Two corners of an isosceles triangle are at $(3, 2)$ $(3 ,2 )$ and $(9, 1)$ $(9 ,1 )$ . If the triangle's area is $12$ $12$ , what are the lengths of the triangles sides?

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Answer 1 · 2017-12-08T10:25:23

Measure of the three sides are (6.0828, 3.6252, 3.6252)

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Length $a = \sqrt{{(9 - 3)}^{2} + {(1 - 2)}^{2}} = \sqrt{37} = 6.0828$ $a = sqrt((9-3)^2 + (1-2)^2) = sqrt 37 = 6.0828$

Area of $Delta = 12$
$:. h = (Area) / (a/2) = 12 / (6.0828/2) = 6 / 3.0414 = 1.9728$

$side b = sqrt((a/2)^2 + h^2) = sqrt((3.0414)^2 + (1.9728)^2)$
$b = 3.6252$

Since the triangle is isosceles, third side is also $= b = 3.6252$

Measure of the three sides are (6.0828, 3.6252, 3.6252)

Two corners of an isosceles triangle are at (3 ,2 )(3,2)(3 ,2 ) and (9 ,1 )(9,1)(9 ,1 ). If the triangle's area is 12 1212 , what are the lengths of the triangles sides?