Chapter 12.2, Problem 32WE

(a)

To determine

To check: The pyramid with greater volume.

Answer to Problem 32WE

Both the pyramids have equal volume.

Explanation of Solution

Given information:

Different pyramids are inscribed in two identical cubes, as shown below:

  

The formula for the volume of pyramids is one third of product of base area and height.

The base of both the pyramids is of same dimensions. So, the base area for both the pyramids is equal. The height of both the pyramids is also the same. Both the base area and height of pyramids are equal.

Therefore, both the pyramids have equal volume.

(b)

To determine

To check: The pyramid that has the greater total area.

Answer to Problem 32WE

The total area of pyramid $F-ABCD$ has greater total area.

Explanation of Solution

Given information:

Different pyramids are inscribed in two identical cubes, as shown below:

  

The total area of pyramid is the sum of lateral area and area of base. The area of base for both the pyramids is same.

The lateral area of first pyramid is the sum of the area of triangles $AFB$ , $BFC$ , $CFD$ and $DFA$ .

The area of triangle $AFB$ is equal to $\frac{1}{2}\left(1\right)\left(1\right)=\frac{1}{2}$ .

The area of triangle $AFB$ and $BFC$ is equal.

So, the area of triangles $AFB$ and $BFC$ is equal to $\frac{1}{2}$ .

The height of triangle $CFD$ is the length of diagonal of square with side $1$ . So, the length of diagonal is equal to $\sqrt{2}$ .

So, The area of triangle $CFD$ is equal to $\frac{1}{2}\left(\sqrt{2}\right)\left(1\right)=\frac{\sqrt{2}}{2}$ . The area of triangle $CFD$ and $DFA$ is equal.

So, the area of triangles $CFD$ and $DFA$ is equal to $\frac{\sqrt{2}}{2}$ .

Simplify the lateral area of first pyramid.

  $\begin{array}{c}{A}_1=\frac{1}{2}+\frac{1}{2}+\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\\ =2\times \frac{1}{2}+2\times \frac{\sqrt{2}}{2}\\ =1+\sqrt{2}\\ =2.41\end{array}$

The formula for the lateral area of second pyramid is equal to $\frac{1}{2}pl$ . The perimeter of the base of pyramid is the perimeter of square with side $1$ . So, the perimeter of the base is equal to $4\left(1\right)=4$ .

Simplify the slant height of the pyramid:

  $\begin{array}{c}\sqrt{1^2+{\left(\frac{1}{2}\right)}^2}=\sqrt{1+\frac{1}{4}}\\ =\sqrt{\frac{5}{4}}\\ =\frac{\sqrt{5}}{2}\end{array}$

Substitute $\frac{\sqrt{5}}{2}$ for $l$ and $4$ for $p$ in the formula for lateral area.

  $\begin{array}{c}\text{A}=\frac{1}{2}pl\\ =\frac{1}{2}\left(4\right)\left(\frac{\sqrt{5}}{2}\right)\\ =\sqrt{5}\\ =2.23\end{array}$

The lateral area of first pyramid is greater than the lateral area of second pyramid.

Therefore, the total area of pyramid $F-ABCD$ has greater total area.