Chapter 12.2, Problem 12PSC

a.

To determine

To calculate: The total surface area of a regular octahedron with edge 6 mm.

Answer to Problem 12PSC

The total surface area of a regular octahedron is $72\sqrt{3}.$ .

Explanation of Solution

Given Information:

Each edge of a regular octahedron $\left(a\right)=6$

Formula used:

Area of an equilateral triangle $\left(A\right)=\frac{\sqrt{3}}{4}{a}^2$

  $"a"$ is the length of side of an equilateral triangle

Calculation:

Each face of a regular octahedron is an equilateral triangle.

So, Area of one face of a regular octahedron $\left(A_1\right)=$ Area of an equilateral triangle

As area of an equilateral triangle $\left(A\right)=\frac{\sqrt{3}}{4}{a}^2$

  $\therefore A_1=\frac{\sqrt{3}}{4}{a}^2$

Where $"a"$ is the edge of a regular octahedron

As $a=6,$ we have

  $A_1=\frac{\sqrt{3}}{4}\times 6^2$

Now, total surface area of a regular octahedron $\left(SA\right)=8\times A_1$

  $\begin{array}{l}\therefore SA=8\times \frac{\sqrt{3}}{4}\times 6^2\\ \therefore SA=72\sqrt{3}\end{array}$

Hence, total surface area of a regular tetrahedron is $72\sqrt{3}.$

b.

To determine

To find: The distance of C from E where both are the vertices of a regular octahedron.

Answer to Problem 12PSC

The distance from C to E is $6\sqrt{2}.$

Explanation of Solution

Given Information:

Edge of a regular tetrahedron $\left(a\right)=6$

Formula used:

Height of a regular tetrahedron $\left(h\right)=\frac{\sqrt{6}}{3}a$

  $"a"$ is the edge of a regular tetrahedron

Calculation:

Distance of point C from E will be obtained by joining E and C.

EC forms diagonal of a square CDEF.

Now, each side of a square CDEF is $\left(s\right)=a=6$

We know that, diagonal of square $\left(d\right)=\sqrt{2}s$

  $\begin{array}{l}\therefore d=\sqrt{2}\times 6\\ \therefore d=6\sqrt{2}\end{array}$

Hence, the distance of C from E is $6\sqrt{2}.$

c.

To determine

To find: The distance of A from B where both are the vertices of a regular octahedron.

Answer to Problem 12PSC

The distance from C to E is $6\sqrt{2}.$

Explanation of Solution

Given Information:

Edge of a regular tetrahedron $\left(a\right)=6$

From part b, $CE=6\sqrt{2}$

Formula used:

Using Pythagoras theorem, we have

  ${\text{Hypotenuse}}^2={\text{Base}}^2+{\text{Altitude}}^2$

Calculation:

Distance of point A from B will be obtained by joining A and B .

Let O be the centre of the square on which the eight triangular faces stand.

Now, $AO,OC\text{and}AC$ form a right angled triangle with $\angle O={90}^{\text{o}}$

Using Pythagoras theorem, we get

  $A{C}^2=O{A}^2+O{C}^2$

Now, $AC=a=6\text{mm}$ and $OC=\frac{CE}{2}=\frac{6\sqrt{2}}{2}=3\sqrt{2}$

  $\begin{array}{l}\therefore 6^2=O{A}^2+{\left(3\sqrt{2}\right)}^2\\ \therefore O{A}^2=36-18\\ \therefore OA=3\sqrt{2}\end{array}$

  $\begin{array}{l}AB=2OA=2\left(3\sqrt{2}\right)\\ \therefore AB=6\sqrt{2}\end{array}$

Hence, the distance from A to B is $6\sqrt{2}.$

d.

To determine

To find: The name of a quadrilateral formed by the shape $ACBE.$

Answer to Problem 12PSC

The shape of quadrilateral $ACBE$ is a square.

Explanation of Solution

Given Information:

Edge of a regular tetrahedron $\left(a\right)=6$

From part b, $CE=6\sqrt{2}$

From part c, $AB=6\sqrt{2}$

In the quadrilateral $ACBE,$ we have

  $AC=BC=BE=AE$ … (Each = 6, being edge of a regular octahedron)

  $AB=CE$ … (Each is equal to $6\sqrt{2}$ )

Therefore, all the sides of a quadrilateral $ACBE$ are equal.

Also, the diagonals $AB\text{and}CE$ are equal and bisect each other.

So, $ACBE$ becomes a square.

Hence, the shape of quadrilateral $ACBE$ is a square.