The plaster cast

The plaster cast has the shape of a regular quadrilateral pyramid. The cover consists of four equilateral triangles with a 5 m side. Calculate its volume and surface area.

Result

V =  29.463 m3
S =  68.301 m2

Solution:

a=5 m S1=a2=52=25 m2  a2=(a/2)2+h12  h1=a2(a/2)2=52(5/2)24.3301 m  h12=h22+(a/2)2  h2=h12(a/2)2=4.33012(5/2)23.5355 m  V=13 S1 h2=13 25 3.535529.462829.463 m3a=5 \ \text{m} \ \\ S_{1}=a^2=5^2=25 \ \text{m}^2 \ \\ \ \\ a^2=(a/2)^2+h_{1}^2 \ \\ \ \\ h_{1}=\sqrt{ a^2 - (a/2)^2 }=\sqrt{ 5^2 - (5/2)^2 } \doteq 4.3301 \ \text{m} \ \\ \ \\ h_{1}^2=h_{2}^2 + (a/2)^2 \ \\ \ \\ h_{2}=\sqrt{ h_{1}^2 - (a/2)^2 }=\sqrt{ 4.3301^2 - (5/2)^2 } \doteq 3.5355 \ \text{m} \ \\ \ \\ V=\dfrac{ 1 }{ 3 } \cdot \ S_{1} \cdot \ h_{2}=\dfrac{ 1 }{ 3 } \cdot \ 25 \cdot \ 3.5355 \doteq 29.4628 \doteq 29.463 \ \text{m}^3
S2=12 a h1=12 5 4.330110.8253 m2  S=4 S2+S1=4 10.8253+2568.301368.301 m2S_{2}=\dfrac{ 1 }{ 2 } \cdot \ a \cdot \ h_{1}=\dfrac{ 1 }{ 2 } \cdot \ 5 \cdot \ 4.3301 \doteq 10.8253 \ \text{m}^2 \ \\ \ \\ S=4 \cdot \ S_{2}+ S_{1}=4 \cdot \ 10.8253+ 25 \doteq 68.3013 \doteq 68.301 \ \text{m}^2



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