Truncated pyramid

Find the volume of a regular 4-sided truncated pyramid if a1 = 14 cm, a2 = 8 cm, and the angle that the side wall with the base is 42 degrees.

Correct answer:

V =  334.9503 cm³

Step-by-step explanation:

$$\begin{gathered} a_1=14\text{cm} \\ {a}_2=8\text{cm} \\ \alpha =42{}^{\circ } \\ \tan \alpha =h_1:(a_2/2) \\ \tan \alpha =h_2:(a_2/2) \\ {h}_1=a_1/2\cdot \tan (\alpha )=14/2\cdot \tan (42°)=6.30283\text{cm} \\ {h}_2=a_2/2\cdot \tan (\alpha )=8/2\cdot \tan (42°)=3.60162\text{cm} \\ {S}_1=a_1^2=14^2=196{\text{cm}}^2 \\ {S}_2=a_2^2=8^2=64{\text{cm}}^2 \\ {V}_1=\frac{1}{3}\cdot S_1\cdot h_1=\frac{1}{3}\cdot 196\cdot 6.3028\doteq 411.7848{\text{cm}}^3 \\ {V}_2=\frac{1}{3}\cdot S_2\cdot h_2=\frac{1}{3}\cdot 64\cdot 3.6016\doteq 76.8345{\text{cm}}^3 \\ V=V_1-V_2=411.7848-76.8345=334.9503{\text{cm}}^3 \end{gathered}$$

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