Chapter 12.2, Problem 11PSC

a.

To determine

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

Answer to Problem 11PSC

The total surface area of a regular tetrahedron is $36\sqrt{3}.$ .

Explanation of Solution

Given Information:

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

Formula used:

Area of one face of a regular tetrahedron $\left(A\right)=\frac{\sqrt{3}}{4}{a}^2$

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

Calculation:

Each face of a regular tetrahedron is an equilateral triangle.

So, Area of one face of a regular tetrahedron $=\frac{\sqrt{3}}{4}{a}^2$

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

As $a=6,$ we have

  $A=\frac{\sqrt{3}}{4}\times 6^2$

Now, total surface area of a regular tetrahedron $\left(SA\right)=4\times$ Area of one face of a tetrahedron

  $\begin{array}{l}\therefore SA=4\times \frac{\sqrt{3}}{4}\times 6^2\\ \therefore SA=36\sqrt{3}\end{array}$

Hence, total surface area of a regular tetrahedron is $36\sqrt{3}.$

b.

To determine

To find: The height of a regular tetrahedron with edge 6.

Answer to Problem 11PSC

The height of a regular tetrahedron is $2\sqrt{6}.$

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:

It is given that, edge $\left(a\right)=6$

Now, we know that

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

  $\begin{array}{l}\therefore h=\frac{\sqrt{6}}{3}\times 6\\ h=\sqrt{6}\times 2\\ h=2\sqrt{6}\end{array}$

Hence, height of a regular tetrahedron is $2\sqrt{6}.$