Quadrilateral pyramid

In a regular quadrilateral pyramid, the side edge is e = 7 dm and the diagonal of the base is 50 cm. Calculate the pyramid shell area.

Result

S =  47.893 dm2

Solution:

e=7 dm u=50 cm=50/10 dm=5 dm  a=u/2=5/2≐3.5355 dm  h2+(a/2)2=e2 h=e2−(a/2)2=72−(3.5355/2)2≐6.7731 dm  S1=a⋅ h2=3.5355⋅ 6.77312≐11.9733 dm2  S=4⋅ S1=4⋅ 11.9733≐47.8931≐47.893 dm2e=7 \ \text{dm} \ \\ u=50 \ cm=50 / 10 \ dm=5 \ dm \ \\ \ \\ a=u / \sqrt{ 2 }=5 / \sqrt{ 2 } \doteq 3.5355 \ \text{dm} \ \\ \ \\ h^2 + (a/2)^2=e^2 \ \\ h=\sqrt{ e^2 - (a/2)^2 }=\sqrt{ 7^2 - (3.5355/2)^2 } \doteq 6.7731 \ \text{dm} \ \\ \ \\ S_{1}=\dfrac{ a \cdot \ h }{ 2 }=\dfrac{ 3.5355 \cdot \ 6.7731 }{ 2 } \doteq 11.9733 \ \text{dm}^2 \ \\ \ \\ S=4 \cdot \ S_{1}=4 \cdot \ 11.9733 \doteq 47.8931 \doteq 47.893 \ \text{dm}^2

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