Quadrilateral 36061

In a regular quadrilateral pyramid, the length of the base edge is a = 8 cm, and the length of the side edge is h = 17 cm. Calculate the surface of the pyramid.

Correct answer:

S =  328.3634 cm²

Step-by-step explanation:

$$\begin{gathered} a=8\text{cm} \\ s=17\text{cm} \\ {u}^2=a^2+a^2 \\ u=\sqrt{2}\cdot a=\sqrt{2}\cdot 8=8\sqrt{2}\text{cm}\doteq 11.3137\text{cm} \\ {h}^2+(u/2{)}^2=s^2 \\ h=\sqrt{s^2-(u/2{)}^2}=\sqrt{17^2-(11.3137/2{)}^2}=\sqrt{257}\text{cm}\doteq 16.0312\text{cm} \\ {S}_1=a^2=8^2=64{\text{cm}}^2 \\ {h}_2=\sqrt{h^2+(a/2{)}^2}=\sqrt{16.0312^2+(8/2{)}^2}=\sqrt{273}\text{cm}\doteq 16.5227\text{cm} \\ {S}_2=\frac{a\cdot h_2}{2}=\frac{8\cdot 16.5227}{2}=4\sqrt{273}{\text{cm}}^2\doteq 66.0908{\text{cm}}^2 \\ S=S_1+4\cdot S_2=64+4\cdot 66.0908=328.3634{\text{cm}}^2 \end{gathered}$$

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