Question 286895
Since the perimeter of JKL is 72, and we know two sides of JKL, we can easily solve for the missing length JL by remembering that 


Perimeter = side1+side2+side3


So {{{72=12+42+JL}}}. Solve for JL to get {{{JL=18}}}



Since ABC~JKL (triangle ABC is similar to triangle JKL), we can set up the ratios:


{{{AB/JK=BC/KL}}} and {{{AC/JL=BC/KL}}}



note: it might help to draw out the triangles to see what's going on.



{{{AB/JK=BC/KL}}} Start with the first equation



{{{AB/12=28/42}}} Plug in BC=28, JK=12, and KL=42



{{{AB/12=2/3}}} Reduce.



{{{AB=(2/3)*12}}} Multiply both sides by 12



{{{AB=24/3}}} Multiply



{{{AB=8}}} Reduce.



So one side of ABC is AB = 8




{{{AC/JL=BC/KL}}} Move onto the second equation



{{{AC/18=28/42}}} Plug in BC=28, JL=18, and KL=42



{{{AC/18=2/3}}} Reduce.



{{{AC=(2/3)*18}}} Multiply both sides by 18



{{{AC=36/3}}} Multiply



{{{AC=12}}} Reduce.



So the last side of ABC is AC = 12



Now just add up the sides AB = 8, BC = 28, and AC = 12 to get the perimeter of ABC: 


P = 8+28+12 = 48



So the perimeter of ABC is 48 units.



Note: the ratio of the perimeter of ABC to the perimeter of JKL is the same as the ratio of any two corresponding sides. This saves us a lot of work, but it's always good to take the long way so we can see what's going on.