Pyramid four sides

In a regular tetrahedral pyramid is a body height 38 cm and a wall height 42 cm. Calculate the surface area of the pyramid; the result round to square centimeters.

Correct result:

S =  4285 cm²

Solution:

$$\begin{gathered} h=38\text{ cm } \\ s=42\text{ cm } \\ {h}^2=s^2-(a\mathrm{/}2{)}^2 \\ a=2\cdot \sqrt{s^2-h^2}=2\cdot \sqrt{42^2-38^2}=16\sqrt{5}\text{ cm}\doteq 35.7771\text{ cm } \\ {S}_1=a^2=35.7771^2=1280{\text{cm}}^2 \\ {S}_2=a\cdot s\mathrm{/}2=35.7771\cdot 42\mathrm{/}2=336\sqrt{5}{\text{cm}}^2\doteq 751.3188{\text{cm}}^2 \\ S=S_1+4\cdot S_2=1280+4\cdot 751.3188=4285{\text{cm}}^2 \end{gathered}$$

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