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Hint:- Apply Pythagoras Theorem.

Let the two legs of the triangle be \[x\] and \[y\].

And let the smaller leg be \[x\].

And the larger leg be \[y\].

As, we are given with the initial length of hypotenuse of the right-angled triangle.

So, according to Pythagoras theorem.

Sum of squares of two sides of a right-angled triangle is equal to the square of its hypotenuse.

So, \[{\left( x \right)^2} + {\left( y \right)^2} = {\left( {3\sqrt {10} } \right)^2}\]

\[ \Rightarrow {\left( x \right)^2} + {\left( y \right)^2} = 90\] (1)

Now, going to the other condition,

Smaller leg is tripled. So, the smaller leg becomes \[3x\].

Larger leg is doubled. So, the larger leg becomes \[2y\].

And, as given in the question, that new hypotenuse is \[9\sqrt 5 \].

So, again applying Pythagora's theorem.

\[{\left( {3x} \right)^2} + {\left( {2y} \right)^2} = {\left( {9\sqrt 5 } \right)^2}\]

\[ \Rightarrow 9{x^2} + 4{y^2} = 405\] (2)

Now, we had to find the value of\[x\] and \[y\] using equation 1 and 2.

So, splitting \[9{x^2}\] from equation 2. We get,

\[ \Rightarrow 5{x^2} + 4({x^2} + {y^2}) = 405\]

Now, putting the value of \[{x^2} + {y^2}\] from equation 1 to above equation we get.

\[

\Rightarrow 5{x^2} + 360 = 405 \\

\Rightarrow 5{x^2} = 405 - 360 = 45 \\

\Rightarrow {x^2} = 9 \\

\]

As, \[x\] and \[y\] are the length of sides of a triangle. So, they can have only positive values.

So, \[x = 3\]

Now, putting the value of \[x\] in equation 1. We get,

\[

\Rightarrow {y^2} + 9 = 90 \\

\Rightarrow {y^2} = 81 \\

\Rightarrow y = 9 \\

\]

Hence, sides of the right-angled triangle are 9cm and 3cm.

Note:-In these types of problems first find all the equations for the given conditions

using Pythagoras theorem. And then find the value of each side by solving the quadratic

equation. And remember that the length of the side should always be positive.

Let the two legs of the triangle be \[x\] and \[y\].

And let the smaller leg be \[x\].

And the larger leg be \[y\].

As, we are given with the initial length of hypotenuse of the right-angled triangle.

So, according to Pythagoras theorem.

Sum of squares of two sides of a right-angled triangle is equal to the square of its hypotenuse.

So, \[{\left( x \right)^2} + {\left( y \right)^2} = {\left( {3\sqrt {10} } \right)^2}\]

\[ \Rightarrow {\left( x \right)^2} + {\left( y \right)^2} = 90\] (1)

Now, going to the other condition,

Smaller leg is tripled. So, the smaller leg becomes \[3x\].

Larger leg is doubled. So, the larger leg becomes \[2y\].

And, as given in the question, that new hypotenuse is \[9\sqrt 5 \].

So, again applying Pythagora's theorem.

\[{\left( {3x} \right)^2} + {\left( {2y} \right)^2} = {\left( {9\sqrt 5 } \right)^2}\]

\[ \Rightarrow 9{x^2} + 4{y^2} = 405\] (2)

Now, we had to find the value of\[x\] and \[y\] using equation 1 and 2.

So, splitting \[9{x^2}\] from equation 2. We get,

\[ \Rightarrow 5{x^2} + 4({x^2} + {y^2}) = 405\]

Now, putting the value of \[{x^2} + {y^2}\] from equation 1 to above equation we get.

\[

\Rightarrow 5{x^2} + 360 = 405 \\

\Rightarrow 5{x^2} = 405 - 360 = 45 \\

\Rightarrow {x^2} = 9 \\

\]

As, \[x\] and \[y\] are the length of sides of a triangle. So, they can have only positive values.

So, \[x = 3\]

Now, putting the value of \[x\] in equation 1. We get,

\[

\Rightarrow {y^2} + 9 = 90 \\

\Rightarrow {y^2} = 81 \\

\Rightarrow y = 9 \\

\]

Hence, sides of the right-angled triangle are 9cm and 3cm.

Note:-In these types of problems first find all the equations for the given conditions

using Pythagoras theorem. And then find the value of each side by solving the quadratic

equation. And remember that the length of the side should always be positive.

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