Question

Two corners of a triangle have angles of `pi / 4` and `pi / 2`. If one side of the triangle has a length of `6`, what is the longest possible perimeter of the triangle?

Answer 1

`12 +6sqrt2`
or
`~~20.49`

Explanation:

okay the total angles in triangle are `pi`

`pi - pi/4 - pi/2`

`(4pi)/4 - pi/4 - (2pi)/4 = pi/4`

so we have a triangle with angles: `pi/4,pi/4,pi/2` so 2 sides have the same length and the other is the hypotenuse.

using the Pythagorean theorem:

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

we know that the hypotenuse is longer than the other 2 sides:

`c = sqrt(a^2 + b^2)`

`c = sqrt(6^2 + 6^2)`

`c = sqrt(36 + 36) = 6sqrt2~~8.49`

so the permitter is:

`6+6+6sqrt2 = 12 +6sqrt2 ~~20.49`