Regular hexagonal prism

Calculate the volume of a regular hexagonal prism whose body diagonals are 24cm and 25cm long.

Correct result:

V =  2636.796 cm3

Solution:

u=24 cm v=25 cm  u1=2a  (u2/2)2+(a/2)2=a2 u22/4+a2/4=a2 u22/4+=3/4 a2  u2=3 a  h2+u12=v2 h2+u22=u2  h2+(2a)2=v2 h2+(3 a)2=u2 4a23a2=v2u2 a2=v2u2 a=v2u2=252242=7 cm  h=v24 a2=2524 72429 cm20.7123 cm  S1=3 a24=3 72421.2176 cm2 S=6 S1=6 21.2176127.3057  V=S h=127.3057 20.7123=2636.796 cm3u=24 \ \text{cm} \ \\ v=25 \ \text{cm} \ \\ \ \\ u_{1}=2a \ \\ \ \\ (u_{2}/2)^2 + (a/2)^2=a^2 \ \\ u_{2}^2/4 + a^2/4=a^2 \ \\ u_{2}^2/4 +=3/4 \ a^2 \ \\ \ \\ u_{2}=\sqrt{ 3 } \cdot \ a \ \\ \ \\ h^2 + u_{1}^2=v^2 \ \\ h^2 + u_{2}^2=u^2 \ \\ \ \\ h^2 + (2a)^2=v^2 \ \\ h^2 + (\sqrt{ 3 } \cdot \ a)^2=u^2 \ \\ 4a^2 - 3a^2=v^2-u^2 \ \\ a^2=v^2-u^2 \ \\ a=\sqrt{ v^2-u^2 }=\sqrt{ 25^2-24^2 }=7 \ \text{cm} \ \\ \ \\ h=\sqrt{ v^2-4 \cdot \ a^2 }=\sqrt{ 25^2-4 \cdot \ 7^2 } \doteq \sqrt{ 429 } \ \text{cm} \doteq 20.7123 \ \text{cm} \ \\ \ \\ S_{1}=\dfrac{ \sqrt{ 3 } \cdot \ a^2 }{ 4 }=\dfrac{ \sqrt{ 3 } \cdot \ 7^2 }{ 4 } \doteq 21.2176 \ \text{cm}^2 \ \\ S=6 \cdot \ S_{1}=6 \cdot \ 21.2176 \doteq 127.3057 \ \\ \ \\ V=S \cdot \ h=127.3057 \cdot \ 20.7123=2636.796 \ \text{cm}^3



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