In △ABC, ∠B = 90°, AB = (2x + 1) cm and BC = (x + 1) cm. If the area of the △ABC is 60 cm², find its perimeter.

Given,

AB = (2x + 1) cm

BC = (x + 1) cm

We know that,

Area of △ABC = $\dfrac{1}{2}$ × base × height

= $\dfrac{1}{2}$ × BC × AB

Substituting the values we get,

⇒ 60 = $\dfrac{1}{2}$ × (x + 1) × (2x + 1)

⇒ 60 × 2 = (2x + 1)(x + 1)

⇒ 120 = 2x² + 3x + 1

⇒ 2x² + 3x + 1 – 120 = 0

⇒ 2x² + 3x – 119 = 0

⇒ 2x² + 17x – 14x – 119 = 0

⇒ x(2x + 17) – 7(2x + 17) = 0

⇒ (x – 7)(2x + 17) = 0

⇒ x – 7 = 0 or 2x + 17 = 0

⇒ x = 7 or 2x = -17

⇒ x = 7 or x = $-\dfrac{17}{2}$

Since, x cannot be negative as length of a side cannot be negative. So, x = 7.

⇒ AB = (2x + 1) = 2 × 7 + 1 = 15 cm

⇒ BC = (x + 1) = 7 + 1 = 8 cm.

In right angled △ABC,

Using Pythagoras Theorem,

⇒ AC² = AB² + BC²

Substituting the values we get,

⇒ AC² = 15² + 8²

⇒ AC² = 225 + 64

⇒ AC² = 289

⇒ AC² = 17²

⇒ AC = 17 cm

Perimeter of △ABC = AB + BC + AC = 15 + 8 + 17 = 40 cm.

Hence, perimeter of △ABC = 40 cm.