Firstly let's find hypotenuse(let it will be "n" of smaller triangle
Let use Pythagorean theorem
[tex]a^{2} + {b}^{2} = {n}^{2} \\ {20}^{2} + {10}^{2} = {n}^{2} \\ n = \sqrt{500} [/tex]
Now we need to find hypotenuse(x) of bigger triangle
[tex] {c}^{2} + {n}^{2} = {x}^{2} \\ {9}^{2} + { \sqrt{500} }^{2} = {x}^{2} \\ x = \sqrt{581} [/tex]
The value of x must be rounded to 1 DP, so
[tex] \sqrt{581} = 24.1039... \\ \sqrt{581} \simeq \: 24.1[/tex]
Answer: x=24.1
