Any tips would be greatly appreciated. (this is from The Big Book, p.100)
you can use the 45:45:90 ratio to see the relationship b/w side and diagonal of a square, it’s \mathrm{(side \times \sqrt2)}.
I’ve been trying to figure this out and I’m stuck on how we know the diagonal of the small square is 1.
I know that the diagonal of the large square is √2 but can’t get the rest.
Hey there. See, if the area of the smaller square, say B is 0.5 times of the larger one, say A, then B= 0.5A, which means the side length for B is 1/sqrt(2) or sqrt(2)/2. Hence, the diagonal of B would be 1. Rest is easy I guess.
I get the answer as 2 - sqrt(2)
Let the area of smaller square be 1 so diagonal is root(2)
The sides of the bigger square will be root (2) to get area 2 . so the diagonal of larger square is 2
So diff is 2 - sqrt(2)
But there’s not a matching answer ?
First, if we look at the given things here,
It is given that the area of the smaller square is 1/2 of the size of the larger square, and the side of the larger square is given.
So we can calculate the area of the larger square and the area of the smaller square.
Once we get the area of the smaller square, we can get the side of the smaller square.
Now, we can directly use the diagonal of the square formula (side * sqrt(2)) or we can divide the square into a 45-45-90 triangle with its diagonal. We know that in a 45-45-90 triangle, the hypotenuse is sqrt(2) * side, or we could use Pythagoras’ theorem to find the answer manually.
Once we apply this to smaller and bigger triangle and subtract them, we’ll get the answer.