Bunuel wrote:

How many tiles of the size and shape shown above are needed to completely cover a rectangular floor measuring 30 feet by 20 feet?

(A) 120

(B) 150

(C) 200

(D) 250

(E) 300

Attachment:

2017-10-04_1121_003.png

I think we must assume that there is a large square with side length = 2, and a missing piece that is also a square, with side length = 1.

We do not know if the figure has right angles. Two adjacent equal sides, without more, cannot render, definitively, these figures as squares.

The "more" needed to be certain that, e.g., the larger figure (with half the right and bottom sides grayed out) is a square (not exhaustive):

1) it is a rectangle (rectangle with equal adjacent sides is a square), or

2) it is a parallelogram with two equal adjacent sides and one right angle at a vertex; or

3) 3 of the four vertices are right angles

Am I missing something? I cannot see a way to prove these figures are squares.

Assume the large figure is a square (which the prompt seems to suggest). Its side length = 2, its uncut area is (2*2) = 4

Asssume also that the large square has a square piece cut out.

The square piece cut out has side length = 1, area = 1

Total area of the tile is 4 - 1 = 3

The tiles are all the same shape. If the tile shown were rotated 180° (or, flipped on the diagonal, had its left lower corner where its right corner is), it would interlock with itself to form a rectangle with length = 3 and width = 2.

Area of rectangle formed by two interlocked tiles =

(2) * (tile with area 3) = 6, OR

Rectangle formed by interlocked tiles has area of LW = (2*3) = 6

Floor area = LW = (30*20) = 600

Floor area / small rectangles' area:

600/6 = 100 tiles of area 6

But it takes TWO tiles to make the rectangular area of 6

So 100 rectangles * 2 tiles per rectangle = 200 tiles

Answer C

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that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"