Quadrilateral pyramid

In a regular quadrilateral pyramid, the height 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 =  273 cm²
V =  225.8 cm³

Step-by-step explanation:

$$\begin{gathered} h=6.5\text{cm} \\ A=42{}^{\circ } \\ \tan A=h:(u/2) \\ u=2\cdot h/\tan A=2\cdot h/\tan 42°=2\cdot 6.5/\tan 42°=2\cdot 6.5/0.900404=14.43796\text{cm} \\ a=u/\sqrt{2}=14.438/\sqrt{2}\doteq 10.2092\text{cm} \\ {S}_1=a^2=10.2092^2\doteq 104.2274{\text{cm}}^2 \\ {h}_2=\sqrt{h^2+(a/2{)}^2}=\sqrt{6.5^2+(10.2092/2{)}^2}\doteq 8.2648\text{cm} \\ {S}_2=a\cdot h_2/2=10.2092\cdot 8.2648/2\doteq 42.1884{\text{cm}}^2 \\ S=S_1+4\cdot S_2=104.2274+4\cdot 42.1884=273{\text{cm}}^2 \end{gathered}$$

$V=\frac{1}{3}\cdot S_1\cdot h=\frac{1}{3}\cdot 104.2274\cdot 6.5=225.8{\text{cm}}^3$

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