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 side wall and the plane of the base.)
S =? , V =?

Result

h =  10.392 cm
S =  936 cm²
V =  1621.2 cm³

Solution:

$a_{1}=18 \ \text{cm} \ \\ a_{2}=6 \ \text{cm} \ \\ A=60 \ ^\circ \ \\ \ \\ S_{1}=a_{1}^2=18^2=324 \ \text{cm}^2 \ \\ S_{2}=a_{2}^2=6^2=36 \ \text{cm}^2 \ \\ t=\tan( A ^\circ \rightarrow\ \text{rad})=\tan( A ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=\tan( 60 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=1.73205 \ \\ b_{1}=a_{1}/2=18/2=9 \ \text{cm} \ \\ b_{2}=a_{2}/2=6/2=3 \ \text{cm} \ \\ \ \\ h_{1}=b_{1} \cdot \ t=9 \cdot \ 1.7321 \doteq 9 \ \sqrt{ 3 } \ \text{cm} \doteq 15.5885 \ \text{cm} \ \\ h_{2}=b_{2} \cdot \ t=3 \cdot \ 1.7321 \doteq 3 \ \sqrt{ 3 } \ \text{cm} \doteq 5.1962 \ \text{cm} \ \\ \ \\ h_{3}=\sqrt{ h_{1}^2 + b_{1}^2 }=\sqrt{ 15.5885^2 + 9^2 }=18 \ \text{cm} \ \\ h_{4}=\sqrt{ h_{2}^2 + b_{2}^2 }=\sqrt{ 5.1962^2 + 3^2 }=6 \ \text{cm} \ \\ \ \\ h=h_{1}-h_{2}=15.5885-5.1962 \doteq 6 \ \sqrt{ 3 } \doteq 10.3923 \doteq 10.392 \ \text{cm}$

$S_{3}=a_{1} \cdot \ h_{3}/2=18 \cdot \ 18/2=162 \ \text{cm}^2 \ \\ S_{4}=a_{2} \cdot \ h_{4}/2=6 \cdot \ 6/2=18 \ \text{cm}^2 \ \\ S=4 \cdot \ (S_{3} - S_{4}) + S_{1} + S_{2}=4 \cdot \ (162 - 18) + 324 + 36=936 \ \text{cm}^2$

$V=h/3 \cdot \ (S_{1}+\sqrt{ S_{1} \cdot \ S_{2} }+S_{2})=10.3923/3 \cdot \ (324+\sqrt{ 324 \cdot \ 36 }+36)=1621.152=1621.2 \ \text{cm}^3$

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