Two corners of an isosceles triangle are at $(6, 6)$ $(6 ,6 )$ and $(2, 7)$ $(2 ,7 )$ . If the triangle's area is $36$ $36$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-12-20T12:02:18

Lengths of the isosceles triangle are 4.1231, 17.5839, 17.5839

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Length of the base $a = \sqrt{{(7 - 6)}^{2} + {(2 - 6)}^{2}} = 4.1231$ $a = sqrt((7-6)^2 + (2-6)^2) = 4.1231$

Given area $= 36 = (\frac{1}{2}) \cdot a \cdot h$ $= 36 = (1/2) * a * h$

$:. h = 36 / (4.1231 /2) = 17.4626$

Length of one of the equal sides of the isosceles triangle is
$b = sqrt((a/2)^2 + h^2) = sqrt((4.1231/2)^2 + (17.4626)^2) = 17.5839$

Lengths of the isosceles triangle are 4.1231, 8.17.5839, 17.5839

Two corners of an isosceles triangle are at (6 ,6 )(6,6)(6 ,6 ) and (2 ,7 )(2,7)(2 ,7 ). If the triangle's area is 36 3636 , what are the lengths of the triangle's sides?