SOLUTION: The length of the longer leg of a right triangle is 4m more than twice the length of the shorter leg. The length of the hypotenuse is 6m more than twice the length of the s

Question 1010866: The length of the longer leg of a right triangle is
4m
more than twice the length of the shorter leg. The length of the hypotenuse is
6m
more than twice the length of the shorter leg. Find the side lengths of the triangle.
Found 2 solutions by fractalier, addingup:
Answer by fractalier(6550) (Show Source): You can put this solution on YOUR website!
Call the length of the shorter leg, x.
Then the longer leg is 2x + 4m.
The hypotenuse is 2x + 6m.
By the Pythagorean Theorem, we have
a^2 + b^2 = c^2
x^2 + (2x + 4m)^2 = (2x + 6m)^2
x^2 + 4x^2 + 16xm + 16m^2 = 4x^2 + 24xm + 36m^2
Now collect like terms and get
x^2 - 8xm - 20m^2
We can factor this into
(x - 10m)(x + 2m) = 0
so that
x = 10m or x = -2m
If we assume m is positive, the solution must be x = 10m.
Then the other sides are
24m and 26m.
Answer by addingup(3677) (Show Source): You can put this solution on YOUR website!
We'll call them L, S and H for long leg, short leg, and hypotenuse:
L= 2S+4
H= 2S+6
--------------------------------------------------
Pythagoras:
(2S+4)^2+S^2= (2S+6)^2
4S+16+S^2= 4S+36 Subtract 4S+16 on both sides:
S^2= 20 Take the square root on both sides:
S= 4.47 This is the short leg
L= 2(4.47)+4= 12.94 This is the long leg
H= 2(4.47)+6= 14.94 This is the hypotenuse
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