In $ \vartriangle ABC $ , $ \angle B = {90^ \circ } $ . Find the sides of the triangle, if AB = x cm, BC = (4x + 4) cm and AC = (4x+5) cm.

Hint: As one of the angles is 90°, we can say that it is a right angled triangle and hence we can use Pythagoras theorem to calculate the value of sides.
According to Pythagora's theorem, square of hypotenuse is equal to the sum of squares perpendicular and base.
Mathematically:
 $ {H^2} = {P^2} + {B^2} $

Complete step-by-step answer:
Drawing the diagram according to the question, we get:

Given: AB = x cm
              BC = (4x + 4) cm
              AC = (4x+5) cm.
Applying Pythagoras theorem on this right angled triangle, we get:
$A{C^2} = A{B^2} + B{C^2}$
Substituting the given values and calculating x:
$
  {(4x + 5)^2} = {x^2} + {\left( {4x + 4} \right)^2} \\
  16{x^2} + 25 + 40x = {x^2} + 16{x^2} + 16 + 32x \\
  {x^2} - 8x - 9 = 0 \\
  {x^2} - 9x + x - 9 = 0 \\
  x(x - 9) + 1(x - 9) = 0 \\
  \left( {x + 1} \right)(x - 9) = 0 \\
 $ $\left[ {\because {{(a + b)}^2} = {a^2} + {b^2} + 2ab} \right]$
The values of x obtained are:
(x + 1) = 0
x = -1
And
(x – 9) = 0
x = 9
As a side cannot take a negative value, the value of x will be 9.
Substituting this value of x in the sides, we get:
: AB = x cm
         = 9 cm
  BC = (4x + 4) cm
        = (4 X 9 + 4)
        =40 cm
    AC = (4x+5) cm.
          = (4 X 4 + 5)
          = 41 cm
Therefore, the sides of the given triangle are 9 cm, 40 cm and 41 cm.

Note: Pythagoras theorem can be applied only on the right angled triangles (one angle measuring 90°). The hypotenuse is always the longest side out of the three sides of a right angled triangle.