Truncated pyramid

The concrete pedestal in the shape of a regular quadrilateral truncated pyramid has a height of 12 cm, the pedestal edges have lengths of 2.4 and 1.6 dm. Calculate the surface of the base.

Correct result:

S =  1843.9289 cm²

Solution:

$$\begin{gathered} h=12\text{ cm } \\ a=2.4\text{ dm}\to \text{ cm}=2.4\cdot 10\text{ cm}=24\text{ cm } \\ b=1.6\text{ dm}\to \text{ cm}=1.6\cdot 10\text{ cm}=16\text{ cm } \\ {S}_1=a^2=24^2=576{\text{cm}}^2 \\ {S}_2=b^2=16^2=256{\text{cm}}^2 \\ x=a\mathrm{/}2-b\mathrm{/}2=24\mathrm{/}2-16\mathrm{/}2=4\text{ cm } \\ {h}_2=\sqrt{x^2+h^2}=\sqrt{4^2+12^2}=4\sqrt{10}\text{ cm}\doteq 12.6491\text{ cm } \\ {S}_3={\displaystyle \frac{a+b}{2}}\cdot h_2={\displaystyle \frac{24+16}{2}}\cdot 12.6491=80\sqrt{10}{\text{cm}}^2\doteq 252.9822{\text{cm}}^2 \\ S=S_1+S_2+4\cdot S_3=576+256+4\cdot 252.9822=1843.9289{\text{cm}}^2 \end{gathered}$$

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