Given: The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm.

To find: the lengths of these sides.

Solution: We know(Hypotenuse)² = (perpendicular)² + (base)² ..... (1)

Given, hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm

Let the base be ‘b’.

⇒ Perpendicular = b – 5Put the known values in (1),

⇒ 252 = b² + (b – 5)²

Apply the formula (x – y)² = x² + y² - 2xy in (b – 5)².

Here x = b and y = 5

⇒ 625 = b² + b² + 25 – 10b

⇒ 625 = 2b²+ 25 – 10b

⇒ 2b²+ 25 – 10b = 625

⇒ 2b2+ 25 – 10b - 625 = 0

⇒ 2b²– 10b - 600 = 0

Take out 2 common of the above equation.

⇒ b² – 5b – 300 = 0

Factorise the above quadratic equation by splitting the middle term. 

⇒ b² – 20b + 15b – 300 = 0 

⇒ b(b – 20) + 15(b – 20) = 0

⇒ (b - 20 ) ( b + 15 ) = 0

⇒ (b - 20 ) = 0 and ( b + 15 ) = 0 

⇒ b = 20 and b=-15

Since the length of any side cannot be negative,

We will ignore -15. ⇒ b = 20 

Perpendicular = 20 – 5 = 15 

Hence,Sides are 15 and 20