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?

Result

S =  64.031 m2

Solution:

a=10 h=4 q=12/4=12=0.5 h1=h2+(a/2)2=42+(10/2)2416.4031 S1=a h1/2=10 6.4031/25 4132.0156 S=q 4 S1=0.5 4 32.015610 4164.031264.031 m2a=10 \ \\ h=4 \ \\ q=1-2/4=\dfrac{ 1 }{ 2 }=0.5 \ \\ h_{1}=\sqrt{ h^2+(a/2)^{ 2 } }=\sqrt{ 4^2+(10/2)^{ 2 } } \doteq \sqrt{ 41 } \doteq 6.4031 \ \\ S_{1}=a \cdot \ h_{1}/2=10 \cdot \ 6.4031/2 \doteq 5 \ \sqrt{ 41 } \doteq 32.0156 \ \\ S=q \cdot \ 4 \cdot \ S_{1}=0.5 \cdot \ 4 \cdot \ 32.0156 \doteq 10 \ \sqrt{ 41 } \doteq 64.0312 \doteq 64.031 \ \text{m}^2



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