Triangular pyramid

The regular triangular pyramid ABCDV has a base edge length of 8 cm and a height of 7 cm. Calculate the pyramid's surface area and volume.

Final Answer:

V =  64.6632 cm³
S =  116.1662 cm²

Step-by-step explanation:

$$\begin{gathered} a=8\text{cm} \\ h=7\text{cm} \\ {h}_2=\sqrt{a^2-(a/2{)}^2}=\sqrt{8^2-(8/2{)}^2}=4\sqrt{3}\text{cm}\doteq 6.9282\text{cm} \\ {h}_1^2=h^2+(h_2/3{)}^2 \\ {h}_1=\sqrt{h^2+(h_2/3{)}^2}=\sqrt{7^2+(6.9282/3{)}^2}\doteq 7.3711\text{cm} \\ {S}_1=a\cdot h_1/2=8\cdot 7.3711/2\doteq 29.4845{\text{cm}}^2 \\ {S}_2=\frac{\sqrt{3}}{4}\cdot a^2=\frac{\sqrt{3}}{4}\cdot 8^2=16\sqrt{3}{\text{cm}}^2\doteq 27.7128{\text{cm}}^2 \\ V=\frac{1}{3}\cdot S_2\cdot h=\frac{1}{3}\cdot 27.7128\cdot 7=64.6632{\text{cm}}^3 \end{gathered}$$

$S=3\cdot S_1+S_2=3\cdot 29.4845+27.7128=116.1662{\text{cm}}^2$

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