Tower

The top of the tower is a regular hexagonal pyramid with base edge 6.1 meters long and a height 11.7 meters. How many m² of the sheet is required to cover the top of the tower if we count 9% of the sheet waste?

Correct answer:

S =  256 m²

Step-by-step explanation:

$$\begin{gathered} a=6.1m \\ v=11.7m \\ {a}^2-(a/2{)}^2=v_2^2 \\ {v}_2=a\frac{\sqrt{3}}{2}=5.283m \\ {h}^2=v^2+v_2^2 \\ h=\sqrt{11.7^2+5.283^2}=12.837m \\ {S}_1=\frac{1}{2}ah=\frac{1}{2}\cdot 6.1\cdot 12.837=39.15m^2 \\ S=6\cdot S_1\cdot (1+\frac{9}{100})=6\cdot 39.15\cdot 1.09=256{\text{m}}^2 \end{gathered}$$

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