# $2$-inch squares are cut from the corners of a $10$-inch square. What is the area of the largest square that fits the remaining space?

$$2$$-inch squares are cut from the corners of this $$10$$-inch square. What is the area in square inches of the largest square that can be fitted into the remaining black colored space? I approached this problem in this way:

The biggest possible square we can get if the square is tilted. Now if i construct a square with the midpoints then the area of that inscribed square is 50 inches. But if a tilted inscribed square goes through the vertices of the small squares then the square can have highest 60 inches of area (the area of the square LJSF in the below figure is 36 inches. With extra area of 4 triangles like $$\triangle ULJ$$ which have base of 6 inches and height of 2 inches.) But i coudn't find a way to construct it So my question is:

1. Is the square having area 60 is possible to construct inside the big square?

2. if not then What is the biggest possible square that can be inscribed in that black region

You'll want something like this: i.e. where the side of the inscribed square goes through the corner points of the little cut out squares. And you find that the area is the middle $$6\times 6$$ square plus exactly half of the area of all the black and blue triangles (since for every blue triangle there is a corresponding black triangle). So, the area is:

$$36 + \frac{4 \cdot 2 \cdot 6}{2}=36+24=60$$

Here's a graphical assist from @Blue: • +1. You can simplify the calculation ever-so-slightly by noting that the blue triangles on, say, the left and right can fill the black triangles at the top and bottom, giving a blue $6\times 10$ rectangle. – Blue Oct 8 '19 at 18:59
• @Blue Yeah, I thought of making a picture of exactly that ... but here my drawing skills failed me :P Or rather: I'd have had to redraw the whole thing and was too lazy :( – Bram28 Oct 8 '19 at 19:02
• I hope you don't mind that I added an image. :) – Blue Oct 8 '19 at 19:51
• @Blue Not at all .. thanks! – Bram28 Oct 8 '19 at 19:53

The blue part has a total area of $$2\times 2\times 6 = 24$$, Thus, the largest area that can still fit is

$$10^2 - 4\times 2^2 - 24 = 60$$ • Can you tell me what is the ratio of the area of the big and small triangle? – sakib Oct 8 '19 at 18:06
• The ratio is $\frac{3+\sqrt 5}{3-\sqrt 5}$. But, you don’t need it for the answe. – Quanto Oct 8 '19 at 18:16