since the triangles are similar, then 32 = x+7 those two angles are too
solve for "x"
now using proportions [tex]\bf \cfrac{\textit{small triangle}}{\textit{large triangle}}\qquad \cfrac{JK}{18}=\cfrac{12}{12+10}[/tex]
solve for JK
[tex]\bf \cfrac{\textit{small triangle}}{\textit{large triangle}}\qquad \cfrac{12}{12+10}=\cfrac{9}{9+KG}\implies 12(9+KG)=(22)9
\\\\\\
108+9KG=198[/tex]
solve for KG
the scale factor of the perimeter of both triangles is just the same as the ratio of the sides, or [tex]\bf \cfrac{12}{12+10}[/tex]
