Rectangular base pyramid

Calculate an area of the shell of the pyramid with a rectangular base of 2.8 m and 1.4 m and height 2.5 meters.

Correct result:

S =  11.2807 m²

Solution:

$$\begin{gathered} h=2.5\text{ m } \\ a=2.8\text{ m } \\ b=1.4\text{ m } \\ {h}_1=\sqrt{h^2+(b\mathrm{/}2{)}^2}=\sqrt{2.5^2+(1.4\mathrm{/}2{)}^2}\doteq 2.5962\text{ m } \\ {h}_2=\sqrt{h^2+(a\mathrm{/}2{)}^2}=\sqrt{2.5^2+(2.8\mathrm{/}2{)}^2}\doteq 2.8653\text{ m } \\ {S}_1={\displaystyle \frac{a\cdot h_1}{2}}={\displaystyle \frac{2.8\cdot 2.5962}{2}}\doteq 3.6346{\text{m}}^2 \\ {S}_2={\displaystyle \frac{b\cdot h_2}{2}}={\displaystyle \frac{1.4\cdot 2.8653}{2}}\doteq 2.0057{\text{m}}^2 \\ S=2\cdot S_1+2\cdot S_2=2\cdot 3.6346+2\cdot 2.0057=11.2807{\text{m}}^2 \end{gathered}$$

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