Five Dimensions

Just as the side length of a square is $\sqrt{area}$ and the side length of a cube is $\sqrt[3]{volume}$ , we can find the side length of our hypercube with $\sqrt[5]{32}=2$ . Setting the center of our hypercube to the relative coordinate (0, 0, 0, 0, 0) then lets us set a vertex coordinate at (1, 1, 1, 1, 1), since 1 is half of the side length. We can finally use the generalized distance formula to find the distance between these two points $\sqrt{{1}^{2}+{1}^{2}+{1}^{2}+{1}^{2}+{1}^{2}}=\boxed{\sqrt{5}}$

a much easier solution to this problem is implied by repeating patterns in the value of the diagonal. For a 2d cube, i.e, a square, the diagonal is sqrt(2) times the side, for a 3d cube its sqrt(3) times the side, for a 4d cube (tesseract) its sqrt(4) or twice the edge (you can figure it out with just a little insight!!!). so for a 5d cube, which we obviously can't picture, the diagonal has to be sqrt(5) times the edge which in this problem would be 2. but the given distance is only half the diagonal, so d=2xsqrt(5)/2=sqrt(5) =>this value is given to be sqrt(a) so the value of a=5

d=2xsqrt(5)/2=sqrt(5) =>this value is given to be sqrt(a) so the value of a=5

Let L = lenght of edge of the hypercube C = distance of vertices to the center (origin) Volume of hypercube is L^5 = 32, so L = 2 Distance of vertices from the center is C = sqrt (a) = L × sqrt (5) / 2 = sqrt (5); therefore a= 5