Two corners of an isosceles triangle are at $(9, 4)$ $(9 ,4 )$ and $(1, 8)$ $(1 ,8 )$ . If the triangle's area is $48$ $48$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-12-08T10:48:53

Measure of the three sides are (8.9443, 11.6294, 11.6294)

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Length $a = \sqrt{{(9 - 1)}^{2} + {(4 - 8)}^{2}} = \sqrt{80} = 8.9443$ $a = sqrt((9-1)^2 + (4-8)^2) = sqrt 80 = 8.9443$

Area of $Delta = 48$
$:. h = (Area) / (a/2) = 48 / (8.9443/2) = 48 / 4.4772 = 10.733$

$side b = sqrt((a/2)^2 + h^2) = sqrt((4.4772)^2 + (10.733)^2)$
$b = 11.6294$

Since the triangle is isosceles, third side is also $= b = 11.6294$

Measure of the three sides are (8.9443, 11.6294, 11.6294)

Two corners of an isosceles triangle are at (9 ,4 )(9,4)(9 ,4 ) and (1 ,8 )(1,8)(1 ,8 ). If the triangle's area is 48 4848 , what are the lengths of the triangle's sides?