Question

Determine the surface of a pyramid with a base of 21 meters cubed and a height of 10 meters, and a pyramid with a base of 24 meters cubed and 8 meters in height.

Answer 1

Task. Determine the surface of a pyramid with a base of 21 meters cubed and a height of 10 meters, and a pyramid with a base of 24 meters cubed and 8 meters in height.

Solution. The answer depends on the form of the base. However it is not clear from the assumptions what is the base: “21 meters cubed” is ambiguous.

So we assume that in both cases the base of the pyramid is a square and the vertex of pyramide is over the center of that square. Then the most probable formulation of the problem is the following: find the area of the surface of the pyramid having volume 21 meters cubed and a height of 10 meters. And similarly in the second case.

1) Notice that the surface of the pyramid consists of a base being a square and 4 faces being equal triangles, see Figure:

Let $S_{b}$ be the area of the base, and $S_{f}$ be the area of one of the faces. Then the area of all the surface of pyramid is

$S=S_{b}+4S_{f}.$

We have that

$V=21m^{3},\qquad h=10m.$

It is known that the volume of the pyramid is

$V=\frac{1}{3}hS_{b},$

where $S_{b}$ is the area of the base. Hence

$S_{b}=\frac{3V}{h}=\frac{3*21}{10}=6.3m^{2}.$

It remains to find $S_{f}$.

The base of the pyramid is a square. Let $a$ be its side. Then

$S_{b}=a^{2}=6.3,$

whence

$a=\sqrt{6.3}m.$

The face of the pyramid is a triangle and $a$ is its base. So we should find the height $l$ of this triangle. From figure we obtain that

$l=\sqrt{h^{2}+(a/2)^{2}}=\sqrt{10^{2}+6.3/4}=\sqrt{101.575}m.$

Therefore the area of the face is

$S_{f}=\frac{1}{2}la=\frac{1}{2}\sqrt{101.575}\sqrt{6.3}=\frac{1}{2}\sqrt{639.9225}=\sqrt{159.980625}m^{2}.$

Hence the area of the pyramid is

$S=S_{b}+4S_{f}=6.3+4\sqrt{159.980625}\approx 56.89m^{2}.$

2) In the second case the arguments are literally the same. We have that

$S_{b}=\frac{3V}{h}=\frac{3*24}{8}=9,$

2

$$a = \sqrt{S_b} = \sqrt{9} = 3,$$ $$l = \sqrt{h^2 + (a/2)^2} = \sqrt{8 + 9/4} = \sqrt{10.25} \, \text{m}.$$ $$S_f = \frac{1}{2} l a = \frac{1}{2} \sqrt{10.25} \cdot 3 = \sqrt{23.0625} \, \text{m}^2,$$

and so the area of the pyramid is

$$S = S_b + 4 S_f = 9 + 4 \sqrt{23.0625} \approx 28.21 \, \text{m}^2.$$