Chapter 7.3, Problem 19PPS

To determine

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

Answer to Problem 19PPS

  $\Delta HJK\sim \Delta NQP$ , $HJ=15,HK=10$ .

Explanation of Solution

Given:

The given length is $HJ,HK$

  

Calculation:

From the above triangle in the figure,

  $\begin{array}{l}\angle H=\angle N\left\{given\right\}\\ \angle J=\angle Q\left\{given\right\}\\ \Delta HJK\sim \Delta NQP\{\text{by AA-aimilarity}}\end{array}$

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

  $\begin{array}{l}\frac{HJ}{NQ}=\frac{HK}{NP}\\ \Rightarrow \frac{4x+7}{12}=\frac{6x-2}{8}\left\{\text{substitutes these value}\right\}\end{array}$ .

  $\begin{array}{l}\therefore 8\left(4x+7\right)=12\left(6x-2\right)\left\{\text{cross-multiplication}\right\}\\ \Rightarrow 32x+56=72x-24\{\text{open the parenthesis}}\\ 32x-72x=-24-56\{\text{simplify}}\\ \Rightarrow -40x=-80\\ \therefore x=\frac{-80}{-40}\text{{divide both sides by}-32\}\\ \Rightarrow x=2\end{array}$

Thus, the value of

  $\begin{array}{l}HJ=4x+7=4\left(2\right)+7=15\\ HK=6x-2=12+6=10\end{array}$

Hence, the similar triangle is $\Delta HJK\sim \Delta NQP$ and the value of the given length

  $HJ=15,HK=10$ .