Question

Need a bit of help with this question:

Answer 1

Firstly let's find hypotenuse(let it will be "n" of smaller triangle

Let use Pythagorean theorem

$a^{2} + {b}^{2} = {n}^{2} \\ {20}^{2} + {10}^{2} = {n}^{2} \\ n = \sqrt{500}$

Now we need to find hypotenuse(x) of bigger triangle

${c}^{2} + {n}^{2} = {x}^{2} \\ {9}^{2} + { \sqrt{500} }^{2} = {x}^{2} \\ x = \sqrt{581}$

The value of x must be rounded to 1 DP, so

$\sqrt{581} = 24.1039... \\ \sqrt{581} \simeq \: 24.1$

Answer: x=24.1