle DEF has length $x$ cm.
• The area of triangle ABC is 60 cm².
• The area of triangle DEF is 15 cm².

To Prove:

$T = p^2 + 2x + 1$

Let's proceed with the solution.

Step 1: Establish the relationship between sides of similar triangles.

Since the triangles are similar, the ratio of their corresponding sides will be the same, and the ratio of their areas will be the square of the ratio of their sides.

Let the ratio of the corresponding sides be $k$: $k = \frac{\text{AB}}{\text{DE}} = \frac{2x + 10}{x}$

Step 2: Set up the ratio of the areas.

Since the triangles are similar: $\frac{\text{Area of } ABC}{\text{Area of } DEF} = k^2$ Substitute the given areas: $\frac{60}{15} = k^2$ Simplify this: $4 = k^2$ $k = 2 \quad (\text{since } k > 0)$

Step 3: Express the ratio in terms of $x$.

$k = 2 = \frac{2x + 10}{x}$

Step 4: Solve the equation for $x$.

Multiply both sides by $x$: $2x = 2x + 10$ Subtract $2x$ from both sides: $0 = 10$

There seems to be an issue with the setup or the values provided in the problem, as $0 = 10$ is not a valid equation. I might need to recheck the problem's conditions or assumptions. Could you verify if there is any other context or information available that could help solve this?

Would you like to explore any part of this problem further, or do you want to try another problem?