The sides of a right-angled triangle, containing the right angle, are 3(x + 1) cm and (2x - 1) cm. If the area of the triangle is 30 cm², find the lengths of the sides of the triangle.

Let ABC be the right angle triangle with base = (2x - 1) cm and height = 3(x + 1) cm.

Area of right angle triangle = $\dfrac{1}{2} \times$ base × height

Substituting values we get :

$\Rightarrow 30 = \dfrac{1}{2} \times BC \times AB \\[1em] \Rightarrow 30 = \dfrac{1}{2} \times (2x - 1) \times 3(x + 1) \\[1em] \Rightarrow 60 = (2x - 1)(3x + 3) \\[1em] \Rightarrow 60 = 6x^2 + 6x - 3x - 3 \\[1em] \Rightarrow 60 = 6x^2 + 3x - 3 \\[1em] \Rightarrow 60 = 3(2x^2 + x - 1) \\[1em] \Rightarrow 20 = 2x^2 + x - 1 = 0 \\[1em] \Rightarrow 2x^2 + x - 1 - 20 = 0 \\[1em] \Rightarrow 2x^2 + x - 21 = 0 \\[1em] \Rightarrow 2x^2 + 7x - 6x - 21 = 0 \\[1em] \Rightarrow x(2x + 7) - 3(2x + 7) = 0 \\[1em] \Rightarrow (x - 3)(2x + 7) = 0 \\[1em] \Rightarrow x - 3 = 0 \text{ or } 2x + 7 = 0 \\[1em] \Rightarrow x = 3 \text{ or } 2x = -7 \\[1em] \Rightarrow x = 3 \text{ or } x = -\dfrac{7}{2}.$

Since, side cannot be negative.

∴ x = 3 cm.

AB = 3(x + 1) = 3(3 + 1) = 3 × 4 = 12 cm,

BC = (2x - 1) = (2 × 3 - 1) = 6 - 1 = 5 cm.

In right angle triangle ABC,

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 12² + 5²

⇒ AC² = 144 + 25

⇒ AC² = 169

⇒ AC = $\sqrt{169}$ = 13 cm.

Hence, length of sides of triangle are 5, 12 and 13 cm.