Three positive integers $a$, $b,$ and $x$ form an O'Hara triple $(a,b,x)$ if $\sqrt{a}+\sqrt{b}=x.$ For example, $(1,4,3)$ is an O'Hara triple because $\sqrt{1}+\sqrt{4}=3.$

If $(a,9,5)$ is an O'Hara triple, determine the value of $a.$
I notice that $9$ is a perfect square, so $\sqrt{9}=3.$ That means I can write $\sqrt{a}+3=5.$ Subtracting $3$ from both sides, I get $\sqrt{a}=2.$ Squaring both sides, I get $a=4.$ 