Quadrilateral 8109

The regular quadrilateral pyramid has a base diagonal of 5√2 cm, and the side edges are 12√2 cm long. Calculate the height of the pyramid and its surface.

Final Answer:

h =  16.5982 cm
S =  192.8541 cm²

Step-by-step explanation:

$$\begin{gathered} u_1=5\cdot \sqrt{2}=5\sqrt{2}\text{cm}\doteq 7.0711\text{cm} \\ s=12\cdot \sqrt{2}=12\sqrt{2}\text{cm}\doteq 16.9706\text{cm} \\ a=u_1/\sqrt{2}=7.0711/\sqrt{2}=5\text{cm} \\ h=\sqrt{s^2-(u_1/2{)}^2}=\sqrt{16.9706^2-(7.0711/2{)}^2}=16.5982\text{cm} \end{gathered}$$

$$\begin{gathered} S_1=a^2=5^2=25{\text{cm}}^2 \\ {h}_2=\sqrt{s^2-(a/2{)}^2}=\sqrt{16.9706^2-(5/2{)}^2}\doteq 16.7854\text{cm} \\ {S}_2=\frac{a\cdot h_2}{2}=\frac{5\cdot 16.7854}{2}\doteq 41.9635{\text{cm}}^2 \\ S=S_1+4\cdot S_2=25+4\cdot 41.9635=192.8541{\text{cm}}^2 \end{gathered}$$

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