Box dimensions from the diagonal and one edge

Text

In a rectangular box the diagonal is $13$ cm, one dimension is $12$ cm and, of the other two, one is $\dfrac34$ of the other. Find the three dimensions and the total surface area.

Solution

From the diagonal $a^2+b^2+c^2=13^2=169$. With $a=12$: $b^2+c^2=169-144=25$. Setting $b=\dfrac34c$: $\ \dfrac{9}{16}c^2+c^2=\dfrac{25}{16}c^2=25\Rightarrow c^2=16\Rightarrow c=4,\ b=3.$ Dimensions $12,4,3$. Total surface area: $S=2(12\cdot4+12\cdot3+4\cdot3)=2(48+36+12)=192\ \text{cm}^2.$ $\boxed{12,\ 4,\ 3\ \text{cm};\ \ S=192\ \text{cm}^2}$