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.

Correct result:

S =  47.8931 dm²

Solution:

$$\begin{gathered} e=7\text{ dm } \\ u=50\text{ cm}\to \text{ dm}=50\mathrm{/}10\text{ dm}=5\text{ dm } \\ a=u\mathrm{/}\sqrt{2}=5\mathrm{/}\sqrt{2}\doteq 3.5355\text{ dm } \\ {h}^2+(a\mathrm{/}2{)}^2=e^2 \\ h=\sqrt{e^2-(a\mathrm{/}2{)}^2}=\sqrt{7^2-(3.5355\mathrm{/}2{)}^2}\doteq 6.7731\text{ dm } \\ {S}_1={\displaystyle \frac{a\cdot h}{2}}={\displaystyle \frac{3.5355\cdot 6.7731}{2}}\doteq 11.9733{\text{dm}}^2 \\ S=4\cdot S_1=4\cdot 11.9733=47.8931{\text{dm}}^2 \end{gathered}$$

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