The hypotenuse of right angled triangle …

Let the smaller side of the right triangle be x cm and the larger side be y cm .
Then, Using Pythagoras Theorem, we get
x² + y² ={tex}(3\sqrt5)^2{/tex} 

{tex}\implies x^2+y^2=9 (5){/tex}
{tex}\Rightarrow{/tex} {tex}x^2 + y^2 = 45 {/tex} ...............(i)
And second condition is if the smaller side is tripled and the larger side be doubled, the new hypotenuse is 15 cm.
Therefore,  {tex}(3x)^2 + (2y)^2 = 15^2{/tex}

{tex}\Rightarrow{/tex} {tex}9x^2 + 4y^2 = 225{/tex}  .................(ii).
From equation (i), we get {tex} y^2 = 45 - x^2{/tex}
Putting {tex} y^2 = 45 - x^2{/tex} in equation (ii), we get
{tex}9x^2 + 4(45 - x^2) = 225{/tex}
{tex}\Rightarrow{/tex}{tex}5x^2 + 180 = 225{/tex}

{tex}\Rightarrow{/tex} {tex}5x^2 = 45{/tex}

{tex}\Rightarrow{/tex} {tex}x^2 = 9{/tex}

{tex}\Rightarrow{/tex} x = ±3.
But, length of a side cannot be negative. Therefore, x = 3.
Putting x = 3 in (i), we get
9 + y² = 45

{tex}\Rightarrow{/tex} y² = 36

{tex}\Rightarrow{/tex} y = 6
Hence, the length of the smaller side is 3 cm and the length of the larger side is 6 cm.