Colorful Prisms

As shown in the picture, the surface area of the bases doesn't change in either formation. In other words, the increased surface area comes from the additional lateral areas.

Let $x$ = the side length of the hexagon, $h$ = height of each prism = $\sqrt{3}$ , and $B$ = the area of the hexagon base = $\dfrac{3\sqrt{3}(x^2)}{2}$ .

Now in the first formation, the lateral surface of hexagonal prisms consists of 12 rectangular areas of sides $x$ and $h$ . On the other hand, in the latter formation, the lateral surface of triangular prisms will consist of 18 rectangular areas; there are 6 rectangles in each of the 3 prisms.

Hence, from the 12.5% increase,

$\dfrac{\text{surface area of 3 triangular prisms}}{\text{surface area of 2 hexagonal prisms}}$ = $\dfrac{4B + 18xh}{4B + 12xh} = 1.125$

Then, $4B + 18xh = 4.5B + 13.5xh$

$4.5xh = 0.5B$

$9xh = 9\sqrt{3}(x) = B = \dfrac{3\sqrt{3}(x^2)}{2}$ .

$3x = \dfrac{x^2}{2}$

Thus, $x = \boxed{6}$ $cm.$

Let $x$ be the side length of the hexagonal prism, then the side length of the triangular prism is $2x$ . The surface area of one hexagonal prism is equal to the perimeter of the base multiplied by the height plus twice the area of the base. We have $6x\sqrt{3}+2\left(\dfrac{3}{2}\sqrt{3}x^2\right)=6x\sqrt{3}+3\sqrt{3}x^2$ . So the surface area of the two hexagonal prisms is $12x\sqrt{3}+6\sqrt{3}x^2$ . The surface area of one triangular prism is equal to the perimeter of the base multiplied by the height plus twice the area of the base. We have $6x\sqrt{3}+2\left(\dfrac{\sqrt{3}}{4}(2x)^2\right)=6x\sqrt{3}+2\sqrt{3}x^2$ . So the surface area of the three triangular prisms is $18x\sqrt{3}+6\sqrt{3}x^2$ . From the problem statement, we have

$1.125(12x\sqrt{3}+6\sqrt{3}x^2)=18x\sqrt{3}+6\sqrt{3}x^2$

Dividing both sides by $1.125$ , we get

$12x\sqrt{3}+6\sqrt{3}x^2=16x\sqrt{3}+\dfrac{16}{3}\sqrt{3}x^2$

Simplifying further, we get $\boxed{x=6~cm}$ .

let $x$ be the side length of the hexagon

Surface area of one hexagon $=6x\sqrt{3} + (6x^2)\frac{\sqrt{3}}{2}=6x\sqrt{3} + 3x^2\sqrt{3}$

Surface area of two hexagons $=12x\sqrt{3} + 6x^2\sqrt{3})$

Surface area of one equilateral triangle $=6x\sqrt{3} + (4x^2)\frac{\sqrt{3}}{2}=6x\sqrt{3} + 2x^2\sqrt{3}$

Surface area of three equilateral triangles $=18x\sqrt{3} + 6x^2\sqrt{3}$

$(1.125)(12x\sqrt{3} + 6x^2\sqrt{3}) =$ $18x\sqrt{3} + 6x^2\sqrt{3}$

$0.75x=4.5$

$x=6cm$