Two corners of an isosceles triangle are at $(9 ,4 )$ and $(3 ,8 )$ . If the triangle's area is $48$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-12-05T10:25:30

Three sides of the triangle are $color(blue)(6.4031, 15.3305, 15.3305)$

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Length $a = sqrt((3-9)^2 + (8-4)^2) = sqrt41 = 6.4031$

Area of $Delta = 48$
$:. h = (Area) / (a/2) = 48 / (6.4031/2) = 48/3.2016 = 14.9925$

$side b = sqrt((a/2)^2 + h^2) = sqrt((3.2016)^2 + (14.9925)^2)$
$b = 15.3305$

Since the triangle is isosceles, third side is also $= b = 15.3305$

Two corners of an isosceles triangle are at (9 ,4 )(9 ,4 ) and (3 ,8 )(3 ,8 ). If the triangle's area is 48 48 , what are the lengths of the triangle's sides?