Quadrilateral 5130

There is a regular quadrilateral pyramid with the base edge length a = 3 cm and with the length of the side edge h = 8 cm.
Please calculate its surface area and volume.

Correct answer:

S =  56.1487 cm²
V =  23.1409 cm³

Step-by-step explanation:

$$\begin{gathered} a=3\text{cm} \\ h=8\text{cm} \\ u=\sqrt{2}\cdot a=\sqrt{2}\cdot 3=3\sqrt{2}\text{cm}\doteq 4.2426\text{cm} \\ v=\sqrt{h^2-(u/2{)}^2}=\sqrt{8^2-(4.2426/2{)}^2}\doteq 7.7136\text{cm} \\ {h}_2=\sqrt{v^2+(a/2{)}^2}=\sqrt{7.7136^2+(3/2{)}^2}\doteq 7.8581\text{cm} \\ {S}_1=a^2=3^2=9{\text{cm}}^2 \\ {S}_2=a\cdot h_2/2=3\cdot 7.8581/2\doteq 11.7872{\text{cm}}^2 \\ S=S_1+4\cdot S_2=9+4\cdot 11.7872=56.1487{\text{cm}}^2 \end{gathered}$$

$V=S_1\cdot v/3=9\cdot 7.7136/3=23.1409{\text{cm}}^3$

Try another example