Chapter 7.3, Problem 20PPS

To determine

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

Answer to Problem 20PPS

  $\Delta AFC\sim \Delta DFB$ , $DB=5,CB=15$ .

Explanation of Solution

Given:

The given length is $DB,BC$

  

Calculation:

From the above triangle in the figure,

  $\begin{array}{l}\angle ACF=\angle DBF\left\{given\right\}\\ \angle AFC=\angle DFB\left\{each{90}^{\circ }\right\}\\ \Delta AFC\sim \Delta DFB\{\text{by AA-aimilarity}}\end{array}$

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

  $\begin{array}{l}\frac{AC}{DB}=\frac{CF}{BF}\\ \Rightarrow \frac{20}{2x+1}=\frac{12}{2x-1}\left\{\text{substitutes these value}\right\}\end{array}$ .

  $\begin{array}{l}\therefore 20\left(2x-1\right)=12\left(2x+1\right)\left\{\text{cross-multiplication}\right\}\\ \Rightarrow 20x-20=24x+12\text{}\{\text{open the parenthesis}}\\ 40x-24x=12+20\{\text{simplify}}\\ \Rightarrow 16x=32\\ \therefore x=\frac{32}{16}\text{{divide both sides by}16\}\\ \Rightarrow x=2\end{array}$

Thus, the value of

  $\begin{array}{l}DB=2x-1=2\left(2\right)+1=5\\ CB=12+2x+1=12+2\left(2\right)-1=15\end{array}$

Hence, the similar triangle is $\Delta AFC\sim \Delta DFB$ and the value of the given length $DB=5,CB=15$ .