Digging a pit

The pit has the shape of a regular quadrilateral truncated pyramid. The edges of the bases are 14m and 10m long. The sidewalls form an angle of 135° with a smaller base. Determine how many m³ of soil were excavated when digging the pit?

Correct result:

V =  290.6667 m³

Solution:

$$\begin{gathered} a=14\text{ m } \\ b=10\text{ m } \\ \alpha =135-90=45^{\circ } \\ x=(a-b)\mathrm{/}2=(14-10)\mathrm{/}2=2\text{ m } \\ \tan \alpha =x:h \\ h=x\mathrm{/}\tan {\alpha }^{\circ }=x\mathrm{/}\tan 45^{\circ }=2\mathrm{/}\tan 45^{\circ }=2\mathrm{/}1=2 \\ {S}_1=a^2=14^2=196{\text{m}}^2 \\ {S}_2=b^2=10^2=100{\text{m}}^2 \\ S=S_1+\sqrt{S_1\cdot S_2}+S_2=196+\sqrt{196\cdot 100}+100=436{\text{m}}^2 \\ V={\displaystyle \frac{1}{3}}\cdot h\cdot S={\displaystyle \frac{1}{3}}\cdot 2\cdot 436={\displaystyle \frac{872}{3}}=290.6667{\text{m}}^3 \end{gathered}$$

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