Find two consecutive integers whose squares sum to 61.

Let the first integer be $x$.
The next consecutive integer is $x+1$.

The sum of their squares is 61.

Subtract 61 from both sides:

We need two numbers that multiply to -30 and add to 1. These numbers are 6 and -5.

Set each factor to zero:

If $x = 5$, the integers are 5 and $5+1 = 6$.
Check: $5^2 + 6^2 = 25 + 36 = 61$.

If $x = -6$, the integers are $-6$ and $-6+1 = -5$.
Check: $(-6)^2 + (-5)^2 = 36 + 25 = 61$.

The two integers are either $\boxed{\text{5 and 6}}$ or $\boxed{\text{-6 and -5}}$.

Let the length of the rectangle be $L$ cm.
The width is 6 cm shorter than the length, so the width $W = (L-6)$ cm.

The area of a rectangle is given by $A = L \times W$.
Given $A = 27 \text{ cm}^2$.

Rearrange into a standard quadratic form:

We need two numbers that multiply to -27 and add to -6. These numbers are -9 and 3.

Set each factor to zero:

Since length cannot be negative, we take $L = 9$ cm.

The perimeter of a rectangle is given by $P = 2(L+W)$.

The perimeter of the rectangle is $\boxed{\text{24 cm}}$.