hexagonal pyramid

Question

The regular hexagonal pyramid trunk shown in the figure has a $ 5 $ cm side edge and base areas $ 54 \sqrt {3} $ $cm ^ 2$ and $ 6 \sqrt {3} $ $cm ^ 2$. Calculate your volume.

My ''solution'':

I think it would be $ a-5 <3 + 4 $

Answer 1

Observe that the truck is the difference between the two pyramids with their base areas given by $S_1= 54\sqrt3$ and $S_2=6\sqrt3$. Let $s$ be the side length of the large pyramid. Then, with $a=5$,

$$\frac{S_1}{S_2}=\frac{s^2}{(s-a)^2}=9\implies s = \frac{15}{2}$$

Also, the area of the large pyramid,

$$S_1 = 6\cdot\frac{\sqrt3}{4}OA^2\implies OA = 6$$

The height of the large pyramid is $h=\sqrt{s^2-OA^2}=\frac92$ and the its volume $V_1=\frac13 hS_1 = 81\sqrt3$. Then, use the ratio

$$\frac{V_1}{V_2}=\left(\frac{S_1}{S_2}\right)^{3/2}=27$$

to obtain the volume of the truck,

$$V=V_1-V_2 = \left(1-\frac1{27}\right)\cdot 81\sqrt3=78\sqrt3$$

Answer 2

The highlighted quadrilateral $OO'A'A$ in the image hints at the solution. Note that $\overline{O'A'}=l'=2$ and $\overline{OA}=l=6$

A line through $A'$ parallel to $OO'$ intersecting $OA$ at $X$ gives $\overline{XA}=4$, and with $\overline{AA'}=5$ we have $\overline{A'X}=k=3$ (Pythagorean theorem).

$OO'$ and $AA'$ meet at $Z$. Triangles $OAZ$ and $XAA'$ are similar, giving us $\overline{OZ}=\frac 92$ the full height of the pyramid.

$$V=\frac 92\times\frac{54\sqrt 3}{3}-\frac 32\times\frac{6\sqrt 3}{3}\\V=81\sqrt 3-3\sqrt 3=78\sqrt 3$$