How long are the legs and the hypotenuse of the triangle if the length of the hypotenuse is an integer?

Question

the hypotenuse of a right triangle is three times the length of a leg. the sum of the sides of a triangle is between 6 and 8. given this conditions, how am i going to solve for the sides?

Answer 1

let one leg be x
hypotenus = 3x

other leg = sqrt(9x^2-x^2) = sqrt(8) x = 2 sqrt(2) x
sum = x+3x+2xsqrt(2)
= 4x + 2xsqrt(2)
3x is integer and x(4+2sqrt(2)) is between 6 and 8
so 3x(4+2sqrt(2)) is between 18 and 24

so 3x is between 18/(4+2sqrt(2)) and 24/(4+23sqrt(2))
between 18/6.8and 24/6.8
only possible value is 3
hypotenuos = 3

Answer 2

Let one leg be x
hypotenuse = 3x
so other leg = √[(3x)^2-x^2)], according to Pythogoras theorem
other leg = √(8x^2) = 2√2 x
sum of sides = x + 2√2 x = 3√2 x = 6 to 8
or √2 x = 2 to 8/3
or x = √2 to 8/3 √2
hypotenuse = 3x = 3√2 to 8/ √2
= 4.242 to 5.618
hypotenuse is an integer
the only integer between 4.242 to 5.618 is 5
so hypotenuse = 5
one leg = 5/3
other leg = (2√2 )(5/3) = 10√2/3

Answer 3

Side a = 2
Side b = 5.656854249 (Square root of 32)
Hypotenuse = 6

You said that the hypotenuse had to be an integer, not the sides.

Answer 4

one side = X
second side = Y
hypoteneuese = 3X
sum of sides: X+Y

An integer is simply a number, either positive or negative.