Two corners of an isosceles triangle are at $(2, 4)$ $(2 ,4 )$ and $(3, 8)$ $(3 ,8 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-12-05T13:23:41

Measure of the three sides are (4.1231, 31.1122, 31.1122)

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Length $a = \sqrt{{(3 - 2)}^{2} + {(8 - 4)}^{2}} = \sqrt{17} = 4.1231$ $a = sqrt((3-2)^2 + (8-4)^2) = sqrt 17 = 4.1231$

Area of $Delta = 64$
$:. h = (Area) / (a/2) = 64 / (4.1231/2) = 64 / 2.0616 = 31.0438$

$side b = sqrt((a/2)^2 + h^2) = sqrt((2.0616)^2 + (31.0438)^2)$
#b = 31.1122

Since the triangle is isosceles, third side is also $= b = 31.1122$

Two corners of an isosceles triangle are at (2 ,4 )(2,4)(2 ,4 ) and (3 ,8 )(3,8)(3 ,8 ). If the triangle's area is 64 6464 , what are the lengths of the triangle's sides?