Check the picture below.
let's recall that a rhombus diagonals intersect at right-angles and that in a rhombus, slanted as it may be, the sides are all equal.
so we can find the area of one the triangles in the rhombus by simply using a base of 3 and a height of 4, and since there are four of those triangles, 4 times that much is the area of one of the rhombic faces.
as you see in the picture, if we use the pythagorean theorem, we end up with a hypotenuse of 5, which is the length of the base sides on this prism, which also has four rectangular faces standing up vertically, each face is a 5x12 rectangle.
[tex]\bf \stackrel{\textit{area of one triangle in the rhombus}}{\cfrac{1}{2}(3)(4)\implies 6}\qquad \stackrel{\textit{all 4 triangles}}{4(6)\implies 24}\qquad \stackrel{\textit{area of both rhombuses}}{2(24)\implies 48} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large area of the prism}}{\stackrel{\textit{area of the rectangles}}{4(5\cdot 12)}~~+~~\stackrel{\textit{area of the rhombuses}}{48}\implies 240+48\implies 288 }[/tex]
