Quadrilateral pyramid

The height of a regular quadrilateral pyramid is 6.5 cm, and the angle between the base and the side wall is 42°. Calculate the surface area and volume of the body—round calculations to 1 decimal place.

Correct answer:

S =  489 cm²
V =  451.7 cm³

Step-by-step explanation:

$$\begin{gathered} h=6.5\text{cm} \\ \alpha =42{}^{\circ } \\ \tan \alpha =h:(a/2) \\ a=2\cdot h/\tan \alpha =2\cdot h/\tan 42°=2\cdot 6.5/\tan 42°=2\cdot 6.5/0.900404=14.43796\text{cm} \\ {S}_1=a^2=14.438^2\doteq 208.4548{\text{cm}}^2 \\ {h}_2=\sqrt{h^2+(a/2{)}^2}=\sqrt{6.5^2+(14.438/2{)}^2}\doteq 9.7141\text{cm} \\ {S}_2=a\cdot h_2/2=14.438\cdot 9.7141/2\doteq 70.1259{\text{cm}}^2 \\ S=S_1+4\cdot S_2=208.4548+4\cdot 70.1259=489{\text{cm}}^2 \end{gathered}$$

$V=\frac{1}{3}\cdot S_1\cdot h=\frac{1}{3}\cdot 208.4548\cdot 6.5=451.7{\text{cm}}^3$

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