Chapter 12.5, Problem 17PSC

a.

To determine

To calculate: The volume of tetrahedron if each edge measures $6$ .

Answer to Problem 17PSC

The volume of tetrahedron is $25.45$ .

Explanation of Solution

Given information:

  $e=6.$

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Area of equilateral triangle: $A=\frac{s^2}{4}\sqrt{3}$

s = side of equilateral triangle Volume of pyramid $=\frac{1}{3}B\times h$

B = Base area of pyramid and h = height of pyramid

Calculation:

  

The slant height forms two ${30}^{\circ }{60}^{\circ }{90}^{\circ }\Delta \text{'}s$

  $\begin{array}{l}tan30^o=\frac{oppositeside}{adjacentside}\\ \frac{1}{\sqrt{3}}=\frac{3}{Slantheight}\end{array}$

Slant height $=3\sqrt{3}$

In the base, the triangle formed is a ${30}^{\circ }{60}^{\circ }{90}^{\circ }\Delta \text{'}s$

  $x=\frac{3}{\sqrt{3}}$

Side h can be calculated by applying Pythagoras Theorem.

In right angled triangle, we get

  $\begin{array}{l}{\left(slantheight\right)}^2={\left(x\right)}^2+{\left(h\right)}^2\\ {\left(3\sqrt{3}\right)}^2={\left(\sqrt{3}\right)}^2+{\left(h\right)}^2\\ 27=3+{\left(h\right)}^2\\ h^2=24\\ h=\sqrt{24}=2\sqrt{6}\end{array}$

Volume of pyramid $=\frac{1}{3}B\times h$

  $\begin{array}{l}B=\frac{s^2}{4}\sqrt{3}\\ B=\frac{{\left(6\right)}^2}{4}\sqrt{3}\\ B=9\sqrt{3}\end{array}$

Volume of pyramid $=\frac{1}{3}\times 9\sqrt{3}\times 2\sqrt{6}$

Volume of pyramid $=6\sqrt{18}=18\sqrt{2}=25.45$

b.

To determine

To find: The volume of tetrahedron if each edge measures $s$ .

Answer to Problem 17PSC

The volume of tetrahedron is $\frac{s^3}{12}\sqrt{3}$ .

Explanation of Solution

Given information:

  $e=s.$

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Area of equilateral triangle: $A=\frac{s^2}{4}\sqrt{3}$

s = side of equilateral triangle Volume of pyramid $=\frac{1}{3}B\times h$

B = Base area of pyramid and h = height of pyramid

Calculation:

  

The slant height forms two ${30}^{\circ }{60}^{\circ }{90}^{\circ }\Delta \text{'}s$

  $\begin{array}{l}tan30^o=\frac{oppositeside}{adjacentside}\\ \frac{1}{\sqrt{3}}=\frac{\frac{s}{2}}{Slantheight}\end{array}$

Slant height $=\frac{s}{2}\sqrt{3}$

In the base, the triangle formed is a ${30}^{\circ }{60}^{\circ }{90}^{\circ }\Delta \text{'}s$

Side h can be calculated by applying Pythagoras Theorem.

  $h=\frac{s}{3}\sqrt{6}$

Volume of pyramid $=\frac{1}{3}B\times h$

  $B=\frac{s^2}{4}\sqrt{3}$

Volume of pyramid $=\frac{1}{3}\times \left(\frac{s^2}{4}\sqrt{3}\right)\left(\frac{s}{3}\sqrt{6}\right)$

Volume of pyramid $=\frac{1}{3}\times \left(\frac{s^3}{12}\right)\sqrt{18}$

Volume of pyramid $=\frac{1}{3}\times \left(\frac{s^3}{12}\right)3\sqrt{2}$

Volume of pyramid $=\frac{s^3}{12}\sqrt{2}$