Two corners of an isosceles triangle are at $(2 ,4 )$ and $(4 ,7 )$ . If the triangle's area is $8$ , what are the lengths of the triangle's sides?

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Answer 1 · 2018-05-31T09:34:15

Other two sides are $color(purple)(bar (AB) = bar (BC) = 4.79$ long

Area of triangle $A_t = (1/2) b h$

$h = (A_t * 2) / (b)$

Given $A_t = 8, (x_a,y_a) = (2,4), (x_c, y_c) = (4,7)$

$b = bar(AC) = sqrt((4-2)^2 + (7-4)^2) = sqrt(13)$

$h = (2 * 8) / sqrt(13)= 4.44$

Since it’s an isosceles triangle,

$bar (AB) = bar (BC) = sqrt (h^2 + (c/2)^2)$

$=> sqrt((16/sqrt(13))^2 + (sqrt(13)/2)^2)$

$color(purple)(bar (AB) = bar (BC) = 4.79$

Two corners of an isosceles triangle are at (2 ,4 )(2 ,4 ) and (4 ,7 )(4 ,7 ). If the triangle's area is 8 8 , what are the lengths of the triangle's sides?