Pythagorean theorem#2

Relevant wiki: Pythagorean Theorem

Suppose, thee sides of the triangle is $a,b,$ and $c$ . So we can use Pythagorean theorem .

Here $a = ?, b = 8$ and $c = \sqrt{113}$

Now,

$a^2 + b^2 = c^2$

$\Rightarrow a^2 + 8^2 = (\sqrt{113})^2$

$\Rightarrow a^2 + 64 = 113$

$\Rightarrow a^2 = 113 - 64$

$\Rightarrow a = \sqrt{49}$

Hence $a = 7$

Since $a = 7$ and $a$ is half of the length of $x,$ we can multiply to find $x$

$x = a \times 2$

Hence $x = 7 \times 2 = \boxed{14}$

First name the triangle,then apply apollonius theorem to the triangle. the side which is bisecting the base is median so,(side)^2+(side)^2=2( median^2 + half of base^2 ) Using Apollonius theorem, you will be able to find the base.

$x=2\sqrt{(\sqrt{113})^2-8^2}=2\sqrt{113-64}=2\sqrt{49}=2(7)=14$