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Let $k=\lceil \sqrt{N}\rceil$ be the least integer no less than $\sqrt{N}.$ If we could use real numbers for the sides then the least perimeter would be $4\sqrt{N}$ coming from a square. But we will have $E(N)$ an even integer no smaller than $4\sqrt{N}$. Let with $k=\lceil E(N) \sqrt{N}\rceil$ be the least integer no less than $\sqrt{N}.$ ge 4\sqrt{N}$.

A short answer is that the lower bound is sharp: $E(N)$ is the least even integer no less than $4\sqrt{N}.$ This means that $E(N)=4k-2$ when $(k-1)^2 \lt N \le (k-1)k$ and $E(N)=4k$ when $(k-1)k \lt N \le k^2$. The tile is unique when $N=k^2,k^2-k-1$ and $k^2-k$ but otherwise there can sometimes be many tiles, especially for $N=k^2+1$ and $N=k(k-1)+1$.

At the end is a tile of area $N=1025$ with perimeter $130$. It arises from appropriately removing $8+10+2+1=21$ cells from the corners of a $31 \times 32$ rectangle. The polyomino and bounding rectangle both have perimeter $130.$ The tiles with perimeter $E(N)$ will be exactly things the polyominoes obtained like this: Start with a rectangle of area $ab \ge N$ and perimeter $2a+2b=E(N)$, then appropriately remove $ab-N$ cells from one or more of the corners. In the case of $N=1025$ the bounding rectangle could be $28 \times 37$ with $11$ cells removed or even $28 27 \times 37$ 38$ less one corner cell. Any dimensions $(a,b)=(32-j,33+j)$ with $ab \ge N$ would be possible.

The bounding box of a polyomino $P$ is the minimal rectangle which completely contains it (so all 4 sides of the box share an edge with $P$.) If the area of $P$ is $N$ and the bounding box has dimensions $a \times b$ b,$ then clearly, $ab \le ge N$. Also $E(N) \ge 2a+2b.$ This is because when we walk around the boundary of $P$, an edge at a time, we go up at least $a$ times, down at least $a$ times, and left and right at least $b$ times each. We will assume that $a \le b$ since we can always rotate $P.$

Depending on $N$ there can may be many or few choices of $a,b$ with $N \le ab$ and $2a+2b$ minimal. 2a+2b=E(N)$. Consider again $N=1025=25 \cdot 401$ 401.$ If $ab \ge N=1025$ N=1025,$ then $2a+2b \ge 130$ since the minimum over real values is $4\sqrt{1025} \gt 128.$ But we need an even integer. So any of $(a,b)=(32-j,33+j)=(32,33),(31,34),(30,35),(29,36),(28,37),(27,38)$ are possible. $27 \cdot 38=1026$ is big enough but $26 \cdot 39=1014$ would not be. In general, to have area at least $N$ in a box with $2a+2b =4k+2$, we have $(a,b)=(k-j,k+1+j)$ with area $N \le k^2+k-j^2-j$ so $0 \le j \le \frac{-1+\sqrt{4(k^2-+k-N)+1}}{2}.$ frac{-1+\sqrt{4(k^2+k-N)+1}}{2}.$ The calculations for $(a,b)=(k-j,k+j)$ when $E(n)=4k$ are similar.

Even though we could get area exactly $1025$ with a $25 \times 401$ rectangle, the perimeter of $851$ is much worse than $130.$ Below is Now we have arrived at a blue polyomino $P$ with area $1025$ which fits in a $32 \times 33$ bounding box. The $8+10+2+1=21$ black squares are in the bounding box but are not part of $P$.

So

For any $N$, the possible tiles polyominoes with area $N$ and perimeter $E(n)$ will be anything in an appropriate those obtained from a bounding box rectangle of area $ab \ge N$ with a total of $ab-N$ cells removed in this fashion from some or all the corners.
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Your

The question isnot very clear to me. However : Consider all the question of when a set of polyominoes tile a rectangle is not always easy. Much is known for small ones $P$ made with $N$ unit squares (thanks to lots cells). Let $E(N)$ be the least perimeter (i.e. number of computer time and clever ideasexternal edges) but much is notamong them. Here What is $E(N)$ and which polyominoes attain it?

If we could use real numbers for the sides then the least perimeter would be $4\sqrt{N}$ coming from a diagram showing square. But we will have $E(N)$ an even integer no smaller than $4\sqrt{N}$. Let $k=\lceil \sqrt{N}\rceil$ be the least integer no less than $\sqrt{N}.$

A short answer is that $10$ copies of E(N)$ is the Y-pentominoleast even integer no less than $4\sqrt{N}.$ This means that $E(N)=4k-2$ when $(k-1)^2 \lt N \le (k-1)k$ and $E(N)=4k$ when $(k-1)k \lt N \le k^2$. The tile is unique when $N=k^2,k^2-k-1$ and $k^2-k$ but otherwise there can sometimes be many tiles, especially for $N=k^2+1$ and $N=k(k-1)+1$. At the end is a tile of area $N=1025$ with perimeter $130$. It arises from appropriately removing $8+10+2+1=21$ cells from the corners of a $5 31 \times 10$ box. 32$ rectangle.

Would you say that The polyomino and bounding rectangle both have perimeter $130.$ The tiles with perimeter $E(N)$ will be exactly things like thispentomino has . In the case of $8$ sides N=1025$ the bounding rectangle could be $28 \times 37$ with $11$ cells removed or even $14$? Either way I am not sure how you are getting that 28 \times 37$ less one corner cell. Any dimensions $N$ pentominoes have (a,b)=(32-j,33+j)$ with $4N$ sides, unless they are all rectanglesab \ge N$ would be possible. Can Note that the transition from $15$ Y-pentominoes fill E(N)=4k-2$ to $E(N)=4k$ happens at $N=k^2-k+1$ since $4\sqrt{k^2-k}=2\sqrt{4k^2-4k} \lt 4k-2$ but $4\sqrt{k^2-k+1}=2\sqrt{4k^2-4k+4} \gt 4k-2.$ It is clear why the transition form $4k$ to $4k+2$ is at $N=k^2+1.$

The bounding box of a polyomino $P$ is the minimal rectangle ? It turns out that which completely contains it (so all 4 sides of the answer box share an edge with $P$.) If the area of $P$ is no and also no for $25,35$ N$ and any other odd number under the bounding box has dimensions $45.$ It turns out that a \times b$ then clearly, $45$ ab \le N$. Also $E(N) \ge 2a+2b.$ This is because when we walk around the boundary of them can tile $P$, an edge at a time, we go up at least $15 \a$ times15$ square, down at least $a$ times, and left and right at least $b$ times each. If you look on that site with this diagram you We will find many nice tilings. Also assume that $a \le b$ since we can always rotate $P.$

Depending on $N$ there can be many open questionsor few choices of $a,b$ with $N \le ab$ and $2a+2b$ minimal. It Consider $N=1025=25 \cdot 401$ If $ab \ge N=1025$ then $2a+2b \ge 130$ since the minimum over real values is unknown if $4\sqrt{1025} \gt 128.$ But we need an odd number of copies even integer. So any of this $14$-omino alt text can ever tile a rectangle(a,b)=(32-j,33+j)=(32,33),(31,34),(30,35),(29,36),(28,37),(27,38)$ are possible. The $27 \cdot 38=1026$ is big enough but $26 \cdot 39=1014$ would not be. In general, to have area at least which can tile $N$ in a box with $2a+2b =4k+2$, we have $(a,b)=(k-j,k+1+j)$ with area $N \le k^2+k-j^2-j$ so $0 \le j \le \frac{-1+\sqrt{4(k^2-+k-N)+1}}{2}.$ The calculations for $(a,b)=(k-j,k+j)$ when $E(n)=4k$ are similar.

Even though we could get area $1025$ with a $25 \times 401$ rectangle, the perimeter of $851$ is much worse than $396$ 130.$ Below is a blue polyomino $P$ with area $1025$ which fits in a $32 \times 33$ bounding box. The $8+10+2+1=21$ black squares are in the bounding box but are not part of $P$.

So the possible tiles will be anything in an appropriate bounding box with $ab-N$ cells removed in this fashion from some or all the corners.
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Your question is not very clear to me. However the question of when a set of polyominoes tile a rectangle is not always easy. Much is known for small ones (thanks to lots of computer time and clever ideas) but much is not. Here is a diagram showing that $10$ copies of the Y-pentomino can tile a $5 \times 10$ box.

alt text.

Would you say that this pentomino has $8$ sides or $14$? Either way I am not sure how you are getting that $N$ pentominoes have $4N$ sides, unless they are all rectangles. Can $15$ Y-pentominoes fill a rectangle? It turns out that the answer is no and also no for $25,35$ and any other odd number under $45.$ It turns out that $45$ of them can tile a $15 \times 15$ square. If you look on that site with this diagram you will find many nice tilings. Also many open questions. It is unknown if an odd number of copies of this $14$-omino alt text can ever tile a rectangle. The least which can tile a rectangle is $396$ .