202405071349 Exercise 13.2.1

A right pyramid, $12\,\mathrm{cm}$ high, stands on a rectangular base $6\,\mathrm{cm}$ by $10\,\mathrm{cm}$ . Calculate (a) the length of an edge of the pyramid; (b) the angles the triangular faces made with the base; (c) the volume of the pyramid.

_{Extracted from A. Godman & J. F. Talbert. (1973). Additional Mathematics Pure and Applied in SI Units.}

Roughwork.

Commit my visualization to drawing.

A right pyramid is a pyramid where the base is circumscribed about the circle and the altitude of the pyramid meets at the circle’s center.

_{Wikipedia on Pyramid (geometry)}

The pyramid above has a polygonal base, here the rectangle $QRST$ , and an apex $P$ , here the common vertex of triangles $\triangle PTQ$ , $\triangle PQR$ , $\triangle PRS$ , and $\triangle PST$ . The altitude is based on the origin $O$ . To suit our coordinates to this problem, we write

$\begin{aligned} P & =P(0,0,12) \\ Q & =Q(3,5,0) \\ R & =R(-3,5,0) \\ S & =S(-3,-5,0) \\ T & =T(3,-5,0) \\ \end{aligned}$

such that

$\begin{aligned} a & = c = 10 \\ b & = d = 6 \\ e & = f = g = h \\ & = \surd \big\{ (12)^2 + \big( \sqrt{(6)^2+(10)^2}/2\big)^2 \big\} \\ & = \sqrt{178} \\ & = 13.3417\quad\textrm{(4 d.p.)} \\ \end{aligned}$

For the edges of its base, write

$\begin{aligned} a=\overline{TQ} & :\begin{cases} x=3 \\ |y|\leqslant 5 \\ z=0 \\ \end{cases} \\ b=\overline{QR} & :\begin{cases} |x|\leqslant 3 \\ y=5 \\ z=0 \\ \end{cases} \\ c=\overline{RS} & :\begin{cases} x=-3 \\ |y|\leqslant 5 \\ z=0 \\ \end{cases} \\ d=\overline{ST}& :\begin{cases} |x|\leqslant 3 \\ y=-5 \\ z=0 \\ \end{cases} \\ \end{aligned}$

and for, the lateral, edge $e=\overline{PQ}$ :

$\begin{aligned} \frac{x-3}{0-3} & = \frac{y-5}{0-5} = \frac{z-0}{12-0} \\ \end{aligned}$

edge $f=\overline{PR}$ :

$\begin{aligned} \frac{x-(-3)}{0-(-3)} & = \frac{y-5}{0-5} = \frac{z-0}{12-0} \\ \end{aligned}$

edge $g=\overline{PS}$ :

$\begin{aligned} \frac{x-(-3)}{0-(-3)} & = \frac{y-(-5)}{0-(-5)} = \frac{z-0}{12-0} \\ \end{aligned}$

and edge $h=\overline{PT}$ :

$\begin{aligned} \frac{x-3}{0-3} & = \frac{y-(-5)}{0-(-5)} = \frac{z-0}{12-0} \\ \end{aligned}$

For lateral surface $A$ enclosed by edges $a$ , $h$ , and $e$ , write

$A(x,y,z):\begin{cases} (x,y,z)\in [0,3]\times [-5,5]\times [0,12] \\ \textrm{s.t. }\displaystyle{\frac{|y|}{5}\leqslant \frac{x}{3}=1-\frac{z}{12}} \\ \end{cases}$

for lateral surface $B$ by edges $b$ , $e$ , and $f$ , write:

$B(x,y,z):\begin{cases} (x,y,z)\in [-3,3]\times [0,5]\times [0,12] \\ \textrm{s.t. }\displaystyle{\frac{|x|}{3}\leqslant \frac{y}{5}=1-\frac{z}{12}} \\ \end{cases}$

for lateral surface $C$ by edges $c$ , $f$ , and $g$ , write:

$C(x,y,z):\begin{cases} (x,y,z)\in [-3,0]\times [-5,5]\times [0,12] \\ \textrm{s.t. }\displaystyle{\frac{|y|}{5}\leqslant -\frac{x}{3}=1-\frac{z}{12}} \\ \end{cases}$

for lateral surface $D$ by edges $d$ , $g$ , and $h$ , write:

$D(x,y,z):\begin{cases} (x,y,z)\in [-3,3]\times [-5,0]\times [0,12] \\ \textrm{s.t. }\displaystyle{\frac{|x|}{3}\leqslant -\frac{y}{5}=1-\frac{z}{12}} \\ \end{cases}$