Chapter 7.3, Problem 18PPS

To determine

To identify: the similar triangle and find the value of the given length.

Answer to Problem 18PPS

  $\Delta WUZ\sim \Delta WUY$ , $WZ=30,UZ=18$ .

Explanation of Solution

Given:

The given length is $WZ,UZ$

  

Calculation:

From the above triangle in the figure,

  $\begin{array}{l}\angle WUZ=\angle WUY\left\{given\right\}\\ \angle ZWU=\angle WYU\left\{given\right\}\\ \Delta WUZ\sim \Delta WUY\{\text{by AA-aimilarity}}\end{array}$

Now, in $\angle WUY$ , apply Pythagoras theorem, then

  $\begin{array}{l}W{U}^2+U{Y}^2=W{Y}^2\\ \Rightarrow W{U}^2+{32}^2={40}^2\left\{\text{substitutes these value}\right\}\\ \therefore W{U}^2+1024=1600\\ W{U}^2=576\\ \Rightarrow WU=\sqrt{576}\{\text{taking square root both sides}}\\ WU=24\end{array}$

Since, the triangles are similar, therefore the corresponds sides are proportional. Thus,

  $\begin{array}{l}\frac{ZU}{WU}=\frac{ZW}{WY}\\ \Rightarrow \frac{x+6}{24}=\frac{3x-6}{40}\left\{\text{substitutes these value}\right\}\end{array}$ .

  $\begin{array}{l}\therefore 40\left(x+6\right)=24\left(3x-6\right)\left\{\text{cross-multiplication}\right\}\\ \Rightarrow 40x+240=72x-144\{\text{open the parenthesis}}\\ 40x-72x=-144-240\{\text{simplify}}\\ \Rightarrow -32x=-384\\ \therefore x=\frac{-384}{-32}\text{{divide both sides by}-32\}\\ \Rightarrow x=12\end{array}$

Thus, the value of

  $\begin{array}{l}WZ=3x-6=3\left(12\right)-6=30\\ UZ=x+6=12+6=18\end{array}$

Hence, the similar triangle is $\Delta WUZ\sim \Delta WUY$ and find the value of the given length $WZ=30,UZ=18$ .