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?

1 Answer
May 20, 2018

#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#