PlatformSA

SOLUTION 1:

The top and bottom surfaces of the platform each are $1ft \times 4ft = 4ft^2 \rightarrow 8ft^2$ for both.
The two long edge surfaces of the platform each are $\frac{1}{4}ft \times 4ft = 1ft^2 \rightarrow 2ft^2$ for both.
The two short edge surfaces of the platform each are $\frac{1}{4}ft \times 1ft = \frac{1}{42}ft^2 \rightarrow \frac{1}{2}ft^2$ for both.

So the total SA is $8ft^2 + 2ft^2 + \frac{1}{2}ft^2 = 10.5ft^2$

SOLUTION 2:

But my favorite way to see it is to calculate the difference between the original SA of the cube ( $6ft^2$ ) and the new SA. The original cube foot of wood has $SA = 6ft^2$ because it has six faces, each $1 ft^2$ . Imagine painting it blue and then cutting it up to make the platform.

1) For each of the 3 cuts, 2 new $1ft^2$ surfaces are created/revealed by the cut. That's 6 new $ft^2$ total.

2) Then, when those 4 pieces are joined together in three places on $1 \times \frac{1}{4} \hspace{2mm} ft^2$ edges, that similarly means 6 of those faces are covered up. $6 \times \frac{1}{4} \hspace{2mm} ft^2 = 1.5 ft^2$ lost.

$6ft^2 \hspace{2mm} original + 6ft^2 \hspace{2mm} gained - 1.5ft^2 \hspace{2mm} lost = 10.5 ft^2$