Let [tex]x[/tex] be the length of the shorter leg.
The other leg is 1m longer, so its length is [tex]x+1[/tex]
The hypothenuse is 9m longer, so its length is [tex]x+9[/tex]
The pythagorean theorem states that the sum of the squares of the legs is the square of the hypothenuse, so we have
[tex]x^2+(x+1)^2=(x+9)^2[/tex]
Expanding the squares gives
[tex]x^2+x^2+2x+1=x^2+18x+81[/tex]
Move all to the left hand side:
[tex]x^2-16x-80=0[/tex]
This equation has solutions [tex]x=-4[/tex] and [tex]x=20[/tex]
We can't accept the first solution, because it would lead to the side lengths
[tex]x=-4,\quad x+1=-3,\quad x+9=5[/tex]
And we can't have negative side lengths.
The other solution is fine, because it leads to the side lengths
[tex]x=20,\quad x+1=21,\quad x+9=29[/tex]
So, the side lengths are 20 (shorter leg), 21 (longer leg), 29 (hypothenuse)
Answer:
20 m , 21 m and 29 m.
Step-by-step explanation:
Let the length of the shorter leg be x m. Then the longer leg = x + 1 m and the hypotenuse = x + 9 m. So, by Pythagoras:
(x + 9)^2 = x^2 + (x + 1)^2
x^2 + 18x + 81 = x^2 + x^2 + 2x + 1
18x + 81 = x^2 + 2x + 1
x^2 - 16x - 80 = 0
(x - 20)(x + 4) = 0
x = 20, -4 (as we are dealing with lengths of sides we ignore the negative root).
So the sides have length 20, 21 and 29 (answer).
