Two corners of an isosceles triangle are at #(8 ,2 )# and #(7 ,5 )#. If the triangle's area is #64 #, what are the lengths of the triangle's sides?

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Answer 1 · 2018-05-12T04:10:33

#color(brown)("Length of triangle sides " 3.16, 40.51, 40.51#

#A = (8,2), C = (7,5) A_t = 64#

#bar (AC) = b = sqrt((8-7)^2 + (2-5)^2) = sqrt10 = 3.16#

#A_t = 64 = (1/2) * b * h = (1/2) * sqrt10 * h#

#h = (2 * 64) / sqrt(10) = 128 / sqrt10#

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

#a = sqrt((sqrt10/2)^2 + (128/sqrt10)^2)#

#a = sqrt((10/4) + (16384/10)) = 40.51 " units"#