The hypotenuse of a right angled triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side, find the sides of the triangle.

Let the shortest side be x metre

According to question,

Third side = (x + 1)m

Hypotenuse = (2x - 1)m

For right angled triangle,

(Hypotenuse)² = (Perpendicular)² + (Base)²

$\therefore (2x - 1)^2 = (x + 1)^2 + x^2 \\[1em] \Rightarrow 4x^2 + 1 - 4x = x^2 + 1 + 2x + x^2 \\[1em] \Rightarrow 4x^2 + 1 - 4x = 2x^2 + 2x + 1 \\[1em] \Rightarrow 4x^2 - 2x^2 - 4x - 2x + 1 - 1 = 0 \\[1em] \Rightarrow 2x^2 - 6x = 0 \\[1em] \Rightarrow 2x(x - 3) = 0 \\[1em] \Rightarrow 2x = 0 \text{ or } x - 3 = 0 \\[1em] x = 0 \text{ or } x = 3$

Since, side's length cannot be equal to 0 , hence x ≠ 0.

∴ x = 3, (x + 1) = 4, (2x - 1) = 5

Hence, the sides of triangle are Shortest side = 3 m Hypotenuse = 5 m Third side = 4 m