If the difference between the two sides of a right-angled triangle is 2 cm and the area of the triangle is 24 cm²; find the perimeter of the triangle.

ABC is a right angled triangle with a right angle at B.

The difference between the two perpendicular sides is 2 cm.

The area of the triangle is 24 cm².

Let the lengths of BC and AB be a and b, respectively.

From the given condition:

a - b = 2

∴ b = a - 2

Area of triangle = $\dfrac{1}{2}$ x base x height

⇒ $\dfrac{1}{2}$ x BC x AB = 24

⇒ $\dfrac{1}{2}$ x a x (a - 2) = 24

⇒ a x (a - 2) = 24 x 2

⇒ a² - 2a = 48

⇒ a² - 2a - 48 = 0

⇒ a² - 8a + 6a - 48 = 0

⇒ a(a - 8) + 6(a - 8) = 0

⇒ (a - 8)(a + 6) = 0

⇒ a = 8 or -6

Since length cannot be negative, a = 8 cm.

Now, substituting a = 8 in b = a - 2:

b = 8 - 2 = 6 cm

By using Pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ AC² = BC² + AB²

= 8² + 6²

= 64 + 36

= 100

⇒ AC = $\sqrt{100}$

= 10 cm

Perimeter of the triangle = Sum of all sides of the triangle

= AC + BC + AB

= 10 + 8 + 6 cm

= 24 cm

Hence, the perimeter of the triangle is 24 cm.