The triangle is isosceles. The height of the triangle is therefore given by $h = \sqrt{10^2 - \left( \dfrac{12}{2} \right)^2} = \sqrt{64} = 8$
Now, the area of the triangle is $\dfrac{bh}{2} = \dfrac{12*8}{2} = \dfrac{96}{2} = 48$
We have that the area of the rectangle is the same as the area of the triangle, namely $48$. We also have the width of the rectangle: $4$.
The length of the rectangle therefore is:
$l = \dfrac{48}{4} = 12$
The perimeter of the rectangle then becomes:
$2l + 2w = 2*12 + 2*4 = 32$
The answer is:
$\boxed{\textbf{(D)}\ 32}$

1. Calculate the area of the triangle using Heron's formula.

Heron's formula states that the area $A$ of a triangle with side lengths $a$, $b$, and $c$ is given by:
$$A = \sqrt{s(s-a)(s-b)(s-c)}$$
where $s$ is the semi-perimeter of the triangle:
$$s = \frac{a + b + c}{2}$$

For our triangle with side lengths 10, 10, and 12:
$$s = \frac{10 + 10 + 12}{2} = 16$$

Now, substitute $s$, $a$, $b$, and $c$ into Heron's formula:
$$A = \sqrt{16(16-10)(16-10)(16-12)} = \sqrt{16 \cdot 6 \cdot 6 \cdot 4}$$
$$A = \sqrt{16 \cdot 6^2 \cdot 4} = \sqrt{16 \cdot 36 \cdot 4} = \sqrt{16 \cdot 144} = \sqrt{2304} = 48$$

2. Determine the dimensions of the rectangle.

The area of the rectangle is equal to the area of the triangle, which is 48. Given that the width of the rectangle is 4, we can find the length $l$ of the rectangle using the area formula for a rectangle:
$$\text{Area} = \text{width} \times \text{length}$$
$$48 = 4 \times l$$
$$l = \frac{48}{4} = 12$$

3. Calculate the perimeter of the rectangle.

The perimeter $P$ of a rectangle is given by:
$$P = 2 \times (\text{length} + \text{width})$$
Substituting the values we found:
$$P = 2 \times (12 + 4) = 2 \times 16 = 32$$

The final answer is $\boxed{32}$