Let the smaller perpendicular side = x cm

and the larger perpendicular side = y cm

Now, given length of the hypotenuse = 3√10

=> x² + y² = (3√10)²

=> x² + y² = 9*10

=> x² + y² = 90 ..............1

Again, given if the smaller side is tripled and the longer side is doubled, new hypotenuse will be 9√5 cm

=> (3x)² + (2y)² = (9√5)²

=> 9x² + 4y² = 81*5

=> 9x² + 4y² = 405

=> 5x² + 4x² + 4y² = 405

=> 5x² + 4(x² + y² ) = 405

=> 5x² + 4 * 90 = 405         {from equation 1}

=> 5x² + 360 = 405

=> 5x² = 405 - 360

=> 5x² = 45

=> x² = 45/5

=> x² = 9

=> x = ±√9

=> x = ±3

Since x can not be negative

So, x = 3

From equation 1, we get

     3² + y² = 90

=> 9 + y² = 90

=> y² = 90 - 9

=> y² = 81

=> y = ±√81

=> y = ±9

Since, y can not be negative

So, y = 9

Hence, the two other sides of the triangle are 3 cm and 9 cm.