Rectangular base pyramid

Calculate an area of the shell of the pyramid with a rectangular base of 2.8 m and 1.4 m and height 2.5 meters.

Result

S =  11.281 m2

Solution:

h=2.5 m a=2.8 m b=1.4 m  h1=h2+(b/2)2=2.52+(1.4/2)2≐2.5962 m h2=h2+(a/2)2=2.52+(2.8/2)2≐2.8653 m  S1=a⋅ h12=2.8⋅ 2.59622≐3.6346 m2 S2=b⋅ h22=1.4⋅ 2.86532≐2.0057 m2  S=2⋅ S1+2⋅ S2=2⋅ 3.6346+2⋅ 2.0057≐11.2807≐11.281 m2h=2.5 \ \text{m} \ \\ a=2.8 \ \text{m} \ \\ b=1.4 \ \text{m} \ \\ \ \\ h_{1}=\sqrt{ h^2 +(b/2)^2 }=\sqrt{ 2.5^2 +(1.4/2)^2 } \doteq 2.5962 \ \text{m} \ \\ h_{2}=\sqrt{ h^2 +(a/2)^2 }=\sqrt{ 2.5^2 +(2.8/2)^2 } \doteq 2.8653 \ \text{m} \ \\ \ \\ S_{1}=\dfrac{ a \cdot \ h_{1} }{ 2 }=\dfrac{ 2.8 \cdot \ 2.5962 }{ 2 } \doteq 3.6346 \ \text{m}^2 \ \\ S_{2}=\dfrac{ b \cdot \ h_{2} }{ 2 }=\dfrac{ 1.4 \cdot \ 2.8653 }{ 2 } \doteq 2.0057 \ \text{m}^2 \ \\ \ \\ S=2 \cdot \ S_{1}+2 \cdot \ S_{2}=2 \cdot \ 3.6346+2 \cdot \ 2.0057 \doteq 11.2807 \doteq 11.281 \ \text{m}^2

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