Tetrahedral pyramid

Determine the surface of a regular tetrahedral pyramid when its volume is V = 120 and the angle of the sidewall with the base plane is α = 42° 30'.

Final Answer:

S =  200.6434

Step-by-step explanation:

$$\begin{gathered} V=120 \\ A=42+\frac{30}{60}=\frac{42\cdot 60}{60}+\frac{30}{60}=\frac{2520}{60}+\frac{30}{60}=\frac{2520+30}{60}=\frac{2550}{60}=\frac{85}{2}=42\frac{1}{2}=42.5{}^{\circ } \\ {A}_1=A°\to \text{rad}=A°\cdot \frac{\pi }{180}=42.5°\cdot \frac{3.1415926}{180}=0.74176=17\pi /72 \\ \tan A=h/(a/2) \\ h=a/2\cdot \tan A \\ V=\frac{1}{3}{a}^{2}h \\ V=\frac{1}{6}{a}^3\tan A \\ a=\sqrt[3]{6\cdot V/\tan (A_1)}=\sqrt[3]{6\cdot 120/\tan 0.7418}\doteq 9.2277 \\ h=a/2\cdot \tan (A_1)=9.2277/2\cdot \tan 0.7418\doteq 4.2278 \\ \sin A_1=h:w \\ w=h/\sin (A_1)=4.2278/\sin 0.7418\doteq 6.258 \\ {S}_1=a\cdot w/2=9.2277\cdot 6.258/2\doteq 28.8733 \\ S=a^2+4\cdot S_1=9.2277^2+4\cdot 28.8733=200.6434 \end{gathered}$$

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