In the right triangle shown, `AC = `

and
`AC``BC = `

. What is `BC``AB`

?

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

We want to find `c`

; let `a = `

and `BC``b = `

.`AC`

So `c^2 = `

.`BC`^2 + `AC`^2 = `AB2`

Then, `c = \sqrt{`

.`AB2`}

Simplifying the radical gives `c = `

`formattedSquareRootOf(AB2)`.

In the right triangle shown,

and
`bside` = `AC``AB = `

. What is `AB`

?`aside`

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

We want to find

; let `a`

and `b` = `AC``c = `

.`AB`

So

.`a`^2 = c^2 - `b`^2 = `AB`^2 - `AC`^2 = `BC2`

Then,

.`a` = \sqrt{`BC2`}

Simplifying the radical gives `a` = `formattedSquareRootOf(BC2)`.

We know `a^2 + b^2 = c^2`

.