randRange(3, 8) randRange(3, 8) AC * AC + BC * BC

In the right triangle shown, `AC = AC` and `BC = BC`.

What is `AB`?

betterTriangle( BC, AC, "A", "B", "C", BC, AC, "?" );
AB2

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}`.

randRange(3, 5) randRange(6, 10) AB * AB - AC * AC shuffle([["a", "?", "BC", sqrt(BC2)], ["b", AC, "AC", AC]]) legs[0] legs[1]

In the right triangle shown, `bside = AC` and `AB = AB`.

What is `aside`?

betterTriangle( alen, blen, "A", "B", "C", alabel, blabel, AB );
BC2

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`.