Pyramid roof

2/4 of area of ​​the roof shaped regular tetrahedral pyramid with base edge 10 m and height of 4 m is already covered with roofing. How many square meters still needs to be covered?

Correct result:

S =  64.0312 m²

Solution:

$$\begin{gathered} a=10 \\ h=4 \\ q=1-2\mathrm{/}4={\displaystyle \frac{1}{2}}=0.5 \\ {h}_1=\sqrt{h^2+(a\mathrm{/}2{)}^2}=\sqrt{4^2+(10\mathrm{/}2{)}^2}=\sqrt{41}\doteq 6.4031 \\ {S}_1=a\cdot h_1\mathrm{/}2=10\cdot 6.4031\mathrm{/}2=5\sqrt{41}\doteq 32.0156 \\ S=q\cdot 4\cdot S_1=0.5\cdot 4\cdot 32.0156=10\sqrt{41}=64.0312{\text{m}}^2 \end{gathered}$$

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