Hexagonal pyramid

Calculate the volume and surface area of a regular hexagonal pyramid with a base edge a = 30 m and a side edge b = 50 m.

Correct answer:

V =  31176.9145 m³
S =  6630.995 m²

Step-by-step explanation:

$$\begin{gathered} a=30\text{ m } \\ b=50\text{ m } \\ n=6 \\ {S}_0={\displaystyle \frac{\sqrt{3}}{4}}\cdot a^2={\displaystyle \frac{\sqrt{3}}{4}}\cdot 30^2=225\sqrt{3}{\text{m}}^2\doteq 389.7114{\text{m}}^2 \\ {S}_1=n\cdot S_0=6\cdot 389.7114=1350\sqrt{3}{\text{m}}^2\doteq 2338.2686{\text{m}}^2 \\ h=\sqrt{b^2-a^2}=\sqrt{50^2-30^2}=40\text{ m } \\ V={\displaystyle \frac{1}{3}}\cdot S_1\cdot h={\displaystyle \frac{1}{3}}\cdot 2338.2686\cdot 40=31176.9145{\text{m}}^3=3.118\cdot 10^4{\text{m}}^3 \end{gathered}$$

Try calculation via our triangle calculator.

$$\begin{gathered} o=n\cdot a=6\cdot 30=180\text{ m } \\ {h}_2=\sqrt{b^2-(a\mathrm{/}2{)}^2}=\sqrt{50^2-(30\mathrm{/}2{)}^2}=5\sqrt{91}\text{ m}\doteq 47.697\text{ m } \\ {S}_2={\displaystyle \frac{a\cdot h_2}{2}}={\displaystyle \frac{30\cdot 47.697}{2}}\doteq 715.4544{\text{m}}^2 \\ S=S_1+n\cdot S_2=2338.2686+6\cdot 715.4544=6630.995{\text{m}}^2 \end{gathered}$$

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