Quadrangular pyramid

Calculate the surface area and volume of a regular quadrangular pyramid:
sides of bases (bottom, top): a1 = 18 cm, a2 = 6cm
angle α = 60 ° (Angle α is the angle between the sidewall and the base plane.)
S =? , V =?

Correct result:

h =  10.3923 cm
S =  936 cm²
V =  1621.2 cm³

Solution:

$$\begin{gathered} a_1=18\text{ cm } \\ {a}_2=6\text{ cm } \\ A=60^{\circ } \\ {S}_1=a_1^2=18^2=324{\text{cm}}^2 \\ {S}_2=a_2^2=6^2=36{\text{cm}}^2 \\ t=\tan A^{\circ }=\tan 60^{\circ }=1.732051=1.73205 \\ {b}_1=a_1\mathrm{/}2=18\mathrm{/}2=9\text{ cm } \\ {b}_2=a_2\mathrm{/}2=6\mathrm{/}2=3\text{ cm } \\ {h}_1=b_1\cdot t=9\cdot 1.7321=9\sqrt{3}\text{ cm}\doteq 15.5885\text{ cm } \\ {h}_2=b_2\cdot t=3\cdot 1.7321=3\sqrt{3}\text{ cm}\doteq 5.1962\text{ cm } \\ {h}_3=\sqrt{h_1^2+b_1^2}=\sqrt{15.5885^2+9^2}=18\text{ cm } \\ {h}_4=\sqrt{h_2^2+b_2^2}=\sqrt{5.1962^2+3^2}=6\text{ cm } \\ h=h_1-h_2=15.5885-5.1962=6\sqrt{3}=10.3923\text{ cm} \end{gathered}$$

$$\begin{gathered} S_3=a_1\cdot h_3\mathrm{/}2=18\cdot 18\mathrm{/}2=162{\text{cm}}^2 \\ {S}_4=a_2\cdot h_4\mathrm{/}2=6\cdot 6\mathrm{/}2=18{\text{cm}}^2 \\ S=4\cdot (S_3-S_4)+S_1+S_2=4\cdot (162-18)+324+36=936{\text{cm}}^2 \end{gathered}$$

$V=h\mathrm{/}3\cdot (S_1+\sqrt{S_1\cdot S_2}+S_2)=10.3923\mathrm{/}3\cdot (324+\sqrt{324\cdot 36}+36)=1621.2{\text{cm}}^3$

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