The problem is:
A right circular cone is made from a circular piece of paper of radius $R$ by cutting out a sector of angle $\theta$ radians and gluing the cut edges of the remaining piece together. Find the height $h$ of the cone in terms of $R$ and $\theta$.
In the problem before this one, you solve for $r$, which is the base of the cone. $$ r = R\frac{2\pi - \theta}{2\pi} $$
The method I've gone with is seeing that a right triangle is made from $h$, $r$ and $R$, Use the Pythagorean Theorem to find the value for $h$. Initially, it looks like this: $$ h^2 = R^2 - \left( R\,\frac{2\pi - \theta}{2\pi} \right)^2 $$
When I follow through with the formula, I come to this step, and can't make any further progress: $$ h = R + \left(-R\, \frac{-2\pi + \sqrt{4\pi\theta - \theta^2}}{2\pi} \right) $$
The answer given is:
$$
h = R\, \frac{\sqrt{4\pi\theta - \theta^2}}{2π}
$$
Am I doing something wrong in applying the Pythagorean formula, or just using it incorrectly? I appreciate any advice or help given.
The two expressions are almost equivalent. If you used the other root with a minus sign in front of the square root in your answer, then break up the fraction into two pieces via the pattern $$ \frac{A - B}{C} = \frac{A}{C} - \frac{B}{C}, $$ then there's a cancellation and your answer matches the given answer exactly.
There's an easier way to arrive at the answer (postponing the use of radicals until the end). Factoring $R$ out of the Pythagorean equation yields \begin{align} h^2 &= R^2 \Biggl( 1 - \biggl( \frac{2\pi - \theta}{2\pi} \biggr)^{\!2} \Biggr) \\ &= R^2 \, \frac{(2\pi)^2 - (2\pi - \theta)^2}{(2\pi)^2} \, \\ &= R^2 \, \frac{4\pi\theta - \theta^2}{(2\pi)^2}, \end{align} and so since $h \geq 0$, $$ h = R \, \frac{\sqrt{4\pi\theta - \theta^2}}{2\pi}, $$ in agreement with the given answer.