Let the shortest side (BC) = x m
Then, hypotenuse (AC) = (2x – 1) m
and third side (AB) = (x + 1) m
Now in right ΔABC, We have
AC2 = AB2 + BC2 [By Pythagoras Theorem)
⇒ (2x – 1)2 = (x + 1)2 + (x)2
⇒ 4x2 + 1 – 4x = x2 + 1 + 2x + x2
⇒ 4x2 + 1 – 4x = 2x2 + 2x + 1
⇒ 2x2– 6x = 0
⇒ 2x(x – 3) = 0
⇒ 2x = 0
or x – 3 = 0
x = 0
or x = 3
Since x = 0 is not possible.
So, x = 3
Hence, required sides of the triangle are
AB = x + 1 = 4 m.
BC = x = 3 m
and AC = 2x – 1 = 5 m.