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.

Correct result:

S =  47.893 dm2

#### Solution:

$e=7 \ \text{dm} \ \\ u=50 \ cm \rightarrow dm=50 / 10 \ dm=5 \ dm \ \\ \ \\ a=u / \sqrt{ 2 }=5 / \sqrt{ 2 } \doteq 3.5355 \ \text{dm} \ \\ \ \\ h^2 + (a/2)^2=e^2 \ \\ h=\sqrt{ e^2 - (a/2)^2 }=\sqrt{ 7^2 - (3.5355/2)^2 } \doteq 6.7731 \ \text{dm} \ \\ \ \\ S_{1}=\dfrac{ a \cdot \ h }{ 2 }=\dfrac{ 3.5355 \cdot \ 6.7731 }{ 2 } \doteq 11.9733 \ \text{dm}^2 \ \\ \ \\ S=4 \cdot \ S_{1}=4 \cdot \ 11.9733=47.893 \ \text{dm}^2$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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