Chaya Galloway

2021-01-10

To find: The values of KP, KM, MR, ML, MN and PR.
Given:

$\stackrel{―}{PR}\mid \mid \stackrel{―}{KL}=9,\mathrm{ln}=16,PM=2\left(KP\right)$
KM=KP+PM=3KP

funblogC

Conclusion:
By A-A similarity, it can be proved that:
$\mathrm{△}LMK\sim \mathrm{△}MNK$
Similar triangles have corresponding sides as proportional.
$\frac{MK}{LK}=\frac{NK}{MK}$
$⇒\frac{3KP}{25}=\frac{9}{3KP}$
$⇒KP=5$
$⇒KM=3KP=15$
By A-A similarity, it can be proved that:
$\mathrm{△}LMK\sim \mathrm{△}RMP$
Similar triangles have corresponding sides as proportional.
$\frac{MK}{LK}=\frac{PM}{RP}$
$⇒\frac{15}{25}=\frac{10}{RP}$
$⇒KP=16\frac{2}{3}$
Using Pythagorean Theorem in right angled triangle $\mathrm{△}LMK$:
$Hypotenus{e}^{2}=Bas{e}^{2}+Perpendicula{r}^{2}$
$={25}^{2}={15}^{2}+L{M}^{2}$
$⇒L{M}^{2}=400$
$⇒LM=20$
In similar triangles: $\mathrm{△}LMK\sim \mathrm{△}MNK$
Ratio can be written as:
$\frac{LM}{MK}=\frac{MN}{NK}$
$⇒\frac{20}{15}=\frac{MN}{9}$
$⇒MN=12$
$KP=5,KM=15,MR=13\frac{1}{3},ML=20,MN=12,PR=16\frac{2}{3}$

Do you have a similar question?