Tiling rectangles with Hexomino plus rectangle #2

Question

Inspired by Polyomino T hexomino and rectangle packing into rectangle

See also series Tiling rectangles with F pentomino plus rectangles and Tiling rectangles with Hexomino plus rectangle #1

Previous puzzle in this series: Tiling rectangles with Hexomino plus rectangle #1

Next puzzle in this series: Tiling rectangles with Hexomino plus rectangle #3

The goal is to tile rectangles as small as possible with the given hexomino, in this case number 2 of the 25 hexominoes which cannot tile a rectangle alone. We allow the addition of copies of a rectangle. For each rectangle $a\times b$, find the smallest area larger rectangle that copies of $a\times b$ plus at least one of the given hexomino will tile.

Example shown, with the $1\times 1$ (or the $1\times 2$) you can tile a $2\times 4$ as follows:

Now we don't need to consider $1\times 1$ (or $1\times 2$) further as we have found the smallest rectangle tilable with copies of the hexomino plus copies of $1\times 1$ (or $1\times 2$).

I have tilings for another nine rectangles: Hexomino plus $1\times 3$, $1\times 4$, $1\times 5$, $1\times 6$, $1\times 7$, $1\times 8$, $2\times 2$, $2\times 3$, $2\times 4$

No Computer All of them could be tiled by hand (with significant effort in some cases), so I'm making this a no-computer puzzle. This also means please don't look up answers on the web... if you post an answer it should be because you found it 'by hand'. This does not preclude you from for example using an image program to manipulate shapes on the screen, just from using a computer to search for or automate the arrangement.

Answer 1

Here are the first few in the order I found them.

$2\times2$

$4\times4=16$ square

$1\times3$

$6\times6=36$ square

$2\times3$

$10\times12=120$ rectangle

$2\times4$

$10\times14=140$ rectangle

$1\times4$ (Updated)

$7\times10=70$ rectangle

$1\times7$

$10\times10=100$ square

Answer 2

Here's a smaller solution for $2 \times 3$:

8x12=96

and Jaap's $1 \times 7$ solution was a great help in finding one for $1 \times 6$:

10x12=120

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Here's a (computer-assisted, I admit) generalizable solution for $1 \times n = 4k$. By subdividing the rectangles, you can find solutions for $n$ not divisible by $4$. In those cases, the 'padding' (i.e. areas with just rectangles) don't need to be so large.

When $n$ is divisible by $4$, $1 \times n$ fits in a $2n \times 2n+1$ (left)
When $n$ is even but not divisible by $4$, $1 \times n$ fits in a $2n \times 3n+1$ (right)
When $n$ is odd, $1 \times n$ fits in a $2n \times 5n+1$