Chapter 8.4, Problem 35WE

a)

To determine

To find: the length of the longer diagonal of the quadrilateral.

Answer to Problem 35WE

The length of the longer diagonal is $RT=4\sqrt{2}+4\sqrt{6}$ .

Explanation of Solution

Given Information:

In quadrilateral $QRST$ ,$m\angle R=60,m\angle T=90,QR=RS,ST=8$ and $TQ=8$ .

Calculation:

Consider the following figure,

The figure shows a quadrilateral $QRST$ .

  

The law of sine is stated in the figure given below.

  $\frac{Sin\left(A\right)}{a}=\frac{Sin\left(B\right)}{b}=\frac{Sin\left(C\right)}{c}$

  

  $\frac{a}{Sin\left(A\right)}=\frac{b}{Sin\left(B\right)}=\frac{c}{Sin\left(C\right)}$

Because of symmetry the segment $RT$ is a bisector of angles, $\angle T$ and $\angle R$

The sum of three angles in a triangle is $180°$

This implies that from $\Delta STR$ , the measured of $\angle S$ is,

  $\begin{array}{c}\angle S=180°-\frac{90°}{2}-\frac{60°}{2}\\ =180°-45°-30°\\ =105°\end{array}$

Using the law of sine to get,

  $\begin{array}{c}\frac{RT}{\sin \left(\angle S\right)}=\frac{8}{\sin \left(30°\right)}\\ RT=\frac{8}{\sin 30°}\sin \left(105°\right)\\ =16\sin 105°\end{array}$

Simplify further to get,

  $\begin{array}{c}RT=16\left[\sin 60°\cos 45°+\cos 60°\sin 45°\right]\\ =16\left(\frac{\sqrt{3}}{2}\cdot \frac{1}{\sqrt{2}}+\frac{1}{2}\cdot \frac{1}{\sqrt{2}}\right)\\ =4\sqrt{2}+4\sqrt{6}\end{array}$

Therefore, the length of the longer diagonal is $RT=4\sqrt{2}+4\sqrt{6}$ .

b)

To determine

To find: the ratio of $RT$ to $QS$ .

Answer to Problem 35WE

The required ration is $\frac{1+\sqrt{3}}{2}$ .

Explanation of Solution

Given Information:

In quadrilateral $QRST$ , $m\angle R=60,m\angle T=90,QR=RS,ST=8$ , and $TQ=8$ .

Calculation:

Consider the following figure,

The figure shows a quadrilateral $QRST$ .

  

The law of sine is stated in the figure given below.

  $\frac{Sin\left(A\right)}{a}=\frac{Sin\left(B\right)}{b}=\frac{Sin\left(C\right)}{c}$

  

  $\frac{a}{Sin\left(A\right)}=\frac{b}{Sin\left(B\right)}=\frac{c}{Sin\left(C\right)}$

From the right angles triangle $QTS$ .

By the use of Pythagoras Theorem,

  $\begin{array}{c}QS=\sqrt{8^2+8^2}\\ =8\sqrt{2}\end{array}$

So the ratio of $RT$ to $QS$ is given by,

  $\begin{array}{c}\frac{RT}{QS}=\frac{4\sqrt{2}+4\sqrt{6}}{8\sqrt{2}}\\ =\frac{4\sqrt{2}}{8\sqrt{2}}+\frac{4\sqrt{6}}{8\sqrt{2}}\\ =\frac{1}{2}+\frac{1}{2}\sqrt{\frac{6}{2}}\\ =\frac{1+\sqrt{3}}{2}\end{array}$

Therefore, the required ration is $\frac{1+\sqrt{3}}{2}$ .