The hypotenuse of a right triangle is 310−−√ cm.

Solution:

Let the length of smaller side of right triangle be $=x \mathrm{~cm}$ then larger side be $=y \mathrm{~cm}$

Then, as we know that by Pythagoras theorem

$x^{2}+y^{2}=(3 \sqrt{10})^{2}$

$x^{2}+y^{2}=90$ .......(1)

If the smaller side is triple and the larger side be doubled, the new hypotenuse is $9 \sqrt{5} \mathrm{~cm}$

Therefore,

$(3 x)^{2}+(2 y)^{2}=(9 \sqrt{5})^{2}$

$9 x^{2}+4 y^{2}=405 \ldots \ldots(2)$

From equation (1) we get $y^{2}=90-x^{2}$

Now putting the value of $y^{2}$ in equation (2)

$9 x^{2}+4\left(90-x^{2}\right)=405$

$9 x^{2}+360-4 x^{2}-405=0$

$5 x^{2}-45=0$

$5\left(x^{2}-9\right)=0$

$x^{2}-9=0$

$x^{2}=9$

$x=\sqrt{9}$

$=\pm 3$

But, the side of right triangle can never be negative

Therefore, when $x=3$ then

$y^{2}=90-x^{2}$

$=90-(3)^{2}$

$=90-9$

$=81$

$y=\sqrt{81}$

$=\pm 9$

Hence, length of smaller side of right triangle be $=3 \mathrm{~cm}$ then larger side be $=9 \mathrm{~cm}$