Question

What are the possible lengths for x, the third side of a triangle, if two sides are 21 and 10

Answer 1

To solve we need to use Pythagoras theorem ( $a^{2} + b^{2} = c^{2}$ )

There are 2 possible lengths for x: hypotenuse or one of the 2 shorter sides.

Hypotenuse:
10^{2} + 21^2 = $x^{2}$
100+441= $x^{2}$
$\sqrt{541} = \sqrt{ x^{2} }$
23.3≈x

Shorter leg:
$10^2 + x^2=21^2$
$x^2= 21^2-10^2$
$x^{2}$= 441-100 
$\sqrt{ x^{2} } = \sqrt{341}$
x≈18.47