Quadrangular pyramid

The regular quadrangular pyramid has a base length of 6 cm and a side edge length of 9 centimeters. Calculate its volume and surface area.

Correct answer:

V =  95.247 cm³
S =  137.8234 cm²

Step-by-step explanation:

$$\begin{gathered} a=6\text{ cm } \\ s=9\text{ cm } \\ {S}_1=a^2=6^2=36{\text{cm}}^2 \\ u=\sqrt{2}\cdot a=\sqrt{2}\cdot 6=6\sqrt{2}\text{ cm}\doteq 8.4853\text{ cm } \\ h=\sqrt{s^2-(u\mathrm{/}2{)}^2}=\sqrt{9^2-(8.4853\mathrm{/}2{)}^2}=3\sqrt{7}\text{ cm}\doteq 7.9373\text{ cm } \\ V={\displaystyle \frac{1}{3}}\cdot S_1\cdot h={\displaystyle \frac{1}{3}}\cdot 36\cdot 7.9373=36\sqrt{7}=95.247{\text{cm}}^3 \end{gathered}$$

$$\begin{gathered} h_2=\sqrt{s^2-(a\mathrm{/}2{)}^2}=\sqrt{9^2-(6\mathrm{/}2{)}^2}=6\sqrt{2}\text{ cm}\doteq 8.4853\text{ cm } \\ {S}_2=a\cdot h_2\mathrm{/}2=6\cdot 8.4853\mathrm{/}2=18\sqrt{2}{\text{cm}}^2\doteq 25.4558{\text{cm}}^2 \\ S=S_1+4\cdot S_2=36+4\cdot 25.4558=137.8234{\text{cm}}^2 \end{gathered}$$

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