Quadrilateral pyramid,

A quadrilateral pyramid, which has a rectangular base with dimensions of 24 cm, 13 cm. The height of the pyramid is 18cm.
Calculate
1/the area of the base
2/casing area
3/pyramid surface
4/volume of the pyramid

Correct result:

S₁ =  312 cm²
S =  740.5368 cm²
S₄ =  1052.5368 cm²
V =  1872 cm³

Solution:

$$\begin{gathered} h=18\text{ cm } \\ a=24\text{ cm } \\ b=13\text{ cm } \\ {S}_1=a\cdot b=24\cdot 13=312{\text{cm}}^2 \end{gathered}$$

$$\begin{gathered} h_1=\sqrt{h^2+(b\mathrm{/}2{)}^2}=\sqrt{18^2+(13\mathrm{/}2{)}^2}\doteq 19.1377\text{ cm } \\ {h}_2=\sqrt{h^2+(a\mathrm{/}2{)}^2}=\sqrt{18^2+(24\mathrm{/}2{)}^2}=6\sqrt{13}\text{ cm}\doteq 21.6333\text{ cm } \\ {S}_2={\displaystyle \frac{a\cdot h_1}{2}}={\displaystyle \frac{24\cdot 19.1377}{2}}=6\sqrt{1465}{\text{cm}}^2\doteq 229.6519{\text{cm}}^2 \\ {S}_3={\displaystyle \frac{b\cdot h_2}{2}}={\displaystyle \frac{13\cdot 21.6333}{2}}=39\sqrt{13}{\text{cm}}^2\doteq 140.6165{\text{cm}}^2 \\ S=2\cdot S_2+2\cdot S_3=2\cdot 229.6519+2\cdot 140.6165=740.5368{\text{cm}}^2 \end{gathered}$$

$S_4=S+S_1=740.5368+312=1052.5368{\text{cm}}^2$

$V={\displaystyle \frac{1}{3}}\cdot S_1\cdot h={\displaystyle \frac{1}{3}}\cdot 312\cdot 18=1872{\text{cm}}^3$

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