The pyramid 4s

The pyramid with a rectangular base measuring 6 dm and 8 dm has a side edge of a length of 13 dm. Calculate the surface area and volume of this pyramid.

Final Answer:

S =  222.8492 dm²
V =  192 dm³

Step-by-step explanation:

$$\begin{gathered} a=6\text{dm} \\ b=8\text{dm} \\ e=13\text{dm} \\ d=\sqrt{a^2+b^2}=\sqrt{6^2+8^2}=10\text{dm} \\ h=\sqrt{e^2-(d/2{)}^2}=\sqrt{13^2-(10/2{)}^2}=12\text{dm} \\ {S}_1=a\cdot b=6\cdot 8=48{\text{dm}}^2 \\ {s}_2=\sqrt{h^2+(a/2{)}^2}=\sqrt{12^2+(6/2{)}^2}=3\sqrt{17}\text{dm}\doteq 12.3693\text{dm} \\ {s}_3=\sqrt{h^2+(b/2{)}^2}=\sqrt{12^2+(8/2{)}^2}=4\sqrt{10}\text{dm}\doteq 12.6491\text{dm} \\ {S}_2=\frac{a\cdot s_3}{2}=\frac{6\cdot 12.6491}{2}=12\sqrt{10}{\text{dm}}^2\doteq 37.9473{\text{dm}}^2 \\ {S}_3=\frac{b\cdot s_2}{2}=\frac{8\cdot 12.3693}{2}=12\sqrt{17}{\text{dm}}^2\doteq 49.4773{\text{dm}}^2 \\ S=S_1+2\cdot S_2+2\cdot S_3=48+2\cdot 37.9473+2\cdot 49.4773=222.8492{\text{dm}}^2 \end{gathered}$$

$V=\frac{1}{3}\cdot S_1\cdot h=\frac{1}{3}\cdot 48\cdot 12=192{\text{dm}}^3$

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