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1. not sufficient -> many possibilities (2,4) (3,3) ....
2. also not sufficient, for using Pythagoras we also need at least another side of the triangle

both together 1. l + w = 6 -> l = 6 - w
2. w^2 = d^2 - l^2 = d^2 - (6-w)^2 = 20 - 36 + 12w - w^2 -> w = 3+/-1
-> either 2 or 4, which doesn't matter as l is always the reverse (4 or 2)

-> area is not affected: A = 8

-> C
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C for me

1) l + w = 6
2) d^2 = 20 --> l^2 + w^2 = d^2 = 20

when considering 1 and 2 together, you have two equations for two variables, which means that you can solve for l and w



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