In the right triangle shown, AC = AC and
BC = BC.
What is AB?
We want to find c; let a = BC and b = AC.
\begin{align}
c^2 &= BC^2 + AC^2 \\
c^2 &= BC * BC + AC * AC \\
c^2 &= AB2 \\
c &= \sqrt{AB2}
\end{align}
Simplifying the radical gives c = formattedSquareRootOf(AB2).
So c = formattedSquareRootOf(AB2).
The radical cannot be simplified, so c = 1\sqrt{AB2} or just \sqrt{AB2}.
In the right triangle shown, bside = AC and
AB = AB.
What is aside?
We want to find a; let b = AC and c = AB.
So a^2 = c^2 - b^2
\begin{align}
a^2 &= AB^2 - AC^2 \\
a^2 &= AB * AB - AC * AC \\
a^2 &= BC2 \\
a &= \sqrt{BC2}
\end{align}
Simplifying the radical gives a = formattedSquareRootOf(BC2).
So aside = formattedSquareRootOf(BC2).
The radical cannot be simplified, so aside = 1\sqrt{BC2} or just \sqrt{BC2}.
We know a^2 + b^2 = c^2.