Two corners of an isosceles triangle are at $(2, 4)$ $(2 ,4 )$ and $(8, 5)$ $(8 ,5 )$ . If the triangle's area is $4$ $4$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-12-05T13:01:04

Three sides of the triangle measure $6.0828, 3.3136, 3.3136$ $color(red)(6.0828, 3.3136, 3.3136$

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Length $a = \sqrt{{(8 - 2)}^{2} + {(5 - 4)}^{2}} = \sqrt{37} = 6.0828$ $a = sqrt((8-2)^2 + (5-4)^2) = sqrt 37 = 6.0828$

Area of $Delta = 4$
$:. h = (Area) / (a/2) = 4 / (6.0828/2) = 4 / 3.0414 = 1.3152$

$side b = sqrt((a/2)^2 + h^2) = sqrt((3.0414)^2 + (1.3152)^2)$
$b = 3.3136$

Since the triangle is isosceles, third side is also $= b = 3.3136$

Two corners of an isosceles triangle are at (2 ,4 )(2,4)(2 ,4 ) and (8 ,5 )(8,5)(8 ,5 ). If the triangle's area is 4 44 , what are the lengths of the triangle's sides?