1. ## Rectangles Perimeter??

The pattern below is composed of rectangles. This pattern is used repeatedly to completely cover a rectangular region 12L units long and 10L units wide. How many rectangles of dimension L by W are needed?

- - - - - - -
550!

2. Didn't we just do this somehwere else?

You need three things:

1) Conversion: Looking at the vertical labels in your not-to-scale drawing, one sees that 3W = 2L or W = (2/3)L. This makes the width of your figure L + (2/3)L = (5/3)L.

2) Reassurance: As luck would have it, in a textbook problem, we have 10L / [(5/3)L] = 6 AND 12L / 2L = 6. This means we CAN tile it without cutting anything. This also suggests how to orient the figure in the final construction.

3) Direction: 550 is WAY to many. Just adding the area of the pieces, 550*L*W = 550*L*(2/3)L = [366.66...]L^2 total area. Whereas, (10L)*(12L) = 120L^2. With 550, you can do the whole job THREE TIMES and have a little left over.

3. Hello, fabxx!

An elaboration of TKHunny's solution . . .

The pattern below is composed of rectangles.
The rectangle is used repeatedly to cover a rectangular region $12L \times 10L.$
How many rectangles of dimension $L \times W$ are needed?
Code:
          W         L
* - - - * - - - - - *
|       |           |
|       |           | W
L |       |           |
|       * - - - - - *
|       |           |
* - - - *           | W
|       |           |
|       * - - - - - *
L |       |           |
|       |           | W
|       |           |
* - - - * - - - - - *

As TKHunny pointed out: . $2L = 3W \quad\Rightarrow\quad W = \frac{2}{3}L$

So the width of the rectangle is $\frac{5}{3}L$ . . . its height is $2L.$

Since $6\times \frac{5}{3}L \:=\:10L$, we can put 6 rectangles in a row to cover $10L.$

Since $6 \times 2L \:=\:12L$, we can make 6 rows to cover $12L.$

Therefore, it takes: $6 \times 6 \:=\:{\color{blue}36}$ rectangles to cover the region.

4. 36 of the GROUP rectangles, making it 36*5 = 180 rectangles of size L*W.

As stated previously, 550 >> 180.