A cardboard box measures 12 inches high, 7 inches wide, and 9 inches deep. If another box measuring \(3 \frac{1}{2}\)by \(4 \frac{1}{2}\) inches is to have the same surface area, what will the box’s third dimension need to be?


The original box has panels of \(7 \times 12=84\); \(9 \times 12 = 108\) and \(9 \times 7 = 63\), which add up to 255 square inches for half the panels. 

The new box has panels of   \(3 \frac{1}{2} \times h\), \(4 \frac{1}{2} \times h\), and \(3 \frac{1}{2} \times 4 \frac{1}{2} = 15 \frac{3}{4}\) square inches for half the surface area. 

Adding the new box’s dimensions means that \(8h + 15 \frac{3}{4} = 255\). Then, \(8h=239 \frac{1}{4}\)square inches, and \(h=29 \frac{29}{32}\)inches.

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