Calculate Rectangle Perimeter with 12-Unit Height and Isosceles Triangle

To solve this problem, we will use the information in the diagram and relevant geometric principles.

Firstly, we consider the triangle $\triangle BCD$ within the rectangle:

• $\overline{BC} = 12$ is one leg of the triangle.
• $\overline{BD} = 13$ is the hypotenuse given in the problem.

Using Pythagoras' Theorem $(\overline{BD})^2 = (\overline{BC})^2 + (\overline{CD})^2$:

$13^2 = 12^2 + (\overline{CD})^2$

$169 = 144 + (\overline{CD})^2$

Subtract 144 from both sides to solve for $(\overline{CD})^2$:

$169 - 144 = (\overline{CD})^2$

$25 = (\overline{CD})^2$

Taking the square root gives:

$\overline{CD} = \sqrt{25} = 5$

The rectangle ABCD's other side, $\overline{AB} = \overline{CD} = 5$, since opposite sides of a rectangle are equal.

Using the perimeter formula for a rectangle:

$P = 2 \times (\overline{AB} + \overline{BC})$

Substitute the known lengths:

$P = 2 \times (5 + 12)$

$P = 2 \times 17$

$P = 34$

Therefore, the perimeter of rectangle ABCD is $\textbf{34}$.