Pyramid - angle

Calculate the regular quadrilateral pyramid's surface, the base edge of which is measured 6 cm, and the deviation from the plane of the base's sidewall plane is 50 degrees.

Final Answer:

S =  92.01 cm²

Step-by-step explanation:

$$\begin{gathered} a=6\text{cm} \\ \alpha =50{}^{\circ } \\ h=\frac{a}{2}\cdot \tan \alpha =\frac{a}{2}\cdot \tan 50°=\frac{6}{2}\cdot \tan 50°=\frac{6}{2}\cdot 1.191754=3.57526\text{cm} \\ {h}_2=\sqrt{h^2+(a/2{)}^2}=\sqrt{3.5753^2+(6/2{)}^2}\doteq 4.6672\text{cm} \\ {S}_1=\frac{a\cdot h_2}{2}=\frac{6\cdot 4.6672}{2}\doteq 14.0015{\text{cm}}^2 \\ S=4\cdot S_1+a^2=4\cdot 14.0015+6^2=92.01{\text{cm}}^2 \end{gathered}$$

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