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. Find how many m³ of soil were excavated when digging the pit?

Correct answer:

V =  290.6667 m³

Step-by-step explanation:

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

Try another example