Two corners of an isosceles triangle are at #(4 ,9 )# and #(9 ,3 )#. If the triangle's area is #64 #, what are the lengths of the triangle's sides?

1 Answer
Feb 11, 2016

Answer:

enter image source here The sides are:
Base, #b = bar(AB) = 7.8#
Equal sides, #bar(AC) = bar(BC) = 16.8 #

Explanation:

#A_Delta = 1/2 bh = 64 #
Using the distance formula find b...
#b = sqrt((x_2-x_1)^2 + (y_2 - y_1)^2)#
#x_1 = 4; x_2 = 9; y_1 = 9; y_2 = 3 #
substitute and find h:
#b = sqrt(25 + 36) = sqrt(61) ~~ 7.81#
#h = 2(64)/sqrt(61) = 16.4#
Now using Pythagoras theorem find the sides, #barAC#:
#barAC = sqrt(61/4 + 128^2/61) = sqrt((3,721 + 65,536)/2) = 16.8#