Two corners of an isosceles triangle are at (2 ,4 )(2,4) and (1 ,4 )(1,4). If the triangle's area is 64 64, what are the lengths of the triangle's sides?

1 Answer
Jun 9, 2016

{1,124.001,124.001}{1,124.001,124.001}

Explanation:

Let A = {1,4}A={1,4}, B = {2,4}B={2,4} and C = {(1+2)/2,h}C={1+22,h}
We know that (2-1) xx h/2 = 64(21)×h2=64 solving for hh we have

h = 128h=128.

The side lengths are:

a = norm(A-B) = sqrt((1-2)^2+(4-4)^2) = 1a=AB=(12)2+(44)2=1
b = norm(B-C) =sqrt((2-3/2)^2+(4-128)^2) =124.001b=BC=(232)2+(4128)2=124.001
a = norm(C-A) =sqrt((3/2-1)^2+(128-4)^2) =124.001 a=CA=(321)2+(1284)2=124.001