Two corners of an isosceles triangle are at #(8 ,5 )# and #(9 ,1 )#. If the triangle's area is #12 #, what are the lengths of the triangle's sides?

Question

1800 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-23T02:55:52

#color(maroon)("Lengths of the triangle "a = sqrt 17, b = sqrt(2593 / 68), c = sqrt(2593 / 68)#

#color(red)( B(8,5), C(9,1), A_t = 12#

let #bar(AD) = h#

#bar(BC) = a = sqrt((9-8)^2 + (1-5)^2) = sqrt17#

#Area of triangle " A_t = 12 = (1/2) a *h = (sqrt17 h)/2#

#h = 24 / sqrt17#

#bar(AC) = bar(AB) = b = sqrt((a/2)^2 + h^2)#

#b = sqrt((sqrt17/2)^2 + (24/sqrt17)^2)#

#b = sqrt(17/4 + 576/17) = sqrt(2593/68)#