Hypotenuse on a right triangle in a right triangle, the side $a = 3$ and the subtraction the sides $b - c =\sqrt{3}$, What is the value of the hypotenuse?
Can someone help me? I need understand how to arrive at the answer.
 A: The sides are $3, c, c+\sqrt{3}$. Now consider possible pythagorean pairs.
Either $3$ can be hypotenuse, or $c+\sqrt{3}$


*

*Using 3 as hypotenuse, we get:
$$9 = 2c^2 + 2c\sqrt{3} +3\\
c^2 + c\sqrt{3} -3=0\\
c = \frac{\sqrt 3}{2}$$

*Using $c+\sqrt{3}$ as hypotenuse, you have:
$$c^2 + 2c\sqrt 3+3=c^2+9 \\
c = \frac{1}{\sqrt{3}}$$
A: $c$ can not be the hypotenuse. 
Let $a$ be the hypotenuse.
Thus, $$c^2+(c+\sqrt3)^2=3^2,$$
which gives $c=\frac{\sqrt3}{2}$, which is possible.
Let $b$ be the hypotenuse. 
Thus,
$$(c+\sqrt3)^2=3^2+c^2,$$
which gives $c=\sqrt3$ and $b=2\sqrt3$ and it's possible again.
A: Hint: if $b$ is the hypotenuse then $(b+c)(b-c)=b^2-c^2=9$. So what is $b+c$?
A: As we assue $a,b,c>0$: We know $b=\sqrt{3}+c>c$, hence we know that $c$ cannot be the hypothenuse (= longes side).
There are only two possibilities left. $b$ is the hypotenuse or $a$ is the hypotenuse.
A) $a^2+c^2=b^2=(c+\sqrt{3})^2 \implies 9 +c^2=c^2+2\sqrt 3c + 3 \implies c = \frac{6}{2\sqrt{3}}$
B) $c^2+(c+\sqrt{3})^2=a^2 \implies c^2+c^2+2\sqrt 3 c+3=9 \implies 2c^2+2\sqrt 3 c-6=0$
From $2c^2+2\sqrt 3 c-6=0 \implies c^2+\sqrt 3 c-3=0 \implies c_{1,2}=\frac{-\sqrt{3}\pm\sqrt{3-4(-3)}}{2}$
A: Since $b=c+\sqrt{3}$ it cannot be the hypotenuse. So we have two possibilities:
$$
c^2+a^2=b^2 \iff c^2+9=(c+\sqrt{3})^2
$$
or
$$
c^2+b^2=a^2 \iff c^2+(c+\sqrt{3})^2 = 9
$$
solve for $c$ and in the first case $b$ is the hypotenuse, in the second it is $a$.
