Side edges

The regular 4-sided pyramid has a body height of 2 dm, and the opposite side edges form an angle of 70°.
Calculate the surface area and volume of the pyramid.

Correct answer:

S =  12.7621 dm²
V =  2.6149 dm³

Step-by-step explanation:

$$\begin{gathered} h=2\text{ dm } \\ \alpha =70^{\circ }\to \text{ rad}=70^{\circ }\cdot {\displaystyle \frac{\pi }{180}}=70^{\circ }\cdot {\displaystyle \frac{3.1415926}{180}}=1.22173=7\pi \mathrm{/}18 \\ \theta =\alpha \mathrm{/}2=1.2217\mathrm{/}2\doteq 0.6109\text{ rad } \\ \tan \theta =u\mathrm{/}2:h \\ u=2\cdot h\cdot \tan (\theta )=2\cdot 2\cdot \tan (0.6109)\doteq 2.8008\text{ dm } \\ {u}^2=a^2+a^2 \\ a=u\mathrm{/}\sqrt{2}=2.8008\mathrm{/}\sqrt{2}\doteq 1.9805\text{ dm } \\ {h}_2=\sqrt{h^2+(a\mathrm{/}2{)}^2}=\sqrt{2^2+(1.9805\mathrm{/}2{)}^2}\doteq 2.2317\text{ dm } \\ {S}_1=a^2=1.9805^2\doteq 3.9223{\text{dm}}^2 \\ {S}_2={\displaystyle \frac{a\cdot h_2}{2}}={\displaystyle \frac{1.9805\cdot 2.2317}{2}}\doteq 2.2099{\text{dm}}^2 \\ S=S_1+4\cdot S_2=3.9223+4\cdot 2.2099=12.7621{\text{dm}}^2 \end{gathered}$$

$V={\displaystyle \frac{1}{3}}\cdot S_1\cdot h={\displaystyle \frac{1}{3}}\cdot 3.9223\cdot 2=2.6149{\text{dm}}^3$

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