Chapter 8.4, Problem 36WE

To determine

To find: the perimeter of triangle.

Answer to Problem 36WE

The perimeter of triangle is $8+12\sqrt{2}+4\sqrt{6}$ .

Explanation of Solution

Given:

The triangle with dimension is shown below:

  

Calculation:

Draw a perpendicular as shown below:

  

In triangle ABD,

  $\begin{array}{c}\cos \angle ABD=\frac{\text{Base}}{\text{Hypotenuse}}\\ \cos 45°=\frac{BD}{8}\\ \frac{1}{\sqrt{2}}=\frac{BD}{8}\\ BD\approx 4\sqrt{2}\end{array}$

Also, side BD and AD are equal as angles ABD and ADB are equal to 45 degree. Thus, Side AD is $4\sqrt{2}$ .

In triangle BDC,

  $\begin{array}{c}\cos \angle DBC=\frac{\text{Base}}{\text{Hypotenuse}}\\ \cos 60°=\frac{BD}{BC}\\ \frac{1}{2}=\frac{4\sqrt{2}}{BC}\\ BC\approx 8\sqrt{2}\end{array}$

And,

  $\begin{array}{c}\sin \angle DBC=\frac{\text{perpendicular}}{\text{Hypotenuse}}\\ \sin 60°=\frac{DC}{BC}\\ \frac{\sqrt{3}}{2}=\frac{DC}{8\sqrt{2}}\\ DC\approx 4\sqrt{6}\end{array}$

Perimeter of triangle is calculated as follows:

  $\begin{array}{c}P=AB+BC+AD+DC\\ =8+8\sqrt{2}+4\sqrt{2}+4\sqrt{6}\\ =8+12\sqrt{2}+4\sqrt{6}\end{array}$

Thus, the perimeter of triangle is $8+12\sqrt{2}+4\sqrt{6}$ .