Chapter 4.6, Problem 8CYU

(a)

To determine

To prove: $\Delta RQV\cong \Delta STV$

Explanation of Solution

Given information: $\overline{QR}\&\overline{ST}\perp \overline{QT}$ , $\Delta VSR$ is an isosceles triangle with base $SR$ , and $\overline{QT}\parallel \overline{SR}$ .

  

Proof: Since $\Delta VSR$ is an isosceles triangle,

  $\begin{array}{l}\therefore VS=VR\\ \therefore 90°+\angle TSR=180°\\ \therefore \angle VSR=\angle VRS\end{array}$

  $\angle VTS\&\angle TSR$ are adjacent supplementary angles and $\overline{QT}\parallel \overline{SR}$ and $\overline{ST}$ transversal.

  $\therefore \angle VTS=90°\left[\overline{ST}\perp \overline{QT}\right]$

  $\begin{array}{l}\angle VTS+\angle TSR=180°\\ \therefore 90°+\angle TSR=180°\left(\because \angle VTS=90°\right)\\ \therefore \angle TSR=180°-90°\\ \therefore \angle TSR=90°\end{array}$

Similarly, $\angle VQR\&\angle QRS$ are adjacent supplementary angles and $\overline{QT}\parallel \overline{SR}$ and $\overline{RQ}$ transversal.

  $\therefore \angle VQR=90°\left[\overline{ST}\perp \overline{QT}\right]$

  $\begin{array}{l}\angle VQR+\angle QRS=180°\\ \therefore 90°+\angle QRS=180°\left(\because \angle VQR=90°\right)\\ \therefore \angle QRS=180°-90°\\ \therefore \angle QRS=90°\end{array}$

Therefore, $\angle VTS=\angle VRQ$

  $\begin{array}{l}\angle TVS=180°-90°-\angle VST\\ \angle RVQ=180°-90°-\angle VRQ\\ \therefore \angle TVS=\angle RVQ\end{array}$

We can see that $\angle TVS=\angle RVQ$ , $VS=VR$ and $\angle VST=\angle VRQ$ ,

Therefore, by angle side angle congruence postulate,

  $\Delta RQV\cong \Delta STV$

Hence, proved.

(b)

To determine

To find: The distance between $\overline{QR}\&\overline{ST}$ .

Answer to Problem 8CYU

The distance between $\overline{QR}$ and $\overline{ST}$ is $3$ .

Explanation of Solution

Given information: $VR=2.5m$ and $QR=2m$ .

  

Calculation: Since $\overline{RS}\perp \overline{ST}$ and $\overline{RS}\perp \overline{QR}$ and distance between $\overline{QR}$ and $\overline{ST}$ is $QT$ ,

  $\begin{array}{l}\therefore \angle TSR=90°\\ \therefore \angle QRS=90°\end{array}$

Now,

  $\begin{array}{l}VT=QV\left[\because \Delta RQV\cong STV\right]\\ QT=VT+QV\\ QT=2QV\end{array}$

By using Pythagoras Theorem in $\Delta QVR$ ,

  $\begin{array}{l}V{R}^2=Q{V}^2+Q{R}^2\\ QV=\sqrt{V{R}^2-Q{R}^2}\\ QV=\sqrt{{2.5}^2-2^2}\\ QV=1.5\end{array}$

Therefore, the required distance is

  $\begin{array}{l}QT=2QV\\ QT=2\times 1.5\\ QT=3\end{array}$