Help!!

For the first two, you can use Pythagorean Theorem. This states that \({a}^{2}+{b}^{2}={c}^{2}\), where \(a\) and \(b\) are the legs, and \(c\) is the hypotenuse. 

Problem 1: We already have sides \(a = 10\) and \(c = \sqrt{170}\). We can move the formula around for it to fit this problem better. We can change it to \({b}^{2}={c}^{2}-{a}^{2}\). When subsituting our values in, we can get \({b}^{2}={\sqrt{170}}^{2}-{10}^{2}\). When we square both values, we can get \({b}^{2} = 170-100\), so \({b}^{2} = 70\). We can solve this to be \(b = \sqrt{70}\) or \(x = \sqrt{70}\).

Problem 2: We already have sides \(b = 5\sqrt{2}\) and \(c = 13\). We can move the formula around for it to fit this problem better. We can change it to \({a}^{2}={c}^{2}-{b}^{2}\). When plugging in our values, we can get \({a}^{2} = {13}^{2} - ({5\sqrt{2})}^{2}\). By solving this, we can get \({a}^{2} = 119\) which equals \(a = \sqrt{119}\) or \(x = \sqrt{119}\).

- Daisy

Guest, your last 4 answers are correct!