Chapter 12.5, Problem 25PPS

To determine

To find the volume of the given octagonal pyramidal greenhouse.

Answer to Problem 25PPS

  $V=32.16$ cubic feet (approximately)

Explanation of Solution

Given:

  

Height, ( $h$ ) $=5$ feet

Octagonal base:

Side, ( $a$ ) $=2$ feet

Formula used:

Volume of pyramid, $V=\frac{1}{3}Bh$

where $B$ is the area of the base and $h$ is the height of the pyramid.

Area of octagon $=2(1+\sqrt{2})a^2$ where $a$ is measure of the side of the octagon.

Calculation:

Since the given pyramid is a octagonal pyramid, so it has a octagonal base.

So, area of the base, $B=2(1+\sqrt{2})a^2$ ,

  $\Rightarrow B==2(1+\sqrt{2}){(2)}^2$

  $\Rightarrow B\approx 19.31$ square feet

Now for the volume ( $V$ ) of the pyramid,

  $V=\frac{1}{3}Bh$

Now put the values of $B=19.31$ and $h=5$ ,

  $\Rightarrow \frac{1}{3}\times 19.5\times 5$

  $\Rightarrow \frac{96.5}{3}$

  $\Rightarrow V\approx 32.16$ cubic feet

Conclusion:

Therefore, the volume of the given octagonal pyramid is $32.16$ cubic feet (approximately).