Quadrilateral 7815

The area of the mantle of a regular quadrilateral pyramid is equal to twice the area of its base. Calculate the pyramid's volume if the base edge's length is 20 dm.

Correct answer:

V =  2309.4011 dm³

Step-by-step explanation:

$$\begin{gathered} a=20\text{dm} \\ {S}_1=a^2=20^2=400{\text{dm}}^2 \\ {S}_2=2\cdot S_1=2\cdot 400=800{\text{dm}}^2 \\ {S}_2=4\cdot a\cdot h_2/2 \\ {h}_2=S_2/a/2=800/20/2=20\text{dm} \\ {h}_2^2=h^2+(a/2{)}^2 \\ h=\sqrt{h_2^2-(a/2{)}^2}=\sqrt{20^2-(20/2{)}^2}=10\sqrt{3}\text{dm}\doteq 17.3205\text{dm} \\ V=\frac{1}{3}\cdot S_1\cdot h=\frac{1}{3}\cdot 400\cdot 17.3205=2309.4011{\text{dm}}^3 \end{gathered}$$

Try another example