An 8x8 square is allowed to slide around in a 10 x 18 rectangle, as shown below. The top vertex of the square always stays on the top edge of the rectangle, and the bottom vertex of the square always stays on the bottom edge of the rectangle. What is the area of the region inside the rectangle that the square can never cover?

ThamHamkid Jun 21, 2023

#1**0 **

The area of the region inside the rectangle that the square can never cover is the area of a rectangle with dimensions 10 x 8, minus the area of the square.

The area of the rectangle is 10 x 8 = 80. The area of the square is 8 x 8 = 64.

Therefore, the area of the region inside the rectangle that the square can never cover is 80 - 64 = 16.

Guest Jun 21, 2023

#2**+1 **

You would be right if the problem said that the square is stationary inside the rectangle. That's wrong because the square can slide, and it would cut a swath as it slides from one end of the rectangle to the other. Besides that, your arithmetic is wrong because you called the rectangle 10x8 but the problem says 10x18.

Guest Jun 21, 2023

#4**0 **

I can't post pictures, so you'll have to visualize this along with me.

The top and bottom corners of the square are secured to the top and bottom of the rectangle, respectively. They can slide along the length of the rectangle, but they can't come loose.

Push the square all the way to the left, so that its left corner is touching the left side of the rectangle. You have just formed two triangles, whose area can never be covered by the square. If we can figure out the area of those triangles, we'll be on our way to making an A in geometry.

Notice that both triangles are right triangles, and they both have a hypotenuse that's 8 units long. There's only a single configuration that accomplishes that, therefore the triangles are congruent.

Take that bottom triangle and, hinging it on its sharpest-pointed angle, rotate it counterclockwise until its hypotenuse lies exactly on top of the hypotenuse of the top triangle. You have just formed a rectangle, and we already know the length of its diagonal. Here's the magic step. Look it up if you don't believe me.

The area of a rectangle is equal to half the square of its diagonal: A = (1/2)(d^{2}). We can calculate the area of the small rectangle that you so cleverly formed, thus: A = (1/2)(8^{2}) = (1/2)(64) = 32. Now double that because we have to do the same thing when we slide the square to the other end.

The question was *"What is the area of the region inside the rectangle that the square can never cover?"*

The answer is, **64 square units**. Hmm, that's the same as the area of the square. I wonder if it's a coincidence.

~ edited to correct a couple of typos ~

_{.}

Bosco Jun 22, 2023

#5**+1 **

The top and bottom vertices of the square, (as seen in the diagram), remain in contact with the top and bottom sides of the rectangle respectively, so the orientation of the square will be the same as it slides from one end of the rectangle to the other.

Slide the square to the left so that the left hand vertex of the square comes into contact with the left hand side of the rectangle.

Two triangles are formed and their interiors will be will be areas of the rectangle that can't be covered by the square.

The two triangles are congruent. (Call one of the angles of one of the triangles theta, and work out, in terms of theta, the other three angles of the two triangles. That will show that they are similar. They have the same hypotenuse so they are congruent.)

Now slide the triangle to the right so that the right hand vertex of the square comes into contact with the right hand side of the rectangle. Again two triangles are formed and they will congruent with the two triangles on the left.

The total area of the rectangle that can't be covered by the square will be the combined area of the four triangles, which will be four times the area of a single triangle because of the congruency. Notice that the second dimension of the rectangle, the 18, is irrelevant,(so long as it's big enough for the rectangle to contain the square).

The hypotenuse of the triangle(s) is 8, let the lengths of the other two sides be x and y, then

\(x^{2}+y^{2}=64, \\ x + y = 10.\)

Solving simultaneously,

\(x = 5 \pm \sqrt{7}, \quad y = 5 \mp\sqrt{7},\)

so the area of a single triangle will be

\((5+\sqrt{7})(5-\sqrt{7})/2=(25-7)/2=9,\)

meaning that the total area that can't be covered by the square is 36.

Tiggsy Jun 22, 2023

#6**0 **

Thank You guys for your help! It turned out the correct answer was 36 sq units.

ThamHamkid Jun 22, 2023