Answer:
[tex]18,24\text{ and 30}[/tex]
Explanation:
Here, we want to get the legs of the triangle
Let the shorter length be x cm
The longer leg will be (x + 6) cm
The hypotenuse length will be (x + 12) cm
According to Pythagoras' theorem. the square of the length of the hypotenuse equals the sum of the squares of the lengths of the two other sides
Thus:
[tex]\begin{gathered} (x\text{ + 12\rparen}^2\text{ = \lparen x\rparen}^2\text{ + \lparen x + 6\rparen}^2 \\ x^2+24x\text{ + 144 = x}^2+12x\text{ + 36 + x}^2 \\ x^2+24x+144\text{ = 2x}^2+12x\text{ + 36} \\ 2x^2-x^2+12x-24x\text{ + 36-144 = 0} \\ x^2-12x\text{ -108 = 0} \end{gathered}[/tex]
Solving the quadratic equation, we have it that:
[tex]\begin{gathered} x^2-18x+6x-108\text{ = 0} \\ x(x-18)+6(x-18)\text{ = 0} \\ (x+6)(x-18)\text{ = 0} \\ x\text{ = -6 or 18} \end{gathered}[/tex]
We discard x = -6
This would give us a side length with 0 length
Now, using x = 18:
we have the side lengths as:
18, 18 + 6 and 18 + 12 :
18, 24 and 30
