The lengths of two sides of a right triangle containing the right angle differ by 2 cm. If 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 area of the triangle is 24 cm².

Let the lengths of BC and AB be x and y, respectively.

Given,

The difference between the two perpendicular sides is 2 cm.

x - y = 2

∴ y = x - 2

Area = $\dfrac{1}{2}$ × base × height

⇒ 24 = $\dfrac{1}{2}$ × BC × AB

⇒ 24 = $\dfrac{1}{2}$ × x × (x - 2)

⇒ x × (x - 2) = 48

⇒ x² - 2x = 48

⇒ x² - 2x - 48 = 0

⇒ x² - 8x + 6x - 48 = 0

⇒ x(x - 8) + 6(x - 8) = 0

⇒ (x - 8)(x + 6) = 0

⇒ (x - 8) = 0 or (x + 6) = 0

⇒ x = 8 or x = -6

Since length cannot be negative, ∴ x = 8 cm.

y = x - 2 = 8 - 2 = 6 cm.

Thus, AB = 6 cm and BC = 8 cm.

By using Pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ AC² = BC² + AB²

⇒ AC² = 8² + 6²

⇒ AC² = 64 + 36

⇒ AC² = 100

⇒ AC = $\sqrt{100}$

⇒ AC = 10 cm.

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

= AB + BC + AC

= 10 + 8 + 6

= 24 cm.

Hence, perimeter of a triangle = 24 cm.