MO = 12 and PR = 3
Solution:
Given [tex]\triangle M N O \sim \Delta P Q R[/tex].
Perimeter of ΔMNO = 48
Perimeter of ΔPQR = 12
MO = 12x and PR = x + 2
If two triangles are similar, then the ratio of corresponding sides is equal to the ratio of perimeter of the triangles.
[tex]$\Rightarrow \frac{\text{Perimeter of }\triangle MNO}{\text{Perimeter of }\triangle PQR} =\frac{MO}{PR}[/tex]
[tex]$\Rightarrow \frac{48}{12} =\frac{12x}{x+2}[/tex]
Do cross multiplication.
[tex]$\Rightarrow 48({x+2})= 12(12x)[/tex]
[tex]$\Rightarrow 48x+96= 144x[/tex]
Subtract 48x from both sides.
[tex]$\Rightarrow 48x+96-48x= 144x-48x[/tex]
[tex]$\Rightarrow 96= 96x[/tex]
Divide by 96 on both sides, we get
⇒ 1 = x
⇒ x = 1
Substitute x = 1 in MO an PR.
MO = 12(1) = 12
PR = 1 + 2 = 3
Therefore MO = 12 and PR = 3.
