Chapter 7.5, Problem 43SPR

To determine

To Find: The values of $KP,KM,MR,ML,MN$ and $PR$ .

Answer to Problem 43SPR

  $KP=5,KM=15,MR=13\frac{1}{3},ML=20,MN=12,PR=16\frac{2}{3}$

Explanation of Solution

Given:

  

  $\overline{PR}\left|\right|\overline{KL},KN=9,LN=16,PM=2(KP)$

  $KM=KP+PM=3KP$

By $A-A$ similarity, it can be proved that:

  $\Delta LMK\sim \Delta MNK$

Similar triangles have corresponding sides as proportional.

  $\begin{array}{l}\frac{MK}{LK}=\frac{NK}{MK}\\ \Rightarrow \frac{3KP}{25}=\frac{9}{3KP}\\ \Rightarrow KP=5\end{array}$

  $\Rightarrow KM=3KP=15$

By $A-A$ similarity, it can be proved that:

  $\Delta LMK\sim \Delta RMP$

Similar triangles have corresponding sides as proportional.

  $\begin{array}{l}\frac{MK}{LK}=\frac{PM}{RP}\\ \Rightarrow \frac{15}{25}=\frac{10}{RP}\\ \Rightarrow KP=16\frac{2}{3}\end{array}$

Using Pythagorean Theorem in right angled triangle $\Delta LMK$ :

  $\begin{array}{l}Hypotenus{e}^2=Bas{e}^2+Perpendicula{r}^2\\ \Rightarrow {25}^2={15}^2+L{M}^2\\ \Rightarrow L{M}^2=400\\ \Rightarrow LM=20\end{array}$

In similar triangles: $\Delta LMK\sim \Delta MNK$

Ratio can be written as:

  $\begin{array}{l}\frac{LM}{MK}=\frac{MN}{NK}\\ \Rightarrow \frac{20}{15}=\frac{MN}{9}\\ \Rightarrow MN=12\end{array}$

  $KP=5,KM=15,MR=13\frac{1}{3},ML=20,MN=12,PR=16\frac{2}{3}$