Tiling rectangles with U pentomino plus rectangles

Question

Inspired by Polyomino Z pentomino and rectangle packing into rectangle

Also in this series: Tiling rectangles with F pentomino plus rectangles

Tiling rectangles with N pentomino plus rectangles

Tiling rectangles with T pentomino plus rectangles

Tiling rectangles with V pentomino plus rectangles

Tiling rectangles with W pentomino plus rectangles

Tiling rectangles with X pentomino plus rectangles

The goal is to tile rectangles as small as possible with the U pentomino. Of course this is impossible, so 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 U-pentomino will tile. Example shown, with the $1\times 1$, you can tile a $2\times 3$ as follows:

Now we don't need to consider $1\times 1$ any longer as we have found the smallest rectangle tilable with copies of U plus copies of $1\times 1$.

There are at least 6 more solutions. I tagged it 'computer-puzzle' but you can certainly work some of these out by hand. The larger ones might be a bit challenging.

Answer 1

Here is a way to tile a

6x13 = 78

rectangle with U pentominoes and 1x4 rectangles, which is an improvement over @athin's 9x10 solution:

As a bonus, here are two suboptimal solutions, one of which is asymmetric:

link to two 11x8 = 88 solutions

For 1x5:

12x20 = 240

for 1x6:

14x24 = 336

and for 3x4:

19x40 = 760

Answer 2

Beside Riley's 2 solutions:

$ 1 \times 4 $

Area: $10 \times 9 = 90$

$ 2 \times 3 $

Area: $10 \times 7 = 70$

Answer 3

Here's two to get this one started

$3\times 4$ using a $1\times 2$

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$7\times 4$ using a $1\times 3$

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