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catetos: a, b ; a+b=28……..(1)
pitagoras: a^2+b^2=h^2
(a+b)^2=a^2+b^2+2*a*b
28^2=h^2+2*a*b
784=20^2+2*a*b
784-400=2*a*b
384=2*a*b
a*b=192 ......(2)
de (1): a=28-b …....(3)
en (2): a*b=192
(28-b)*b=192
28*b-b^2=192
b^2-28b+192=0
b. -12
b. -16
b=12 o 16
si b =12 entonces de (1) a =28-b a=16
si b =16 entonces de (1) a =28-b a=12
respuesta: los catetos son : 12 y 16