Question 924439
There are two solutions to this problem, we are given
a + b +c = 70 and a^2 +b^2 +c^2 = 1682
note that a^2 +b^2 = c^2(right triangle), therefore we have
c^2 +c^2 = 1682
2c^2 + 1682
c^2 = 841
c = 29
note we are dealing only with positive roots
now we have
a +b +29 =70
a +b = 41
a^2 +b^2 +841 = 1682
a^2 +b^2 = 841
now solve a +b = 70 for b and substitute b in a^2 +b^2 = 841
a^2 +(41-a)^2 = 841
a^2 +1681 -82a +a^2 = 841
2a^2 -82a +840 = 0
a^2 -41a +420 = 0
factor this equation
(a-21)*(a-20) = 0
a =21 or 20
our two solutions are
a=21, b=20, c=29
a=20, b=21, c=29