Truncated 43851

The pit is a regular truncated 4-sided pyramid, with 14 m and 10 m base edges and a depth of 6m. Calculate how much m³ of soil was removed when we dug this pit.

Final Answer:

V =  872 m³

Step-by-step explanation:

$$\begin{gathered} a=14\text{m} \\ b=10\text{m} \\ h=6\text{m} \\ {S}_1=a^2=14^2=196{\text{m}}^2 \\ {S}_2=b^2=10^2=100{\text{m}}^2 \\ S=S_1+\sqrt{S_1\cdot S_2}+S_2=196+\sqrt{196\cdot 100}+100=436{\text{m}}^2 \\ V=\frac{1}{3}\cdot h\cdot S=\frac{1}{3}\cdot 6\cdot 436=872{\text{m}}^3 \end{gathered}$$

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