Eight squares are packed into a rectangle as shown. If the smallest square has a side length of $4$, find the side length of the biggest square. Show hint

Try to use the information in the diagram to write an equation you can solve.

The diagram tells us the relationship between side lengths of the squares. For example, the top left square’s side length and the smallest square’s must add up to the bottom left square’s side length because of the shared line: Since we’re trying to find the side length of the biggest square, we might as well give it a variable: $x$. Then we can express all the other squares’ side lengths in terms of $x$, starting with the top left square’s, which has to be $x-4$. (You might have decided to give a variable name to a different square’s side length instead, for example the top left square; that’s fine too.)

Now the top square has to have a side length of $x-8$, and the top right square must have the same because they share a side.

The next two small squares have a side length of \[\frac{2\times(x-8) - 4}{2}\] which simplifies to $x-10$, but if it wasn’t clear to you how to find that then you could assign that length a variable, $y$, and then write down an equation like $2y + 4 = 2\times(x-8)$ and solve it for $y$. The bottom right square must have a side length of $2\times(x-10) = 2x - 20$.

Now, how do we use all this to find $x$? There is some information in the diagram that we haven’t used yet. We could look at the central vertical line, giving \[(x) + (4) = (x-10) + (2x - 20)\] or we could equate the two vertical sides of the rectangle, giving \[(x-4) + (x) = (x-8) + (x-10) + (2x-20).\] In either case, solving the equation gives $x = 17$, which is the side length of the biggest square as required.