There are (as with most math problems) at least a couple of different ways you can approach this. I prefer to think in terms of scale factors.
1) The perimeter of the 4:3:5 triangle is 4+3+5 = 12. The middle length of the 4:3:5 triangle is 4. The middle length of ∆ABC is 12, so is a factor of 12/4 = 3 longer. The perimeter of ∆ABC is a factor of 3 longer than 12, so is 3*12 = 36.
The perimeter of ∆ABC is 36 cm.
2) The area of a right triangle is half the product of the leg lenghts. For the 4:3:5 triangle, that is
area = (1/2)*4*3 = 6 units²
∆ABC has leg lengths that are a factor of 3 longer, so its area will be the square of that factor times the area of the 4:3:5 triangle.
area of ∆ABC = 3²×6 = 54 . . . . cm²
The area of ∆ABC is 54 cm².
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We have already worked out the scale factor to be 3, so we know the side lengths of ∆ABC are (4×3):(3×3):(5×3) = 12:9:15. Then the area of ∆ABC is
area ∆ABC = (1/2)(12 cm)(9 cm) = 54 cm².