Pyramid 8

Calculate the volume and the surface area of a regular quadrilateral pyramid with a base side of 9 cm and a side wall with the base has an angle of 75°.

Final Answer:

V =  641.27 cm³
S =  516.1171 cm²

Step-by-step explanation:

$$\begin{gathered} a=9\text{cm} \\ {S}_1=a^2=9^2=81{\text{cm}}^2 \\ s=a/2=9/2=\frac{9}{2}=4.5\text{cm} \\ h=\sqrt{2}\cdot s\cdot \tan 75°=\sqrt{2}\cdot s\cdot \tan 5\pi /12=\sqrt{2}\cdot 4.5\cdot \tan 5\pi /12=23.75063\text{cm} \\ {h}_2=\sqrt{h^2+s^2}=\sqrt{23.7506^2+4.5^2}\doteq 24.1732\text{cm} \\ V=\frac{1}{3}\cdot S_1\cdot h=\frac{1}{3}\cdot 81\cdot 23.7506=641.27{\text{cm}}^3 \end{gathered}$$

$$\begin{gathered} S_2=a\cdot h_2/2=9\cdot 24.1732/2\doteq 108.7793{\text{cm}}^2 \\ S=S_1+4\cdot S_2=81+4\cdot 108.7793=516.1171{\text{cm}}^2 \end{gathered}$$

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