Question 924439: In a right triangle, the perimeter is 70, and the sum of the squares of the sides is 1682. Find the lengths of the three sides. Thanks for your help.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! 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
|
|
|
