Quadrilateral pyramid

Calculate the surface of a quadrilateral pyramid, which has a rectangular base with dimensions a = 8 cm, b = 6 cm and height H = 10 cm.

Result

S =  196.144 cm2

Solution:

h=10 cm a=8 cm b=6 cm  h1=h2+(b/2)2=102+(6/2)2≐109 cm≐10.4403 cm h2=h2+(a/2)2=102+(8/2)2≐2 29 cm≐10.7703 cm  S1=a⋅ h12=8⋅ 10.44032≐4 109 cm2≐41.7612 cm2 S2=b⋅ h22=6⋅ 10.77032≐6 29 cm2≐32.311 cm2  S=2⋅ S1+2⋅ S2+a⋅ b=2⋅ 41.7612+2⋅ 32.311+8⋅ 6≐196.1444≐196.144 cm2h=10 \ \text{cm} \ \\ a=8 \ \text{cm} \ \\ b=6 \ \text{cm} \ \\ \ \\ h_{1}=\sqrt{ h^2 +(b/2)^2 }=\sqrt{ 10^2 +(6/2)^2 } \doteq \sqrt{ 109 } \ \text{cm} \doteq 10.4403 \ \text{cm} \ \\ h_{2}=\sqrt{ h^2 +(a/2)^2 }=\sqrt{ 10^2 +(8/2)^2 } \doteq 2 \ \sqrt{ 29 } \ \text{cm} \doteq 10.7703 \ \text{cm} \ \\ \ \\ S_{1}=\dfrac{ a \cdot \ h_{1} }{ 2 }=\dfrac{ 8 \cdot \ 10.4403 }{ 2 } \doteq 4 \ \sqrt{ 109 } \ \text{cm}^2 \doteq 41.7612 \ \text{cm}^2 \ \\ S_{2}=\dfrac{ b \cdot \ h_{2} }{ 2 }=\dfrac{ 6 \cdot \ 10.7703 }{ 2 } \doteq 6 \ \sqrt{ 29 } \ \text{cm}^2 \doteq 32.311 \ \text{cm}^2 \ \\ \ \\ S=2 \cdot \ S_{1}+2 \cdot \ S_{2} + a \cdot \ b=2 \cdot \ 41.7612+2 \cdot \ 32.311 + 8 \cdot \ 6 \doteq 196.1444 \doteq 196.144 \ \text{cm}^2

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